Naked Science Forum
General Science => General Science => Topic started by: hamdani yusuf on 02/11/2021 03:16:16

What's 0 to the power of 0?

1

I brought this up when I saw a Twitter feed, and the replies showed various answers.
https://twitter.com/pickover/status/1454996588661719041
Let's see which one is the most correct.
We learned it's an undefined value
As 0 to the power 0 looks like 0/0
But for Euler, he considered that 0^0 =1
(in his book "Introduction to Analysis of the Infinite")
There is no consensus among mathematicians about that..
but doesn't the limit from the negative approach negative 1?
No. It approaches a complex number. The real part approaches 1. The imaginary part approaches an infinitesimally small negative number.

Any base raised to zero always equals to one. Also, zero raised to any exponent is zero. But by taking limits closer to zero let's say 0.00000001 base to the power of 0.00000001 is equal to 0.999999 which is closer to one.
But since no value given current information, ie the expression does not determine original limit, so should be indeterminate and leave it undefined?
Correct me if I'm wrong.
In short, from the perspective of mathematics, this is undefined.

I'd show him my phone ad hope he doesn't notice the question at the bottom 
(https://pbs.twimg.com/media/FDHo_xYXEAUIqB2?format=jpg&name=medium)

And here's Wikipedia's answer.
https://en.wikipedia.org/wiki/Exponentiation#Zero_exponent
By definition, any nonzero number raised to the 0 power is 1:[16][1]
b^0 = 1.
This definition is the only possible that allows extending the formula
b^(m+n) = b^m . b^n
to zero exponents. It may be used in every algebraic structure with a multiplication that has an identity.
Intuitionally, b^0 may be interpreted as the empty product of copies of b. So, the equality b^0 = 1 is a special case of the general convention for the empty product.
The case of 0^0 is more complicated. In contexts where only integer powers are considered, the value 1 is generally assigned to 0^0 but, otherwise, the choice of whether to assign it a value and what value to assign may depend on context.
For more details, see Zero to the power of zero.
https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero

Here's an unconventional approach.
0⁰ x must be equal to x. As it means that you multiplied 0 to x zero times. That is you never multiplied 0 to x. Hence x stay as it is. In that case 0⁰ has to be 1 for sure.

(edited to correct inequality direction)
y = x^{x} is welldefined for all x > 0. The limit as you approach from positive side is clearly 1. Sure, the limit as you approach from the negative side, it's not well defined (in the reals), but neither are any of the points for x < 0, so I'm not sure that's a problem.
y = 0^{x} is (boringly) welldefined as 0 for all x > 0. The limit as you approach from positive side is clearly not 1, but again, because the limit as you approach from the negative side is not well defined (in the reals), it doesn't seem that odd to me to have a jump discontinuity for one point before all hell breaks loose.
y = x^{0} is clearly (and boringly) 1 across all reals.
I was taught to think of y = x^{a} as y = 1 × x^{a}, interpreted as 1 times x, a times. So if a = 0, you're multiplying 1 by x zero times, and it doesn't matter what x is.

Here's an unconventional approach.
0⁰ x must be equal to x. As it means that you multiplied 0 to x zero times. That is you never multiplied 0 to x. Hence x stay as it is. In that case 0⁰ has to be 1 for sure.
I'm not sure how unconventional that is. As I say in my previous post (which crossed with yours), this is how I was taught to think of this in school (8th grade, I think?)

Garbage.

Garbage.
Someone's trash is someone else's treasure.

y = x^{x} is welldefined for all x < 0.
What does it mean? Doesn't it mean all negative x?

Here is a video about it.... This exact question starts around 3 minutes...

I'm not sure how unconventional that is. As I say in my previous post (which crossed with yours), this is how I was taught to think of this in school (8th grade, I think?)
Perhaps just my ignorance. Wikipedia article doesn't explain it as obviously.
Many comments used limit to see the pattern, just like in Evan's video.
The video also appear in the Twitter conversation. Here's a response.
Great video, but it doesn't in any way show that 0^0=1, the real answer is undefined. That is about the limit of X^X in R+ as X approaches zero. In fact with with that kind of reasoning you might even argue that the answer is 0 because the same limit on R is 1 (so 0 on average)
But Google's calculator has different answer.
(0.0001)^(0.0001) = 1.00092141  0.000314448745 i

Garbage.
Someone's trash is someone else's treasure.
4^0.5=2
4x(4x0.5)=8
(4x0.5)x(4x0.5)=4
4x(1÷0.5)=8
4^(1÷2)=2
4^1/2=2
Square root 4=2
Square root( 4^1) =2
4^3/4=2.828etc
Quad root (4^3)=2.828etc.
So basically you are looking for a root if the number drops below the whole number. But asking for a root of zero is nonsense for the same mathematical reasons that there are no positive square numbers.
But a zero root of a number is one?

y = x^{x} is welldefined for all x < 0.
What does it mean? Doesn't it mean all negative x?
Sorry, I accidentally put the wrong sign (I meant all positive values). I have edited my post to correct the error. Thanks for pointing it out! :)

Garbage.
Someone's trash is someone else's treasure.
4^0.5=2
4x(4x0.5)=8
(4x0.5)x(4x0.5)=4
4x(1÷0.5)=8
4^(1÷2)=2
4^1/2=2
Square root 4=2
Square root( 4^1) =2
4^3/4=2.828etc
Quad root (4^3)=2.828etc.
...ok.... so?
So basically you are looking for a root if the number drops below the whole number.
If you want to think of it that way, it still kinda works, but that is not the easiest way to think of it. Instead, think of exponents as multiple multiplication (just like multiplication can be thought of as multiple addition)
7 x 3 = 0 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 21
2.7 x 3 = 0 + 3 + 3 + 0.7 x 3 = 6 + 2.1 = 8.1
0 x 3 = 0 + (zero additions of three) = 0
0 x (any number) = 0 + (zero of that number) = 0
0 x 0 = 0
if we then go to exponents:
3^{7} = 1 x 3 x 3 x 3 x 3 x 3 x 3 x 3 = 2187
3^{2.7} = 1 x 3 x 3 x 3^{0.7} = 9 x 2.16 = 19.42
3^{0} = 1 x (zero multiplications by three) = 1
any number ^{0} = 1 x (zero multiplications by that number) = 1
0^{3} = 1 x 0 x 0 x 0 = 0
0^{(any number)} = 1 x 0 (that number of times) = 0
0 ^{0} = 1 x (zero multiplications of zero) = 1
But asking for a root of zero is nonsense for the same mathematical reasons that there are no positive square numbers.
But a zero root of a number is one?
Presumably, you meant to say that there are no negative square numbers? (and that is easily fixed with imaginary numbers, but let's stick with real numbers for now).

0 + 0 = 0.
0 * 0 = 0.
0 / 0 = 1.
0 ^ 0 = Undefined.
0  0 = Nothing.
Ps  Dyscalculia!
👻

0 + 0 = 0.
0 * 0 = 0.
0 / 0 = 1.
0 ^ 0 = Undefined.
0  0 = Nothing.
Ps  Dyscalculia!
👻
I would switch the third and fourth: 0/0 = undefined and 0^{0} = 1

0 + 0 = 0.
0 * 0 = 0.
0 / 0 = 1.
0 ^ 0 = Undefined.
0  0 = Nothing.
Ps  Dyscalculia!
👻
I would switch the third and fourth: 0/0 = undefined and 0^{0} = 1
Let x be a number very close to 0.
let c be a finite nonzero number.
xc/x = c
x^xc = 1
thus
0/0 = undefined
0^0 = 1

Hi.
The original question was:
What's 0 to the power of 0?
This is actually a very good question and it's something that isn't easily resolved.
It also turns out that whenever something like 0^{0} is encountered in science, it has nearly always arisen as a Limt of x^{x} as x→0+ (approaching 0 from the right). This limit is defined and does equal 1. As a consequence of this it has become an unofficial convention that 0^{0} = 1 and regrettably many calculators, like the one you were using, will show you that result.
Too often we start from a false assumption. It is easy to imagine that just because we can write some mathematical term down it must have some numerical value. For example, I have never calculated the value of 10067 + 765409 but I might assume it is some Real number. We also frequently assume that when there are patterns to follow we must be able to extend those patterns. For example, whatever my answer to that sum might be it should be bigger than the first number, 10067.
For exponentiation it's actually much safer if we start by assuming nothing at all. Do not assume a^{b} defines any function from (a,b) → ℜ and don't even assume that the Real numbers exist. Instead start from more basic assumptions (axioms). If we do build up the Real Numbers and develop enough real Analysis to construct the exponential series then we will see that 0^{0} was never defined and indeed it cannot be defined in any consistent way as a Real number.
I could just spit out some chapters from a textbook on Real Analysis or Complex Analysis that talk about the exponential series but I can't do that any better than the textbooks. Instead let's put out some minor problems to consider, which might help to identify just how complicated it is to raise numbers to an exponent:
1. What is 4^{1/2} ? Why?
2. What is (4)^{1/2} ? Is there no real solution?
4. By the rules of indicies we have (x^{2})^{1/2} = x for all x. This seems reasonable but what happens to the LHS and RHS when you let x = 3 ? Don't we obtain +3 = 3 ?
5. 3^{π} cannot be written as the integer root of any integer power. Specifically 3^{π} ≠ (^{a}√3)^{b} for any inetgers a,b. So what is the value of 3^{π}? Could it be a negative number? If that's too easy consider (3)^{π} .
Best Wishes.
PS, yes I know question 3 was missing.

Presumably, you meant to say that there are no negative square numbers? (and that is easily fixed with imaginary numbers, but let's stick with real numbers for now).
yes. I imagined something different.

What is 4^{1/2} ? Why?
y=4^{1/2}
y=2
or
y=2
because 2^{2}=4
and (2)^{2}=4

What is (4)^{1/2} ? Is there no real solution?
(1 . 4)^{1/2} = (1)^{1/2} . (4)^{1/2}
= 2i
or
= 2i
No real solution.

but what happens to the LHS and RHS when you let x = 3 ? Don't we obtain +3 = 3 ?
No.
I found some calculators treat exponentiation of negative value differently. Some treat the negative sign as part of the base, some others treat it as sign of the result. To avoid ambiguity, we can use brackets.

