Naked Science Forum
On the Lighter Side => New Theories => Topic started by: sim on 10/01/2019 11:35:30

1)Mathematics/science end in contradiction  an integer= a noninteger. When mathematics/science end in contradiction it is proven in logic that you can prove anything you want in mathematics ie you can prove Fermat's last theorem and you can disprove Fermats last theorem
http://gamahucherpress.yellowgum.com/wpcontent/uploads/Allthingsarepossible.pdf (http://gamahucherpress.yellowgum.com/wpcontent/uploads/Allthingsarepossible.pdf)
The paper proves 1=0.999...thus maths ends in contradiction
when
1 is a finite number it stops
A finite decimal is one that stops, like 0.157
A nonfinite decimal like 0.888.. does not stop
A nonfinite decimal like 0.999... does not stop
when a finite number 1 = a nonfinite number 0.999.. then maths ends in contradiction
another way
1 is an integer a whole number
0.888... is a noninteger it is not a whole number
0.999... is a noninteger not a whole number
when a integer 1 =a noninteger 0.999... maths ends in contradiction

and you can disprove Fermats last theorem
Go on then...
Incidentally, the 0.999... =1 thing just shows that there are two ways of writing the same number.
Not a problem with maths, a minor problem with our writing system.

when a integer 1 =a noninteger 0.999... maths ends in contradiction
Was that you on Sciforums?
The problem is knowing the difference between an equality and a limit. They are not equivalent!

I can explain why 1 = 0.9999^{.}
0.9999^{.} = 9/10 + 9/100 + 9/1000 + ....
 This is a "Geometric series" = a + a*r + a*r^{2} + a*r^{3} + ...
 In this case, with initial value a=9/10 and ratio between successive terms =1/10.
 The sum of an arbitrary geometric series = a/(1r) = (9/10)/(11/10) = (9/10)/(9/10) = 1
So it is true that 1 = 0.9999^{.}!
0.999 ... a nonfinite number
The paper gets this wrong.
0.9999... is greater than 0, and less than 2 (both of them are finite), so it is definitely finite!
1 number (2) +1 number (2) =1 number (4) .... therefore 1+1=1
The paper is also confused here.
 The addition operator adds the magnitudes of the numbers.
 It does not attempt to count the numbers, since there are an infinite number of ways you can add up numbers to achieve a certain total
if a theory is inconsistent it will contain every sentence of the language
A better expression expression would be to refer to Godel's incompleteness theorem, which can be summarised as "A sufficiently complex system will either be incomplete, or it will be inconsistent".
https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems
There is little debate that science and mathematics are incomplete. Some parts also seem inconsistent, but attempts are made to avoid and or clarify these inconsistencies.
Politics and sociology are also incomplete and often inconsistent...
So with each opponents view being valid so there is no need/point to argue anymore
I argue that the paper as presented is inconsistent...

If u=.999...9 for k positions,
then 1u=.000...1 for k positions.
There is always a remainder >0.
Define u=( (1/10)n, for n=1 to ∞.
The limit of u=1,
meaning u gets closer to the value 1 as n increases without limit.
The limit is an unapproachable boundary which is why u cannot be expressed as an equality.
If u=1 when n=k, the limit statement becomes false.
The repeating sequence of 9's is equated to 1 by rounding the expression, considering the remainder insignificant. This is not a flaw in math concepts, just relaxing the rules for the purpose of approximation.

10 times 0.999... is 9.999....
and 10 times 0.999  0.999 is 9 times 0.999...
And 9.999... 0.999.... = 9
So 9 times 0.999... is 9
so 0.999... is 9/9
since 9/9 is 1
So 0.999... is 1

Wiki will provide a good definition of a 'limit".
u=.999... or 9R
1(.999) =(.001)
1(.999...........9) = (.000..........1)
By definition all positions in u are 9's.
The difference d=(1u) is always a positive number 1/10^{n} at position n.
10uu = 1010d = 99d<1.
Another simple example is x = 1/n.
The limit of x is 0, but x never equals 0.
In your line 1, you multiply the approximation by 10, but not the small difference d.

What is 9.9  0.9 ?
What is the difference between the number of nines after the decimal point in both cases?
What about 9.99  0.9?
And 9.99  0.99?

0.(9) seems to be a number within the interval (0,1) but not in (0,1]
whereas 1 can be found in (0,1] and not in (0,1)
Does it still mean 0.(9)=1 ?

0.(9) seems ...
To whom, and why?