So what is the value of 3^{π}? Could it be a negative number?
A fractional power produces more than one answers. Rational power has finite number of answers. Irrational power has infinite number of answers.
3^{π} has infinite number of answers in complex plane.
In polar complex plane, the magnitude is around 31.544..., but any angle can be the argument.

If that's too easy consider (3)^{π} .
(3)^{π} = (1)^{π} . (3)^{π}
Since any angle can be the answer, then it's the same as previous question.

Hi.
You're doing well there @hamdani yusuf .
You've already got enough to see that real numbers raised to some exponenet aren't always equal to a real number.
Sometimes, you can't find any real number that would be suitable. Sometimes you can find many suitable Real numbers. It's quite natural to extend the scope of the problem to consider complex numbers but then you can sometimes find an infinite set of numbers as a solution.
You already have enough evidence to recognise that 0^{0} was never required to be a unique Real number. It's not even required to be a unique Complex number. You've also presented enough examples to see that it cannot be defined as a real number in any consistent way. If it has to be anything, it could be a bicycle. It's just a collection of mathematical symbols we can write down but it's not representative of any numerical value. ∀6†12 is another set of symbols that doesn't equal or represent any numerical value.
Technically, x^{1/n} is defined to be the positive root wherever there was a choice. So that 9^{1/2} is +3 and nothing else. This is done because it maintains exponentiation as a well defined function for as long as possible. Anyway, using that it would mean that your answer to q. 4 is wrong.
If you were given the expression (x^{2})^{1/2} = x and then set x= 3 then you would have to deduce that +3 = 3.
Obviously we don't really want anything that silly, so the only possibile resolutions are that we give up on considering a^{b} as a well defined function for all a,b ∈ Z OR ELSE accept that the given equation (x^{2})^{1/2} = x was not valid for all x∈ℜ.
Mathematics has taken the second option, the rules of manipulating indicies that we were taught in school do not hold for all real numbers as a base for the exponentiation. A good teacher might have brought that to the attention of their students but it wouldn't matter much anyway: As human beings we want to follow patterns and we want to extend these patterns wherever we can, so we would have ignored any warning.
Thus (x^{m})^{1/n} = x^{m/n} is only a true statement for some values of x. The original question 4 that I presented was a little misleading. The most appropriate response should be "Exponentiation cannot follow this rule even though it seems like it should (because otherwise +3 = 3)".
I needed this to be considered because there are so many misconceptions and false proofs based on using "rules of indicies" even though these rules do not and cannot hold in the system of mathematics we commonly use.
Best Wishes.

If you were given the expression (x^{2})^{1/2} = x and then set x= 3 then you would have to deduce that +3 = 3.
No.
sin(0)=sin(π) doesn't imply 0=π
x^{2}=y^{2} doesn't imply x=y
(x^{2})^{1/2}=(y^{2})^{1/2} doesn't imply x=y

Hi.
If you were given the expression (x2)1/2 = x and then set x= 3 then you would have to deduce that +3 = 3.
No.
I don't know what to say in response to this. I can see that you don't want to accept it and of course you are free to do as you wish.
Perhaps we should go back to one of the earlier posts you wrote.
Quote from: Eternal Student on Yesterday at 17:49:40
What is 4^{1/2} ? Why?
      
Hamdani replied:
y=4^{1/2}
y=2
or
y= 2
because 22=4
and (2)2=4
You can't have y= +2 and also y= 2 because equals (=) is a transitive relation: If y = a and also y= b then a=b. So y= +2 and also y= 2 forces the relationship +2 = 2.
As a consequence, if you decide that 4^{1/2} is equal to a real number then it can only be taken as one number.
In order to keep exponentiation as a well defined function for as long as possible (or if you want to phrase it another way  in order to maintain the consistency of Mathematics so that = is transitive), you are forced to assign only one number to 4^{1/2} or else you must not assign any number to it at all.
By convention we take 4^{1/2} = positive root of 4.
This is slightly different to the example Sin(0) = Sin(π). You can assign more than one value in the domain of a function to the same value in the Range. However, you cannot assign one value in the domain to more than one value in the Range. ("Many to One" is OK but "One to Many" is not).
Best Wishes.

Thank You for the Corrections.
😊
🙏
Just wanted to show you folks a lil something...
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Ps  I use a Poco Android Mobile(model) by MI(company).
& I find it Strange that a built in default Chinese designed Calculator App is in clear defiance of the all mighty & powerful GooGle.
🤔
(The damm thing won't even Uninstall, & no I'm Not into Rooting so save that advice for someone else)

Thank You for the Corrections.
Just wanted to show you folks a lil something...
Screenshot_20211104004112800_com.miui.calculator.jpg (116.43 kB . 1080x2156  viewed 1472 times)
Ps  I use a Poco Android Mobile(model) by MI(company).
& I find it Strange that a built in default Chinese designed Calculator App is in clear defiance of the all mighty & powerful GooGle.
(The damm thing won't even Uninstall, & no I'm Not into Rooting so save that advice for someone else)
Appalling grammer, it should be cannot, not "can't". You also cannot divide by 1 or anything beneath

Appalling gramme, it should be cannot, not "can't". You also cannot divide by 1 or anything beneath
Wrong on so many levels.
I can divide by a half perfectly well.
Five divided by half is ten.
(five divided in half is rather less than ten).
I can divide by one very easily indeed.
x/1=x
There's nothing wrong with "can't" as a word.
And it looks like you can't spell grammar though it may be that you can't abbreviate "programme"
Presumably, you consider these statements of fact to be "repetitive antagonism"

https://www.wolframalpha.com/input/?i=4%5E0.5
For fractional power, we get a principal root and one or more other roots. But the selection of principal root is just a convention. It's neither more nor less true than the other roots.
Here's another example related to the original question here.
https://www.wolframalpha.com/input/?i=%280.01%29%5E0.01

This is slightly different to the example Sin(0) = Sin(π). You can assign more than one value in the domain of a function to the same value in the Range. However, you cannot assign one value in the domain to more than one value in the Range. ("Many to One" is OK but "One to Many" is not).
Try arcsin(x) = 0

Hi again.
But the selection of principal root is just a convention.
Yes and no. It is a convention but it is a widely used and widely recognised one. I think it would be hard to find any credible source in any country of the world that didn't define a fractional power to be the positive or principal root. So that 4^{1/2} = +2 and never 2.
There's another good reason for making this decision. It turns out that we're only going to get the positive root if we use the exponetial series.
The exponential series, Exp(x) has all the properties we want for e^{x}.
Exp(x) =
I mentioned earlier that if you build up the Real numbers from first principles and develop some Real Anlysis then this series is found to be convergent for all x∈ℜ. We can take the definition further and apply it in the Complex plane and then it turns out the Exp(x) function is just fantastic (it has an infinite radius of convergence, it's holomorphic throughout C, it's just ideal and very well behaved).
The key point is that it has the properties we want for exponentiation, in particular Exp(x) . Exp (y) = Exp (x+y). From which all the other usual properties of exponents will follow Exp(x) = 1/Exp(x) and for rational numbers a/b we will have Exp(a/b) = (^{b}√e)^{a}
It is therefore possible to define exponentiation with this series as your prototype. So it's not necessary to imagine that a^{b} means that you start with the base a multiplied by itself a few times, take a suitable root and sometimes invert the final answer. Instead we have an analytical definition for a^{b} it is the number you obtain as the limit of this series Exp ( b. Ln a) and it will be well defined whenever Ln(a) exists.
(The advantage of making the limit of this series your definition of a^{b} is that it remains consistent with algebriac definitions for rational exponents but also allows you to evaluate irrrational exponents and, if you extend into complex analaysis, arbitrary complex exponents).
Anyway, we get two results from this identification. Firstly, Exp(1/2) = sum of terms all of which are positive, so we get a positive answer. This means that Exp(1/2) is finding the positive square root of e and not the negative root. So if we're going to use this analytical definition for exponentiation then 4^{1/2} = +2 and not 2.
The second item of interest is that, despite all the advantages of defining exponentiation based on this series, it's surprisingly uselss for evaluating 0 raised to a power and we have to fall back to algebraic definitions for this. (This post is already long, so I'm stopping here).
You also mentioned this:
Try arcsin(x) = 0
ArcSin is only a well defined function when you restrict the domain of the parent Sine function.
(https://www.intmath.com/analytictrigonometry/svg/svgphpinversetrigofunctions7s6.svg)
Best Wishes.

ArcSin is only a well defined function when you restrict the domain of the parent Sine function.
The original question doesn't restrict the path to the answer to be a function. There's no restriction either for the solution to be in real number.

it will be well defined whenever Ln(a) exists.
When does Ln(a) not exist?

Here's (0.3)^0.3 according to Wolfram Alpha.
https://www.wolframalpha.com/input/?i=%280.3%29%5E0.3
Position in the complex plane
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=83466.0;attach=32472)
I'm not sure why that particular point is selected as the principal root.
Plot of all roots in the complex plane
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=83466.0;attach=32474)

Hi.
I'm losing track of what was being discussed here.
This bit I can do:
When does Ln(a) not exist?
When a = 0.
Best Wishes.

Appalling gramme, it should be cannot, not "can't". You also cannot divide by 1 or anything beneath
Wrong on so many levels.
I can divide by a half perfectly well.
Five divided by half is ten.
(five divided in half is rather less than ten).
I can divide by one very easily indeed.
x/1=x
There's nothing wrong with "can't" as a word.
And it looks like you can't spell grammar though it may be that you can't abbreviate "programme"
Presumably, you consider these statements of fact to be "repetitive antagonism"
[ Invalid Attachment ]
Umm...unfortunately, the designer is Chinese.
Maybe it provides correct grammatical sentence construction if the language is changed/switched to " Mandarin ".
(I ain't interested in figuring that out, i consider it as a default inbuilt Calculator Application...We all can agree courteously that it ain't a Wren & Martin.)
Ps  If anyone wishes to write in to Xiaomi or call their 1800 Number, Please go ahead.
🙏
I feel my Life, and of Others around me gets Alot more easy n comfortable, provided i can just
" Let Go of Things, & Move On ".
(I shall stick to it)
😇

Umm...unfortunately, the designer is Chinese.
Why is that unfortunate?

Umm...unfortunately, the designer is Chinese.
Why is that unfortunate?
At the time when i purchased this android mobile phone...
China was making leaps & bounds progress in almost all sectors.
👍
I viewed China as a Pioneer & Visionary.
& no doubt The device beats all other competitors in design, coverage, computational power & is economically priced.
But then, Unfortunately, things changed pretty quickly pretty drastically internationally.
(Covid19, TikTok Spying, Chinese aggression in South China Sea, Neighborhood disputes, Hong Kong, Taiwan, Afghanistan, CCP moving towards (complete)Totalitarianism.)
I have lost my trust & faith.
👎
There are quite a few inbuilt applications that are uninstallable.
I cannot root the device, orelse warranty snaps, & of course the pandora's jar opens.
I check complete details before updating os & security patches.
Ps  In short, Now i truly understand, what it means & how it feels like, to be sleeping with the enemy.
🤞
(Sure, i can divorce it & remarry, but to carry on flirting with disasters is in my lame genes.)
😉

Appalling gramme, it should be cannot, not "can't". You also cannot divide by 1 or anything beneath
Wrong on so many levels.
I can divide by a half perfectly well.
Five divided by half is ten.
(five divided in half is rather less than ten).
I can divide by one very easily indeed.
x/1=x
There's nothing wrong with "can't" as a word.
And it looks like you can't spell grammar though it may be that you can't abbreviate "programme"
Presumably, you consider these statements of fact to be "repetitive antagonism"
[ Invalid Attachment ]
Umm...unfortunately, the designer is Chinese.
Maybe it provides correct grammatical sentence construction if the language is changed/switched to " Mandarin ".
(I ain't interested in figuring that out, i consider it as a default inbuilt Calculator Application...We all can agree courteously that it ain't a Wren & Martin.)
Ps  If anyone wishes to write in to Xiaomi or call their 1800 Number, Please go ahead.
🙏
I feel my Life, and of Others around me gets Alot more easy n comfortable, provided i can just
" Let Go of Things, & Move On ".
(I shall stick to it)
😇
You asked why I have my tag line, BC knows I do not respond but snipes anywhoo, that is why I have my tag line.

BC knows I do not respond but snipes anywhoo,
You mean "BC still points it out when I says stupid things".
And we know what the tag line is for it's to warn others that they also might suffer the terrible loss; you might put them on your ignore list too.
Do I look bothered?

Hi.
I'm losing track of what was being discussed here.
This bit I can do:
When does Ln(a) not exist?
When a = 0.
Best Wishes.
What's the difference between something that doesn't exist and something that's not defined in math?
Imaginary number was once thought not to exist.

Hi.
What's the difference between something that doesn't exist and something that's not defined in math?
One is a personality and the other is a mermaid?
(I know that's one a bit old. I don't get out much). Anyway.... What did I mean when I said Ln(a) doesn't exist?
In this context I did mean that it isn't defined.
Best Wishes.

Did he snipe?

Did he snipe?
No; he just stated the obvious.

https://www.quora.com/Whatis00thezerothpowerofzero1
This had previously been the subject of some debate in the mathematical community until Donald Knuth set things straight in 1992, so it’s understandable that some confusion lingers, but the modern convention is to define 0^{0}=1 , for good reason.
https://www.quora.com/Howcanweprove00isequaltoone
A definition cannot be “proven” right or wrong. That doesn’t even make sense.
It might appear that the “theorems” x^{0}=1 and 0^{x}=0 contradict each other in a way that forces us to leave 0^{0} undefined. But 0^{x}=0 cannot actually be a theorem, even if we ignore the 0^{0} case: everyone agrees 0^{x} is not defined for any negative x . It can’t be, because that would break the theorem that b^{x}⋅b^{−x}=1 .
So there are good reasons for wanting to define 0^{0}=1 , but does it break anything the way defining 0/0 would? In fact it doesn’t. All the usual laws of exponentiation remain valid with this definition.
That’s what we mean when we say that the choice to define 0^{0}=1 is the right one.

Hi.
This had previously been the subject of some debate in the mathematical community until Donald Knuth set things straight in 1992, so it’s understandable that some confusion lingers, but the modern convention is to define 00=1 , for good reason.
I don't think there's a thing like the Oxford English Dictionary for Mathematics, somewhere that the very latest terminology and conventions are recorded and will be recognised as the definitive reference. So Donald will have a bit of trouble getting everyone to adopt that convention.
I Googled for "official mathematical conventions" and didn't find anything useful. I did find an extremely long and detailed booklet produced by ETS (Education Testing Service) which seems to be used as the standard for many examinations in the United States.
https://www.ets.org/s/gre/accessible/GRE_Math_Conventions_18_point.pdf.
On page 9 of 31 they write: Because all numbers are assumed to be real, some expressions
are not defined. Here are three examples: .........
Example 3: 0^{0} is not defined.
This was produced in 2010, so about 18 years after Donald had apparently set the precedent that 0^{0} will always be 1.
Anyway, defining 0^{0} = 1 is something you can do and it obviously helps to keep things like power series written in a convenient form (just like the convention 0! = 1 ). It's certainly been an unofficial convention for a long time. Just be careful if you write it in an examination where they aren't using that convention and be extremely careful if you ever obtain an expression like 0^{x} ,or x^{y} and need to examine the limiting behaviour as x→0 and/or y→0.
Best Wishes.

Did he snipe?
Did he snipe?
No; he just stated the obvious.
ROFL!
🤣
I find myself rather precariously positioned between you two.
😔
I'd Request you both now to humbly forgive Me & kindly let me move on.
🙏
Throughout life, i have been corrected and put inline on various occasions by various Individuals.
I'm a bit Sensitive, and have sometimes disagreed with the ways in which Knowledge was imparted to me.
But I'm also a bit Sensible, hence mostly been appreciative that Knowledge was passed over to me.
An avid sweets trader was making the rounds. Going from village to village, selling his products.
🍭
As afternoon arrived, he decided to rest under the shades of a banyan tree.
🌳
As he was quite exhausted, soon he fell asleep, dreaming on the sweet profits he was generating.
😴
Only to be abruptly woken up by a Sage. Who seemed to be a venting in anger for reasons unknown.
😠
The Sage went on a rampage, thrashing the Trader left right & centre. Mostly targeting the trader's stomach.
🤛🤜
The Trader, a pious individual, had never before seen or known his molester, & kept on asking for forgiveness & begging for the beating to Stop!
🙏
As the Trader fell to the ground, the Sage trampled upon his stomach.
Immediately the Trader started vomiting, & lo & behold, A lil Snake sneaked out from the Trader's mouth.
😲
The Sage stopped, sat besides the Trader in a consolatory manner & explained...
As he was passing by, he saw the Trader resting with his mouth open.
Right about then, a lil snakelet made it's way right through inside.
Sage further explained, there was no time to wake up the Trader in an orderly manner, & no way to explain to the Trader what had just happened, considering the trader might have had a heart failure by fear, just knowing a snake had crawled inside of him.
Hence the Sage had to be decisive and act in time with brutality.
🐍
When the Trader Realised all that had happened...he fell onto the Sage's feet.
Thou pretty sensitive and in pain from the beatings, but alot more sensible to foresee what he had gained in return for a lil bit of bashing...his Life.
😇
All Sages have different ways of dealing with the Snake of Ignorance within Us.
✌️
& All Traders are not as appreciative. Humans just follow Logic, but are controlled by Emotions.
Ps  Wish to Apologize to the OP for hijacking it & taking it to a place where it clearly does Not belong.
I'm Sorry Yusuf.
I shall stop now.
🙏

On page 9 of 31 they write: Because all numbers are assumed to be real, some expressions
are not defined. Here are three examples: .........
Example 3: 0^{0} is not defined.
Following the premise above,
√(1) would be undefined. Further implications are:
Euler's identity doesn't exist.
Wave equations don't exist.
Argument from authority has no weight in scientific debates.

"What's 0^0 ?"
When will it matter?

Hi.
Argument from authority has no weight in scientific debates.
By argument, I hope you mean discussion. I'm fine with you defining 0^{0} =1 if you want to.
I'm genuinely surprised that there isn't an official list of mathematical conventions and symbols.... but there just doesn't seem to be. There is one for house building regulations, words in the English Language, Electrical engineers have various recognised industry standards etc. However, Maths doesn't seem to have anything like this.
Best Wishes.

"What's 0^0 ?"
When will it matter?
When the answer is necessary to solve our problem.
There were time when negative numbers didn't matter. There were also time when imaginary numbers didn't matter.

By argument, I hope you mean discussion. I'm fine with you defining 0^{0} =1 if you want to.
The word discussion is more suitable to replace the word debate. The argument from authority itself is a widely known logical fallacy.
https://en.wikipedia.org/wiki/Argument_from_authority
In this discussion, one side think that 0^{0} is defined. The other side think that it's undefined. We try to convince the other side that our thought is better than the other. Many mathematical proofs are in the form of reductio ad absurdum, i.e. following the implications of an assumption to the point of absurdity/contradiction, to conclude that the assumption is false, hence we choose its opposite.
Arguing that our position is better because it's supported by an authoritative body can be useful in cases where each sides of the discussion have inadequate capacities to gather and process necessary data to produce correct and reasonable conclusions. Laymen discussion on complex problems like climate change or how to handle a pandemic are often cited as examples. Statements from the authorities can save our time and energy.
But if the goal is to find the fundamental truth, we need to process the raw data ourselves, and determine why a position is better than the other without relying on what the authority said.

Hi again.
The word discussion is more suitable to replace the word debate.
OK.
Perhaps I wasn't making myself clear. I'm not arguing with you in the sense of deliberately being unpleasant, or I certainly wasn't intending to do that anyway. It's always been a discussion as far I'm concerned.
As regards quoting authority, the most important point is that it isn't in my power to change the conventions. There doesn't even seem to be a place or document where current conventions, notation and terminology for Mathematics is recorded and provides a standard for the world. We could write a letter of complaint but I don't know where to send it.
If it's important to you then you have my support. It's a useful convention to have 0^{0} = 1 it makes a fair level of sense and it certainly makes it easier to write out power series like
P(x) = in one neat formula (which is commonly done) instead of writing something like this....
P(x) = a_{o} +
Or resorting to the use of two lines and splitting the definition of P(x) like this....
P(x) =
               
But if the goal is to find the fundamental truth, we need to process the raw data ourselves, and determine why a position is better than the other without relying on what the authority said.
Yes, seems like good advice.
You're actually doing amazingly well working through all the complications of defining what exponentiation should mean.
How (and why) will we define 0 raised to an irrational power, like 0^{π} ? I honestly don't know. Most people would just take a rational sequence {x_{n}} tending to Pi and look at the limit of 0^{xn}  but what would you do and why?
Best Wishes.