0.(9) seems ...
To whom, and why?
To me.
...because, I thought that 0.999... never reaches 1. But you are right, 0.(9) doesn't clearly mean it belongs to (0,1) and not to (0,1].
Another thing, 1/3 is said to be 0.(3), but is it? What if it is again, belonging to (0,1/3) rather than to (0,1/3]?

0.999999...
is the same as 1
it's just 2 ways of writing the same number

0.999999...
is the same as 1
it's just 2 ways of writing the same number
Yes, it may be but your explanation is flawed because of circular reference.
You assume 9.999... is 10x0.999... But you need to demonstrate it first.
9.99999x10=99.99990 not 9.99999 so what happens if we go on with bigger and bigger numbers? You will always have that 0 decimal, you can't get rid of it although the difference between the two numbers is smaller and smaller.

You assume 9.999... is 10x0.999... But you need to demonstrate it first.
I don't think I need to show that you can multiply a number by 10 by moving the decimal point one place to the right.
And I understand your point, but perhaps you might go back and answer my earlier questions; especially this one
What is the difference between the number of nines after the decimal point in both cases?

You assume 9.999... is 10x0.999... But you need to demonstrate it first.
I don't think I need to show that you can multiply a number by 10 by moving the decimal point one place to the right.
And I understand your point, but perhaps you might go back and answer my earlier questions; especially this one
What is the difference between the number of nines after the decimal point in both cases?
By multiplying by 10 you move the decimal point to the right, but in this case you get a number that is written the same way, 9.999...
What is the difference between the number of nines?
The difference is always 1 for any finite number of decimals, but I cannot tell for an infinite number. Does it vanish?

How many nines are there after the decimal point in 0.9999....
How many nines are there after the decimal point in 9.9999....
What is the difference between those two numbers?

I don't know. ∞∞ undetermined.

How many nines are there after the decimal point in 0.9999....
How many nines are there after the decimal point in 9.9999....
What is the difference between those two numbers?
You and like minded viewers are looking at the example with the preconceived notion that .999...equals 1. Any mathematical manipulation will then produce the desired result, supporting your interpretation.
but,
[9+(1d)]  (1d) =9. with d=1/10^n, the remainder that is always>0.
If you read the wiki article on 'limits', you should understand.

0.9999.... however it is signified, means "not quite 1.0" so it can't even be numerically equivalent to 1, let alone an integer.

0.9999.... however it is signified, means "not quite 1.0" so it can't even be numerically equivalent to 1, let alone an integer.
Have you actually read, and understood, the thread?
I don't know. ∞∞ undetermined.
It's precisely the same thing on either side of the "" sign.
And the difference between two identical things is zero.
What is the difference between the number of nines?
Zero as I just pointed out.You and like minded viewers are looking at the example with the preconceived notion that .999...equals 1.
No, I'm not assuming it at all.
I'm showing it.
If you read the wiki article on 'limits', you should understand.
OK, let's look at limits.
What's the limit of the sum of 9 x 10^n where n ranges from 1 to infinity?
Well, it's clearly 9 times
0.111111111...
which is the sum of 0.1 +0.01 +0.001... and so on (in the limit)
And that's the sum of a geometric progression
http://mathematics.laerd.com/maths/geometricprogressionintro.php
with a common ratio of 10
And that sum is 1/(101)
which is 1/9
So the original number i.e. 0.9999... must be 9/9 which is 1.

0.9999.... is equal to exactly 1.
There are many ways of proving this using generally accepted and previously established methods, either algebraically or using limits. If you want to reinvent mathematics such that this is not the case, you may well find that your version of math gives paradoxical or nonsensical answers. (there are still some holes in our system, but overall it works pretty well, and this is not one of those holesif you really want to tug at the very fiber of our system, look at the "axiom of choice" and the "continuum hypothesis.")
I would urge caution for those who have difficulty interpreting numbers that cannot be expressed perfectly decimal notation. Beware of irrational numbers (which cannot be expressed perfectly as a ratio of any two whole numbers), and especially of transcendental numbers (which cannot be expressed perfectly as a polynomial combination of rational numbers). These numbers are all very much real, easily defined (most of 'em), useful (some of 'em), and together vastly outnumber the "wellbehaved" numbers...

u =.1 + .01 + .001 +...+1/10^n
10u = 1 + u – 1/10^n
9u = 1 – 1/10^n
u = 1/9 – (1/10^n)/9
An 'infinite' sequence is never complete since 'infinity' is not a number.
Why did the early mathematicians define a 'limit' instead of declaring an equality?

An 'infinite' sequence is never complete
Then 0.9999.... does not exist and we don't need to worry about it.
So, you can stop now.