"What's 0^0 ?"
When will it matter?
When the answer is necessary to solve our problem.
There were time when negative numbers didn't matter. There were also time when imaginary numbers didn't matter.
If the answer to 0^0 will solve our problem then the solution to the problem will tell us the answer.
However, in the same way that there's no useful problem that involves calculating 1/0 because the answer is undefined there can be no useful problem that involves calculating 0^0.

However, in the same way that there's no useful problem that involves calculating 1/0 because the answer is undefined there can be no useful problem that involves calculating 0^0.
https://www.quora.com/Whatis00thezerothpowerofzero1
And so we now assign 0^0 the value that’s useful, which is 1 . Why is that useful? Because it lets us manipulate exponentials without adding special cases.

If you can devide by zero

If you can devide by zero
Let x be a number very close to 0.
let c be a finite nonzero number.
xc/x = c
x^xc = 1
thus
0/0 = undefined
0^0 = 1

Hi again.
I don't really know what this was:
Let x be a number very close to 0.
let c be a finite nonzero number.
xc/x = c
x^xc = 1
Why does that last line hold? I'm guessing you meant x^(xc) but perhaps it was (x^x).c Anyway x^(xc) is only approximately 1. Take x = 0.1 and c=10, Then x^(xc) = 0.1
Is this an argument based on continuity as x → 0 or something?
Could I also push this question again, please:
How (and why) will we determine 0 raised to an irrational power, like 0^{π} or 0^{e} or 0^{√2} ?
Can we prove that 0^{e} = 0 from basic principles?
Perhaps we don't need to determine the value of these things, what do you think?
I'm not asking to be hostile. I'm just trying to see if you think that continuity is important. It doesn't have to be.
Best Wishes.

Perhaps we don't need to determine the value of these things, what do you think?
My best guess is that there will be no physical system where any measurable parameter is "predicted" to be 0 ^ pi or anything like that.
However, if there was, we could measure the parameter and find the result of the calculation.

Inorder to contribute a lil that i can to the OP...
Is 0 by default positive(+) ?
Or is it simply Neutral?
Could then there be a scenario in which one(person) could have say like a +0 & 0 occasions?
I've never ever heard of + or  zeroes.
& I suppose it's common sense that 0 is Neutral.
But if We ever had such an imaginary concept of +0 & 0...
Then what would come between them?
Singularity?
Ps  Please continue with your Positive discussions and do not go off track.
🙂👍
(Mathematics seems like a Universal Language, it is perhaps the No 1 largest & most spoken dialect on this planet)

Hi again.
My best guess is that there will be no physical system where any measurable parameter is "predicted" to be 0 ^ pi or anything like that.
However, if there was, we could measure the parameter and find the result of the calculation.
Sounds sensible.
I'm not ignoring your reply. It's just that the trhread might lose it's way if I discuss too much.
It seems that you appreciate that there can be a convention like 0! =1 without there being any fundamental requirement for 0! to be defined following from the basic axioms of mathematics. It is just a convention not some deep truth about 0! following from the axioms.
Same thing for 0^{0} or even 0^{π}. Exactly as you stated, we don't expect it to arise in any physically real situation. Limits are a different kettle of fish. Limits can appear, and 0^{0} as a type of limiting form has all sorts of values depending on the exact situation.
       
If you (Hamdani Yusuf) are still considering exponentiation. Then you may still like to cosider the question 
How can we determine 0^{π}...... is continuity important in anything etc. (You don't have to. I'm not in charge).
      
Is 0 by default positive(+) ?
Or is it simply Neutral?
For real numbers, the definition for "x being positive" is that "x is greater than 0".
Similarly "x is negative" means "x is less than 0". These are what we call strict inequalities, if x > 0 then x must be a bit bigger than 0, it cannot be exactly the same as 0.
Now, 0 is not greater than 0. So it is nonpositive.
Similarly 0 is not less than 0. So it is nonnegative.
There are several ways to describe 0. You could also say it is unsigned, although most of the time we don't bother, just say that it is 0 and we will all know it's not positive or neagtive. The only important thing to note is that when people say "this is positive" it isn't exactly the same as saying "this is nonnegative". Nonnegative numbers inlcude all the positive numbers PLUS the number 0.
Anyway, for ordinary mathematics +0 is identical to 0 and they are just plain old 0.
But if We ever had such an imaginary concept of +0 & 0...
Then what would come between them?
Only people with a cold heart?
I can't really say. There are structures like the Hyperreal numbers where positive and neagtive nonzero infinitessimals exists. However, what lies between is not that interesting, there are just more infinitessimals including 0. I'm sure there are more interesting structures.
Best Wishes.

Why does that last line hold? I'm guessing you meant x^(xc) but perhaps it was (x^x).c Anyway x^(xc) is only approximately 1. Take x = 0.1 and c=10, Then x^(xc) = 0.1
Is this an argument based on continuity as x → 0 or something?
I meant x^(cx)
Try a smaller number for x, say 10^{100}. The equation applies for both positive and negative value of x, provided that its absolute value is much smaller than absolute value of c. Hence there's no reason to exclude only when x is exactly 0.

Could I also push this question again, please:
How (and why) will we determine 0 raised to an irrational power, like 0π or 0e or 0√2 ?
Can we prove that 0e = 0 from basic principles?
Perhaps we don't need to determine the value of these things, what do you think?
We need to distinguish between exact 0 and approximate 0.
dx in calculus is approximately 0. It doesn't work when it's exactly 0.
Fractional power is easier to understand when it's in polar format, because we can analyze the magnitude and phase angle separately.
For exact 0, the angle has no meaning. For approximate 0, when the power is irrational, any angle is elligible, including 0. For simplicity, it's usually chosen as principal value.

Is 0 by default positive(+) ?
Or is it simply Neutral?
Neutral

is continuity important in anything
It simplifies things, which improves efficiency. Efficiency is a universal instrumental goal.

Can you multiply by zero, Yx0 is essentially a non acting sum.
If you can devide by zero
Let x be a number very close to 0.
let c be a finite nonzero number.
xc/x = c
x^xc = 1
thus
0/0 = undefined
0^0 = 1
I'm nOt sure what is supposed to mean

Can you multiply by zero, Yx0 is essentially a non acting sum.
Yes. The result is 0.

Can you multiply by zero, Yx0 is essentially a non acting sum.
Yes. The result is 0.
Is it? In what situation do you actually get such a sum? I know formula can render this sum under a condition, but the sum is discounted. Saying Yx0 is essentially a contradiction, 0xY is not a sum you would see as it would be discounted prior. 273 x 0 is essentially getting half way through a sum and then stopping, leaving it open. Is it undefined?

Can you multiply by zero, Yx0 is essentially a non acting sum.
Yes. The result is 0.
Is it? In what situation do you actually get such a sum? I know formula can render this sum under a condition, but the sum is discounted. Saying Yx0 is essentially a contradiction, 0xY is not a sum you would see as it would be discounted prior. 273 x 0 is essentially getting half way through a sum and then stopping, leaving it open. Is it undefined?
Yes. It is.
It's not half way. It's like planning to add 273 but cancel it.

Hi again.
I meant x^(cx)
Try a smaller number for x, say 10100. The equation applies for both positive and negative value of x, provided that its absolute value is much smaller than absolute value of c. Hence there's no reason to exclude only when x is exactly 0.
Thanks, Hamdani. I can see this. So this is using some idea of continuity, the equation holds as you take a limit as x → 0.
We need to distinguish between exact 0 and approximate 0.
But that's not what we are doing when we use limits or continuity ideas to suggest a value for 0^{0}. We are trying to make the behaviour at x=0 as much like the behaviour for other values of x as we possibly can. Specifically, we are keeping the value of the expression within some small difference of the value of the expression when x is just off 0 and we are confident that we can make this error as small as we want provided we take x close enough to 0.
If you really believed that the value at x=0 is significantly distinguished from values we obtain when x is approximately 0 but not 0, then continuity is certainly not what you're going to have. No matter how close we get to x=0 we should be prepared to accept that the value of the expression could be very different when x is precisely 0.
(From Eternal student) ....is continuity important in anything?
Hamdani Yusuf replied: It simplifies things, which improves efficiency. Efficiency is a universal instrumental goal.
OK. So continuity is of some importance to you and it was used in your argument x^(xc) = 1 when x→0.
In which case, why are you so willing to lose continuity for 0^{x} as x→0 ?
0^1 = 0, 0^0.1 = 0, 0^0.01 = 0, ....., 0^(any small positive rational) = 0, ......
If there's any hope for continuity then 0^0 = 0.
Don't get me wrong, there's no reason why continuity of these expressions should be important but we do need to ask the following question: Is it reasonable to imply that continuity is important in some situations and some expressions like the x^(xc) argument but then choose to ignore continuity in some other expressions like 0^{x} ?
0^{x} isn't an amazingly complicated expression or some example I have deliberately cooked up to get another limit. It just does look like one of the sensible expressions we might consider when we're trying to determine 0^{0}. What reason would there be for specifically ignoring that expression and prioritising continuity in an expression like x^(xc)?
There are plenty of more unusual expressions that we can "cookup" (deliberately construct) to show other limits. Do we need some criteria to decide which expressions are more important than others?
Anyway, arguing for a value of 0^{0} based on limits and continuity of apparently well defined expressions involving only exponents that are already well defined (so that will be rational exponents for example) doesn't usually get very far. It just shows that there doesn't have to be a unique value AND/OR that attempting to maintain continuity was not a suitable assumption or approach to tackle the problem. It isn't possible to maintain continuity for every simple expression involving exponents by assigning a single unique value for 0^{0}.
You do seem to have moved on from just using simple ideas based on continuity to something using complex number  but this post is already too long, so I'll stop here for a moment.
Best Wishes.

Hi again.
I can't find much else going on in the forum today, So I'll continue with Hamdani's ideas concerning complex numbers and stuff...
dx in calculus is approximately 0. It doesn't work when it's exactly 0.
I'm not sure what you meant by this or if you intend to develop the idea. The differential dx is not a number of any kind (unless we're using something like nonstandard analysis and Hyperreal numbers). For standard analysis, the differential is just shorthand for expressing something about limits.
I think your main point was made in the previous line: We need to distinguish between exactly 0 and approximately 0. We can probably all agree that calculus does provide some precedent for the usefullness of doing this.
Fractional power is easier to understand when it's in polar format, because we can analyze the magnitude and phase angle separately.
For exact 0, the angle has no meaning. For approximate 0, when the power is irrational, any angle is elligible, including 0. For simplicity, it's usually chosen as principal value.
OK. So you're saying that we can't have exponentiation meaningfully defined just in the Real numbers. It only makes sense in the Complex plane. This is a bit controversial but it's ok, let's go with it.
This helps with determining 0^{π} since it has modulus 0 and no one cares about the argument since that won't actually change the location or value of the complex number anyway, it will be at the origin on an argand diagram so it's just plain old 0. That's great, we at least have a unique definition for 0^{π} but we do seem to have paid a high price for this....
You mentioned in an earlier post that 3^{π} is not uniquely defined as a Complex number. It is an infinite set of values, all with modulus approx. 31.5 but arbitray argument. So we've lost our ability to determine a unique value for most real numbers raised to an irrational exponent.
The best we can hope for is that a convention is applied for 3^{π} so that an argument of 0 is assumed  but there is no fundamental value that is more truthful or follows more naturally from the axioms of mathematics. Is that right?
There's also another minor question I have to ask: How did you determine the modulus of 3^{π} anyway? It's approximately 31.5 but why?
Best Wishes.

Trick or Treat!
1 ➕ +1 = ?
Ps  🎃

We are trying to make the behaviour at x=0 as much like the behaviour for other values of x as we possibly can. Specifically, we are keeping the value of the expression within some small difference of the value of the expression when x is just off 0 and we are confident that we can make this error as small as we want provided we take x close enough to 0.
A picture speaks a thousand words.
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=83466.0;attach=32480)

If you really believed that the value at x=0 is significantly distinguished from values we obtain when x is approximately 0 but not 0, then continuity is certainly not what you're going to have. No matter how close we get to x=0 we should be prepared to accept that the value of the expression could be very different when x is precisely 0.
Yes, in cases where positive x has different value than negative x, like 1/x.
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=83466.0;attach=32482;image)

Don't get me wrong, there's no reason why continuity of these expressions should be important but we do need to ask the following question: Is it reasonable to imply that continuity is important in some situations and some expressions like the x^(xc) argument but then choose to ignore continuity in some other expressions like 0x ?
0x isn't an amazingly complicated expression or some example I have deliberately cooked up to get another limit. It just does look like one of the sensible expressions we might consider when we're trying to determine 00. What reason would there be for specifically ignoring that expression and prioritising continuity in an expression like x^(xc)?
0^x discontinues at x exactly 0.
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=83466.0;attach=32484;image)

I'm not sure what you meant by this or if you intend to develop the idea. The differential dx is not a number of any kind (unless we're using something like nonstandard analysis and Hyperreal numbers). For standard analysis, the differential is just shorthand for expressing something about limits.
For example, in equation y=sin(2x). The gradient at the origin is dy/dx =2 when dx is close to 0, for both positive and negative values of dx. But if dx is exactly 0, dy is also exactly 0, which makes dy/dx undefined.

OK. So you're saying that we can't have exponentiation meaningfully defined just in the Real numbers. It only makes sense in the Complex plane. This is a bit controversial but it's ok, let's go with it.
Let's make it fractional exponentiation can't be meaningfully defined just in the Real numbers.

You mentioned in an earlier post that 3^π is not uniquely defined as a Complex number. It is an infinite set of values, all with modulus approx. 31.5 but arbitray argument. So we've lost our ability to determine a unique value for most real numbers raised to an irrational exponent.
The best we can hope for is that a convention is applied for 3^π so that an argument of 0 is assumed  but there is no fundamental value that is more truthful or follows more naturally from the axioms of mathematics. Is that right?
Right. The reason is consistency with definition for root of unity.
https://en.wikipedia.org/wiki/Root_of_unity
n mathematics, a root of unity, occasionally called a de Moivre number, is any complex number that yields 1 when raised to some positive integer power n.
(https://upload.wikimedia.org/wikipedia/commons/thumb/4/40/One5Root.svg/330pxOne5Root.svg.png)
The 5th roots of unity (blue points) in the complex plane
PS I wonder why superscript formats keeps disappearing whenever I quote previous posts. Is it a bug, or is it a feature?

Hi again.
You've been busy, Hamdani, well done. Looking through your posts it looks like you are actually trying to find good reasons to disregard some expressions and prioritise others. I'm quite impressed with the effort, most people would have given up on the idea already.
It looks like you've decided that a good reason for disregarding an expression is that it doesn't have a limit from the right and the left (and/or these aren't the same). That's actually quite reasonable.
If you're still interested in using that criteria, you should carefully check that there aren't any expressions that could still be a problem.
I might offer this example as a starting point:
[That doesn't always display well, it should be: Limit as x goes to 0 of (10 raised to 1/x) all raised to x ]
It has the general limiting form 0^{0} which is what you wanted and it should have a well defined limit from the left and the right. You can either determine the limit or else plot it with some software or one of the websites offering a graph plotter online.
Best Wishes.

You're right. We need other criteria if we want to define 0^0=1.
In your example, the x cancel each other. So, no matter what the value of x is, the result is the same constant.
It's no longer an exponential function.
lim x>0 (10^(1/x))^x
= lim x>0 (10^(x/x))
= lim x>0 (10^(1))
= 0.1
lim x>99 (10^(1/x))^x
= 0.1
So, if we want to use limit to prove something, we need to make sure that the value of the limit affects the result. This is a criterion we often take for granted, and only realize when it's being violated.

Hi.
@hamdani yusuf .
1. Well done for getting the right limit. That thing tends to 1/10 or 0.1
2. The second part of your post isn't especially clear. Try explaining what you mean and what you would do another way and see if it still meakes sense. Then write it down.
Here's an example, which tries to generalise what you have done: Suppose a person wants to determine the height of the tower of London. They try various things and read various books until finally they find a statement like this "All towers are 100 metres high". Now you might think that they have their answer but as they consider it a bit more they realise that the height of towers never changes no matter where you are in the world or which tower it is. So they decide that this was information they really should NOT use. Since all towers are 100 metres high it just can't be giving any useful information about my tower in London.
This is the line of reasoning you are proposing for not using that expression (10^1/x) ^ x. You seem to be saying that because the limit doesn't change when x→anywhere, it spoils the truth or usefulness of the information for indicating what happens as x→0.
Don't get me wrong.... if you are determined then you will eventually find some reasons for disregarding some expressions and only considering limits that arise from other expressions. However, you need to be a bit guarded against human nature. Ideally the criteria should be sensible and something that could reasonably have been established BEFORE examining all the possible situations. AFTER examining all the possibilties, you can certainly find or convince yourself of reasons for only using those expressions. For example, you can decide to use only those expressions for which the limit will evaluate to +1 (it's certainly going to get the result you wanted).
Mainstream pure mathematics, especially Real Analysis, went a different way to you. It's not better, it's just different. It was possible to define exponentiation based on the exponetial series. It's actually a very similar idea if you pull it all apart. In mainstream mathematics they do define A^{B} to be the limit of something involving only whole number exponents. It's the limit of a suitable sum of terms involving only whole number exponents.
Best Wishes.

It has the general limiting form 0^0 which is what you wanted and it should have a well defined limit from the left and the right.
Except that it doesn't.
lim 10^(1/0) is not 0. It's undefined.
So, your example represents undefined^0 instead of 0^0 as required. May be next time you can find a better example.

Hi.
lim 10^(1/0) is not 0. It's undefined.
You can put modulus signs around x, to have x instead of x if you're worried about left and right limits. I probably did miss that out  sorry, it was late and I was rushing and this LaTex equation format takes me ages to get working. By the time I've spellchecked and read through stuff I rarely get a post finished in less than an hour as it is.
If, for some reason you don't like a modulus, replace with an x^{2} term to get something like this:
and in the next step you can either continue to raise this whole thing to x (which gives you a differing two sided limit, not a limit of +1) or else raise all of the expression to the power of x^{2} to get back to something like the original idea.
May be next time you can find a better example.
Sadly, I've got some other work to do and I'll probably be leaving this thread (and entire forum) alone for a while. It's been nice talking to you and I'm glad you've got some interest in mathematics.
Best Wishes and bye for now.

if we then go to exponents:
3^{7} = 1 x 3 x 3 x 3 x 3 x 3 x 3 x 3 = 2187
3^{2.7} = 1 x 3 x 3 x 3^{0.7} = 9 x 2.16 = 19.42
3^{0} = 1 x (zero multiplications by three) = 1
To the power of 0 is 1 due to the answer = 1 portion of.
This is similar to as division where the answer is one portion of, the answer is equated to a singular.
20÷10equated to 1 is 10 singular portions of 20, if this is apparent? It may seem like I am just talking drivel? 20÷0.5 if 0.5 is equated to a singular the process is to find how many portions of 0.5 are in 1 or how many portions of 10 are in 1.
Vi's a vi, 2 portions of 0.5 in 1 meaning when 10 is the dividend the answer is 2 portions of the dividend or 20.
If you follow a similar method for multiplication 20x 10, 20 is a singular portion, vi's a vi 10 singular portions of 20 is 200.
So
3^2 = 3 x 3 = three singular portions that are 3 each= 9
3^1 = 3x1 = one singular portion that is 3 =3
3^0 =? = nil singular portions that are 3 each=? 0?

if we then go to exponents:
3^{7} = 1 x 3 x 3 x 3 x 3 x 3 x 3 x 3 = 2187
3^{2.7} = 1 x 3 x 3 x 3^{0.7} = 9 x 2.16 = 19.42
3^{0} = 1 x (zero multiplications by three) = 1
To the power of 0 is 1 due to the answer = 1 portion of.
This is similar to as division where the answer is one portion of, the answer is equated to a singular.
20÷10equated to 1 is 10 singular portions of 20, if this is apparent? It may seem like I am just talking drivel? 20÷0.5 if 0.5 is equated to a singular the process is to find how many portions of 0.5 are in 1 or how many portions of 10 are in 1.
Vi's a vi, 2 portions of 0.5 in 1 meaning when 10 is the dividend the answer is 2 portions of the dividend or 20.
If you follow a similar method for multiplication 20x 10, 20 is a singular portion, vi's a vi 10 singular portions of 20 is 200.
So
3^2 = 3 x 3 = three singular portions that are 3 each= 9
3^1 = 3x1 = one singular portion that is 3 =3
3^0 =? = nil singular portions that are 3 each=? 0?
Word salad.

So far, the discussion converges into two answers. First, it's 1. Some reasons have been put forward to support it, e.g. usefulness, simplicity, and desciption of exponentiation. Some matemathicians including Euler and Knuth take this position.
The others like Eternal Student said it's undefined. Some counter examples using limits have been proposed to support it.
The dispute may come from the fact that two zeros in the question give different answers. 0^x=0 for positive x
0^x=undefined for negative x
x^0=1 for both positive and negative x
the question is when x is neither positive nor negative.
Let's campare them with other operations, *, /, + and .
0*x=x*0=0
for positive and negative x
x/0=undefined
0/x=0
0+x=x+0=x
x0=x
0x=x
For multiplication, addition, and subtraction, there's no dispute that the result is 0 when x=0. It's different for division and exponentiation.

For multiplication, addition, and subtraction, there's no dispute that the result is 0 when x=0. It's different for division and exponentiation.
There in lies the answer, given multiplication and division are the reverse of each other.

For multiplication, addition, and subtraction, there's no dispute that the result is 0 when x=0. It's different for division and exponentiation.
There in lies the answer, given multiplication and division are the reverse of each other.
What's the reverse of exponentiation?

What's the reverse of exponentiation?
The inverse is the logarithm, just as division is the inverse of multiplication.
Why is this thread still going? You received your answer in the first page.

What's the reverse of exponentiation?
The inverse is the logarithm, just as division is the inverse of multiplication.
Why is this thread still going? You received your answer in the first page.
Do you stop questioning things after you get first answer? Don't you want to hear some alternatives which might be a better answer, or provide a deeper understanding?
What's your response to Eternal Student's answer?

For multiplication, addition, and subtraction, there's no dispute that the result is 0 when x=0. It's different for division and exponentiation.
There in lies the answer, given multiplication and division are the reverse of each other.
What's the reverse of exponentiation?
What is the process first of all?

0x0 is zero if a pattern is looked to if squares increace by the addition of the prior root and the current root to the prior squareFor example
(2x2)+2+3 =9 =3x3
(3x3)+3+4=16=4x4
And so on
0x0=(1x1) +0+ (1)=0
Whether you can consider 1 to be prior to zero is another matter.

Hi.
Looks ok @Petrochemicals .
I'm not sure how you intend to connect 0 x 0 with 0^0 = 0^{0} which was the original question in this thread but that's probably not too important.
Best Wishes.

Hi.
Looks ok @Petrochemicals .
I'm not sure how you intend to connect 0 x 0 with 0^0 = 0^{0} which was the original question in this thread but that's probably not too important.
Best Wishes.
Short of fiddling in your brain I am powerless to do anything more.

if x=0^0, then
log_{0}(x)=0
log_{0}(x)=ln(x)/ln(0)
ln(x)=0
x=1
https://en.wikipedia.org/wiki/Logarithm#Definition
The logarithm of a positive real number x with respect to base b[nb 1] is the exponent by which b must be raised to yield x. In other words, the logarithm of x to base b is the unique real number y such that (https://wikimedia.org/api/rest_v1/media/math/render/svg/7b51b8dcca3e64714e2f3aba12bb8c68812f77cc)
The logarithm is denoted "log_{b} x" (pronounced as "the logarithm of x to base b", "the baseb logarithm of x", or most commonly "the log, base b, of x").
An equivalent and more succinct definition is that the function log_{b} is the inverse function to the function (https://wikimedia.org/api/rest_v1/media/math/render/svg/873987f9618fe2c30ce4e72cbd1a967ff759c1d4).

Hi.
How do you get line 2?
You let x = 0^0 so x isn't known to be a unique well defined real number. You are assuming it is.
If x = a battleship then Log_{0}(x) = ?
This isn't intended to be a criticism. What you've done is quite common and sensible. It's human nature to start by assuming if 0^0 is a number then ask what number can it be? It's just that we will have already imposed some conditions on 0^0 just by tacitly assuming it is a real number.
A more thorough development of 0^0 would start by describing what you think 0^0 should be just in terms of some numbers which you accept or assume to exist (perhaps the integers or the standard set of Reals if you wish) and elementary operations defined on that set of numbers (so that could be binary operations like + and x). You have somewhat jumped over the first obstacle when you assume 0^0 is a real number.
However, if you do assume 0^0 is a real number then methods such as yours do start to indicate what number it must be. There's a few minor issues in your reasoning, like line 3 where you seem to have divided by Ln(0). Ln(0) is not a real number, it's ∞ in the extended reals but not defined at all in the reals. Perhaps you were only intending to identify 0^0 as an element of the extended reals.
Best Wishes.

if x=0^0, then
log0(x)=0
log0(x)=ln(x)/ln(0)
ln(x)=0
x=1
Let's say that ln(0) is unknown. It can even be a complex number, for the sake of the argument.
log0(x)=0 by definition.
it can only happen if ln(x)=0, which means x=1.

Hi.
log_{0}(x)=0 by definition.
You should ask yourself what or who's definition of a Logarithm you are using.
Here's one example of a definition from Wikipedia:
Definition
The logarithm of a positive real number x with respect to base b is the exponent by which b must be raised to yield x.
So that definition only applies when both of the following things apply:
i) x is a real number.
ii) x > 0.
In my opinion the Wikipedia definition isn't actually perfect either, for example it makes very little sense if the base b = 0  but that's a separate issue.
Best Wishes.

Hi.
log_{0}(x)=0 by definition.
You should ask yourself what or who's definition of a Logarithm you are using.
Here's one example of a definition from Wikipedia:
Definition
The logarithm of a positive real number x with respect to base b is the exponent by which b must be raised to yield x.
So that definition only applies when both of the following things apply:
i) x is a real number.
ii) x > 0.
In my opinion the Wikipedia definition isn't actually perfect either, for example it makes very little sense if the base b = 0  but that's a separate issue.
Best Wishes.
In my example, x=1, which is both real and greater than 0.

Hi.
I'm not sure what you want me to say here, @hamdani yusuf .
I've already said this This isn't intended to be a criticism. What you've done is quite common and sensible.
I'm always quite happy when anyone is showing some interest in Mathematics and more than willing to spend some time in discussion like I've done here.
It's regrettable if I need to disagree with something or suggest a minor problem exists. However, this is where you need to recognise that a good student CAN learn from problems and only the weaker students (like myself, I'm quite a poor student) need to have their actions constantly reinforced. There is a lot of what you've done that is good but hopefully you're a better student than I am. Hopefully, you can consider what might have been a problem.
The issue is that, presumably, you did not know x=1 until after the last line in the work.
So in line 2 you were making an assumption about the properties of x, you were assuming x is real and x>0 .
To phrase this another way, by performing line 2 in your proof you have forced the conditions:
(i) x is real
and (ii) x>0.
This is unless you are using some other definition of the Logarithm function (i.e. not a definition like that found in Wikipedia).
Best Wishes.

Hi again.
This is something I used to do in school and it took me ages to work out where the problems are:
Suppose you have an equation x^{2} = x and you're asked to solve it.
It's reasonable to divide by x on both sides, you obtain:
x = 1 which is perfectly good, it's a totally acceptable solution.
The thing is the mathematics has accepted the tacit assumption you made when you divided by x. It has accepted that x could not have been 0 and it does then lead you to a perfectly valid solution. The only solution now possible is that x=1.
You've done much the same sort of thing when you perform line 2 in your proof. You have tacitly excluded possibilities like x = 0 or even that x is not a real number. The mathematics doesn't stop to question if that exclusion is reasonable, it just does as it is was told to do. You can proceed to perform valid mathematical operations after this but it is, of course, very likely to lead you to a solution where x is real and x ≠ 0.
What you have got is a proof that 0^{0} =1 under certain assumptions:
Assumption 1) That 0^{0} is a real number.
Assumption 2) That 0^{0} > 0.
You also made this comment:
Let's say that ln(0) is unknown. It can even be a complex number, for the sake of the argument.
Ln(0) isn't well defined. By the sound of it, you wish to force the condition that Ln(0) is a complex number. This is similar to the earlier situation. The mathematics will only find a solution given the conditions that were forced whether that is directly stated as an assumption or else tacitly assumed by the mathematical operations you perform.
Best Wishes.

What you have got is a proof that 00 =1 under certain assumptions:
Assumption 1) That 0^{0} is a real number.
Assumption 2) That 0^{0} > 0.
I didn't make those assumptions. The answer was obtained as conclusion from the available definitions in mathematical operations.
I'm open to alternative solutions if strong supports can be provided.
What do we get if we assume the contrary of those statements?

The answer was obtained as conclusion from the available definitions in mathematical operations.
What definition of a Logarithm are you using?
Using your definition can you evaluate the following:
1. Log_{0}(10)
2. Log_{10} (0)
Most definitions for Log_{b}(x) fail to give you a well defined function for some input numbers b and/or x.
Best Wishes.

Hi again.
There is now a considerable amount of discussion about 0^{0} presented on Wikipedia.
https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero
Interestingly, using Google with the phrase "What is 0 to the power of 0" seems to bring up a slightly different set of results than I think it did when this thread was new.
The workings of Google and Wikipedia are a bit mysterious but it is supposed to be influenced by current internet activity and especially interactive sites. So it's possible that some interest in 0^{0} shown in this forum has had a small role in influencing the recent changes in Google results and the Wiki entry. (Ok... I'm probably just unrealistic. More realistically, time spent on the forum is just time spent).
Best Wishes.

What definition of a Logarithm are you using?
Using your definition can you evaluate the following:
1. Log0(10)
2. Log10 (0)
Most definitions for Logb(x) fail to give you a well defined function for some input numbers b and/or x.
Best Wishes.
Reverse of exponential.
1. Log(10)/Log(0) = 1/(~) =0
2. ~
How do you define "undefined" in mathematical sense?
Is following expression defined?
0 . 0
0 / 0
1/0
~
~
√1
√(1)
π
e^π
e^(πi)
arcsine(0)
arcsine(2)

Any number to the zeroth power is equal to 1 by definition.
I saw a fantastic explanation of this in Commercial link removed (http://Spam) some time ago,

Hi.
Reverse of exponential.
That's a personal definition. I guess I did ask what definition you are using but I meant can you site a definition from a textbook or online reference.
Using your definition there are at least two issues:
1. Take b= 0 as the base for the logarithm (or the exponentiation). 0^1 = 0 but also 0^2 = 0 and 0^ 3 = 1 etc.
So which value do you map the value 0 back to when you try to invert that function? Is Log_{0} (0) = 1 , 2 or 3 ? To say that another way, exponentiation is sometimes a manytoone function so that no inverse function exists.
What I'm trying to say is that it isn't as simple as saying "Logarithms are just the reverse of exponentiation", sometimes you can't find a reverse procedure.
2. Exponentiation isn't a trivial operation either. It certainly doesn't follow easily just from the basic field operations of (R, +, x). Algebra will allow a definition for a^{b} where a and b are rational numbers but you need some results from real analysis to ascribe a meaning to a^{b} where a and b could be irrational numbers.
        
You (@hamdani yusuf ) gave this as an evaluation of Log_{0} (10)
Log(10)/Log(0) = 1/(~) =0
So, you're saying that Log_{0} (10) = 0 if I've understood that correctly.
So, by your definition of a Logarithm we have that 0 ^{0} = 10.
How do you reconcile that with the proposal you made that 0^{0} = 1 ?
         
How do you define "undefined" in mathematical sense?
It depends on the situation. A function, f, is well defined if it can be expressed as subset, F, of A X B where A = Domain(f) and B = Range(f) such that (a,b) ∈ F and (a,c) ∈ F implies b=c.
To paraphrase that, a function is a rule that maps things in a domain to things in the range. It might map many things in the domain to one thing in the range but it cannot map one thing in the domain to many things in the range. It can be manytoone but not onetomany.
This probably isn't what you were asking, so I'm going to guess the general nature of the thing you were asking:
1. If we can write down some symbols and some numbers where there is clear mathematical meaning for some sets of numbers, does it follow that there is always a well defined mathematical meaning where we just change the numbers but keep the same symbols in the same places?
Answer: No.
2. If there is some pattern that seems to hold over a good range of values, can we just assume that the pattern does hold and push the numbers to the very limit?
Answer: No, not always.
                   
About the set of example expressions you (Hamdani) gave and asked if they were defined:
It depends on the context. For example ∞ is not a real number, so we would say that it is not defined in the Reals. However, you can use the symbol to mean something else, something that does have some meaning in a different mathematical structure. For example, in the extended Reals, there is an element denoted by the symbol ∞.
The symbol doesn't mean the same thing as it did in the Reals. In the Reals there was no such thing and the axioms for the incorporation of ∞ into the basic algebraic operations of + and x on other elements (finite numbers) in the extended reals do not force any equivalent results to hold in the reals.
                        
Any number to the zeroth power is equal to 1 by definition.....(here's my website)....
That looks like advertising. The best thing about it is that you've clearly shown the website got something wrong and probably isn't worth the subscription fee. Anyway, on the off chance that you were genuinely trying to join the discussion  hello and welcome.
Best Wishes.

That's a personal definition. I guess I did ask what definition you are using but I meant can you site a definition from a textbook or online reference.
An equivalent and more succinct definition is that the function log_{b} is the inverse function to the function (https://wikimedia.org/api/rest_v1/media/math/render/svg/873987f9618fe2c30ce4e72cbd1a967ff759c1d4).
https://en.wikipedia.org/wiki/Logarithm#Definition

Hi.
An equivalent and more succinct definition is that the function logb is the inverse function to the function .
It might have been accidental but this editing has totally corrupted the original meaning. It's like quoting "Yes...I agree" from an extract that read "Yes that is interesting. However I can't agree with your conclusion".
1. It's the last line in a block of text where the first lines were more specific and excluded certain values.
2. There was a hyperlink on the word "inverse function" which has been lost. That link discusses the inability to find an inverse functions in some cases.
               
You also haven't addressed this point:
So, you're saying that Log_{0} (10) = 0 if I've understood that correctly.
So, by your definition of a Logarithm we have that 0^{ 0} = 10.
How do you reconcile that with the proposal you made that 0^{0} = 1 ?
                
NOTE: I am not trying to be rude here. It just takes me hours to keep writing long posts. In the same style as your own posts, it's more useful to keep the points short and repeat anything that was important that was missed.
If you want to have 0^{0} =1 then that's fine by me. I was just assuming you wanted some discussion.
Best Wishes.

1. Take b= 0 as the base for the logarithm (or the exponentiation). 0^1 = 0 but also 0^2 = 0 and 0^ 3 = 1 etc.
So which value do you map the value 0 back to when you try to invert that function? Is Log0 (0) = 1 , 2 or 3 ? To say that another way, exponentiation is sometimes a manytoone function so that no inverse function exists.
What I'm trying to say is that it isn't as simple as saying "Logarithms are just the reverse of exponentiation", sometimes you can't find a reverse procedure.
The thread title is about finding a value for a mathematical expression, regardless if it's a function or not.
√1=1 but also √1=1
I don't think it's a reason to say that √1 is undefined.

It might have been accidental but this editing has totally corrupted the original meaning. It's like quoting "Yes...I agree" from an extract that read "Yes that is interesting. However I can't agree with your conclusion".
1. It's the last line in a block of text where the first lines were more specific and excluded certain values.
2. There was a hyperlink on the word "inverse function" which has been lost. That link discusses the inability to find an inverse functions in some cases.
It just shows that the definition I used here is not just my personal opinion like what you said.

NOTE: I am not trying to be rude here. It just takes me hours to keep writing long posts. In the same style as your own posts, it's more useful to keep the points short and repeat anything that was important that was missed.
If you want to have 00 =1 then that's fine by me. I was just assuming you wanted some discussion.
I think it's better to keep a post focused on one or a few points, which would make it easier to address in a response. Besides, sometimes I have to open the site from a mobile device, which makes long posts harder to read or write. Sometimes the page restarted unexpectedly, which made my writing lost. When it happens, sometimes I just don't bother to rewrite it.

So, you're saying that Log0 (10) = 0 if I've understood that correctly.
So, by your definition of a Logarithm we have that 0^0 = 10.
How do you reconcile that with the proposal you made that 0^0 = 1 ?
Here are some other results.
Log0 (100) = 0
Log0 (1000) = 0
Log0 (1) = 0
Log0 (10) = 0
Apparently, using logarithm is not a correct way to answer the question in the OP.
Simple negative exponent is better for this task, as offered by Mascheroni.
(https://wikimedia.org/api/rest_v1/media/math/render/svg/95a871955aba3ddb741357a9ae1a827355aca7d1)

Hi.
as offered by Mascheroni.
I hadn't actually heard of him (The internet says Lorenzo Mascheroni was an Italian mathematician).
I'm not sure how he got from to = 1. It looks suspiciously like some repeated division by (aa) was done.
i.d.k. maybe he had something else in mind.
Best Wishes.

What's 0 to the power of 0?
0^0 = Owl, no?

∀n∩ℜ, log_{n}0 = ∞ so 0^{0} is undefined by this route.

Hi.
Looks sensible, @alancalverd . There might be one symbol out of place but it's not serious.
∀n∩ℜ
∀n∈ℜ
Best Wishes.

Agreed, and thank you for the correction!

0^0 = Owl, no?
Ah, yes. I see it now said piglet

Hi.
as offered by Mascheroni.
I hadn't actually heard of him (The internet says Lorenzo Mascheroni was an Italian mathematician).
I'm not sure how he got from to = 1. It looks suspiciously like some repeated division by (aa) was done.
i.d.k. maybe he had something else in mind.
Best Wishes.
The equation is found in Euler's book, which attributes it to Mascheroni.
In 1752, Euler in Introductio in analysin infinitorum wrote that a0 = 1[15] and explicitly mentioned that 0^{0} = 1.[16] An annotation attributed[17] to Mascheroni in a 1787 edition of Euler's book Institutiones calculi differentialis[18] offered the "justification"
https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero#As_a_value

∀n∩ℜ, log_{n}0 = ∞ so 0^{0} is undefined by this route.
Is log_{n}0 well defined?
How do you define "undefined" in mathematical sense?

Acording to Mathmetica which I consider the last word on maths 0^0 is inditerminate

Is logn0 well defined?
Yes.Of all the infinities, the real continuum infinity is perhaps the best characterised and the easiest to work with.

Hi.
Is logn0 well defined?
No, not in the Reals. There's no Real number which is assigned to Log_{n} (0).
How do you define "undefined" in mathematical sense?
It depends on the context. In it's simplest usage "undefined" just means "this thing is not defined". In the context of this thread I think the phrase "undefined" has most commonly been used to suggest that no real value is assigned to the expression. However, I'm sure some people have used it in a different context or to convey some other meaning. It may be best if you identitfy the specific sections where the phrase was used.
Best Wishes.

Hi.
Is logn0 well defined?
No, not in the Reals. There's no Real number which is assigned to Log_{n} (0).
How do you define "undefined" in mathematical sense?
It depends on the context. In it's simplest usage "undefined" just means "this thing is not defined". In the context of this thread I think the phrase "undefined" has most commonly been used to suggest that no real value is assigned to the expression. However, I'm sure some people have used it in a different context or to convey some other meaning. It may be best if you identitfy the specific sections where the phrase was used.
Best Wishes.
In your definition, √(1) is undefined. It makes i, as well as Euler's identity, e^{πi}=1, meaningless.

Entirely true, since √1 is not a real number and is therefore not defined within the set of reals. But Euler uses complex numbers, which include an imaginary part, a multiple of i.
I bow to ES's statement: since you can't assign a numerical value to −∞ it isn't a member of ℜ. But it isn't imaginary either!

Entirely true, since √1 is not a real number and is therefore not defined within the set of reals. But Euler uses complex numbers, which include an imaginary part, a multiple of i.
I bow to ES's statement: since you can't assign a numerical value to −∞ it isn't a member of ℜ. But it isn't imaginary either!
So, i is defined, even though it's not a real number? It implies that being a real number is not a necessity for something to be mathematically defined.
Can you assign a numerical value to π?
What makes −∞ different?

True.
You can assign a real number that is infinintesimally less than π, and one that is greater, so you can define π as "that which lies between..." . You can't assign a real number that is infinitesimally greater or less than ∞, so as ES says, it isn't defined on the real number line.
a real number is a value of a continuous quantity that can represent a distance along a line
i is completely defined as √1 but it doesn't lie on the line that passes through 0, 1,  2.995, π^{2.7} , etc. so it isn't a real number.

Hi.
So, i is defined, even though it's not a real number?
As alancalverd said, that is entirely true.
It's not to be given too much importance. The meaning of a word like "unconstitutional" is defined in some system (we might call that system the English language, for example) but it is not defined in the real number system.
Which leads us to the next comment. There are a whole load of things that are defined but aren't real numbers. There's even a whole load of things that are "mathematically defined" but aren't real numbers.
It implies that being a real number is not a necessity for something to be mathematically defined.
Yes, entirely true. There are many axiom systems, or many different systems of mathematics. It is quite possible for something to be "mathematically defined" but not be a real number. The set of all rotations in 3dimensional space is a structure called a group under the group operation • which is just the composition of two rotations. It's a perfectly good mathematical structure, all the objects in it are mathematically defined etc.  you can do things with that structure that we would all consider as being mathematics. However, none of the objects in that structure are real numbers.
I should think it is possible to make everything perfectly well defined in SOME system but that system isn't always useful. (Actually this is debatable  if the system is built from standard set theory then there are problems defining things like the set of all sets which do not contain themselves  but let's just assume that in general you can make everything well defined in some system of mathematics). Let's just bring this back to the main topic of the thread: There may be some system in which the expression 0^{0} is defined, if that's what you wanted to know. However, it's not the conventional real number system and where the symbol 0 and the notation for exponentiation a^{b} is used to produce the expression 0^{0}. In the real number system, the expression 0^{0} is deliberately left "undefined": It is a meaningless expression, it's just a collection of symbols put together. There is a convention that 0^{0} =1 which is very useful for things like writing out power series in the form because you don't have to keep writing the constant term a_{0} separately from the sigma notation when x could be 0. (But that's just a convention for ease of notation).
Why is it deliberately left undefined? See earlier discussions. There are problems maintaining all the properties we may want from the exponential function (like continuity) and/or all the usual rules of manipulating indices in an algebraic expression.
Can you just add the definition 0^{0} =1 ? Sure you can, just declare it and the job is done. It doesn't make the aforementioned problems go away but there's no law that says you can't extend the definition of a^{b} like that and no special squad of mathematicians, let's call them the continuity police, that will come knocking on your door to punish you for that decision.
However, since the problems haven't actually just gone away, you're going to have to be very careful when working with expressions like 0^{0} and not assume all the usual rules of algebraic manipulation and/or analytical conditions like continuity apply. So you're still almost always going to have to treat the appearance of an expression like 0^{0} as a special or separate case from the appearance of most other exponential expressions like a^{b}. You gained nothing useful just by adding the definition 0^{0} =1. What we would really want is that all the properties of exponents we've come to expect for a^{b} continue to hold when a and b are both 0 and that does seem to be impossible (it produces inconsistencies) in the Real Number system. You can't have all the usual properties holding you have to pick and choose just a few. I agree that setting 0^{0} = 1 probably allows the greatest amount of these properties to hold but the important thing is that it's not all of them.
Best Wishes.

I agree that setting 0^{0} = 1 probably allows the greatest amount of these properties to hold but the important thing is that it's not all of them.
OK, deal.
I see the reluctance from accepting to define a value of a mathematical operation is related to stability of the result when the operands are perturbed. Here are my findings for mathematical operations with defined results:
0 * 0 is extremely stable. The operands can be perturbed heavily without changing the result (e.g. by multiplying them with some constants). Adding one of the operands with a finite constant doesn't change the result. Adding both operands with a finite constant does change the result.
1 * 0 is very stable. The first operand can be perturbed heavily without changing the result. Multiplying the second operand with a finite constant doesn't change the result, but adding it with a finite constant does change the result.
1 * 1 is less stable. Perturbing one of the operands with a finite constant does change the result. But multiplying them with a constant infinitesimally close to 1, or adding them with a constant infinitesimally close to 0 doesn't change the result.

1 * 1 is less stable. Perturbing one of the operands with a finite constant does change the result. But multiplying them with a constant infinitesimally close to 1, or adding them with a constant infinitesimally close to 0 doesn't change the result.
Oh yes it does!
(1 * Δ)^{2} = Δ^{2} ≠ 1
(1+Δ) ^{2} = 1 + 2Δ + Δ^{2} ≠ 1.

Hi.
I see the reluctance from accepting to define a value of a mathematical operation is related to stability of the result when the operands are perturbed.
I don't really know what you mean by "stability". It sounds like you're considering continuity when you consider infinitessimal perturbations.
Best Wishes.

Not sure that stability is ever a problem when dealing with real numbers since their functions are inherently continuous. It can be a problem when modelling a continuous function with discretised variables but that's computing, not maths!

1 * 1 is less stable. Perturbing one of the operands with a finite constant does change the result. But multiplying them with a constant infinitesimally close to 1, or adding them with a constant infinitesimally close to 0 doesn't change the result.
Oh yes it does!
(1 * Δ)^{2} = Δ^{2} ≠ 1
(1+Δ) ^{2} = 1 + 2Δ + Δ^{2} ≠ 1.
Ok. They don't change the result much.

Hi.
I see the reluctance from accepting to define a value of a mathematical operation is related to stability of the result when the operands are perturbed.
I don't really know what you mean by "stability". It sounds like you're considering continuity when you consider infinitessimal perturbations.
Best Wishes.
Yeah it does.
0^0 is more stable than 1*1 when the perturbation is in form of multiplication with the operands.
The first operand is stable against addition, but the second operand is not.

Acording to Mathmetica which I consider the last word on maths 0^0 is inditerminate
What's the consideration behind its answer? Argument from authority is not a strong one in a scientific discussion.

Ok. They don't change the result much.
Squeeze the trigger slowly and you'll appreciate the enormous difference between "not much" and "unstable".

1 * 1 is less stable. Perturbing one of the operands with a finite constant does change the result. But multiplying them with a constant infinitesimally close to 1, or adding them with a constant infinitesimally close to 0 doesn't change the result.
Oh yes it does!
(1 * Δ)^{2} = Δ^{2} ≠ 1
(1+Δ) ^{2} = 1 + 2Δ + Δ^{2} ≠ 1.
Ok. They don't change the result much.
Squeeze the trigger slowly and you'll appreciate the enormous difference between "not much" and "unstable".
How can 1*1 unstable?

It isn't. Only you suggested it was.

It isn't. Only you suggested it was.
I said 1*1 is less stable than 0*0.

Hi.
This kind of "stability" idea just isn't worried about too much, to the best of my knowledge. I think that's why you (Hamdani) aren't getting many good replies about the issue. The term "stability" is usually applied to other things in mathematics and I'm not sure that you (Hamdani) need to develop a new set of ideas for the stability of basic field operations like addition and multiplication. There is already a simple idea called continuity which seems to offer all we need.
However, if you want to develop ideas for this sort of stability, then I don't suppose there's any law against it.
Here's something to think about if you are developing some ideas for stability of basic field operations like + and x on different choices of input numbers:
You mentioned that 0 x 0 was very stable but that depends on how you measure stability. Surely you'd want to consider a proportional or percentage change? A change of +/ 1 unit in an expected answer of 100 units is nothing to worry about but a change of +/ 1 unit in an answer of 0.01 units is a tenthousand percent error. Anyway, using this notion of proportional difference, 0 x 0 and also 0 + 0 isn't stable, it's terrifyingly unstable. Perturb the input numbers even a tiny amount and you generate a proportional difference that is unbounded. No other choice of input numbers is that bad.
Best Wishes.

A simple example of a trigger instability is the equation y = 1/x. As x passes through zero its value switches between +∞ and ∞.

You mentioned that 0 x 0 was very stable but that depends on how you measure stability.
0 * 0 is extremely stable. The operands can be perturbed heavily without changing the result (e.g. by multiplying them with some constants). Adding one of the operands with a finite constant doesn't change the result. Adding both operands with a finite constant does change the result.
Perturbation by multiplication already implies ratio. Percentage change simply converts perturbation from addition to multiplication.
Anyway, using this notion of proportional difference, 0 x 0 and also 0 + 0 isn't stable, it's terrifyingly unstable. Perturb the input numbers even a tiny amount and you generate a proportional difference that is unbounded. No other choice of input numbers is that bad.
Can you show how?

A simple example of a trigger instability is the equation y = 1/x. As x passes through zero its value switches between +∞ and ∞.
Yes. I understand if that instability makes people think that 1/0 is undefined.
https://www.wolframalpha.com/input?i=1%2F0
Wolframalpha says that 1/0 is complex infinity instead of undefined.
1/0=1/0=i/0=i/0=complex infinity. In Riemann's sphere, they are located on the same point.
There are seven indeterminate forms involving 0, 1, and infinity:
https://mathworld.wolfram.com/Indeterminate.html
(https://mathworld.wolfram.com/images/equations/Indeterminate/NumberedEquation1.svg)
If complex infinity is allowed as well, then six additional indeterminate forms result:
(https://mathworld.wolfram.com/images/equations/Indeterminate/NumberedEquation2.svg)

Hi.
Anyway, using this notion of proportional difference, 0 x 0 and also 0 + 0 isn't stable, it's terrifyingly unstable. Perturb the input numbers even a tiny amount and you generate a proportional difference that is unbounded. No other choice of input numbers is that bad.
Can you show how?
0 + 0 = 0 = the expected answer for 0 + 0.
Perturb the input values slightly and consider δ + ε where δ, ε are arbitrarily small but positive.
Then δ + ε ≠ 0, instead δ + ε = something small but positive. The percentage change from the expected answer, 0, is then undefined:
(something positive) / 0 x 100%
 the division by 0 is a problem.
Best Wishes.

0 + 0 = 0 = the expected answer for 0 + 0.
Perturb the input values slightly and consider δ + ε where δ, ε are arbitrarily small but positive.
Then δ + ε ≠ 0, instead δ + ε = something small but positive. The percentage change from the expected answer, 0, is then undefined:
(something positive) / 0 x 100%
 the division by 0 is a problem.
Best Wishes.
Perturbation by multiplication already implies ratio. Percentage change simply converts perturbation from addition to multiplication.

Hi.
Perturbation by multiplication already implies ratio. Percentage change simply converts perturbation from addition to multiplication.
? I've no idea what you neant there.
Best Wishes.