Naked Science Forum
On the Lighter Side => New Theories => Topic started by: Yahya A.Sharif on 08/08/2019 17:06:14

We have problems in physics that involve simple arithmetic such as division of two numbers, for example the speed v of an object at stationary equals zero, the time it elapses t is zero and the distance it ravels is also zero, however 0/0 is not defined and although its speed has a defined value which is zero.
Gravity at infinity equals zero, and that is according to the equation:
F=GMm/r^2 , although the value infinity for r can't be added to the equation , but as r approaches infinity F approaches zero.
Even though gravity at infinity equals zero but the expression : F=GMm/(infinity) is not defined.
Redfining simple arithmetic expressions:
It acceptable that:
1+1=2, that is if we have one apple and we put another apple beside it then we will have a total of 2 apples.
4/2=2, that is if we have 4 apples and we want to divide them between two people , then we will give each person 2 apples.
Let's define other expressions with the same logic, these expressions are the subject of speaking:
4/0=0 , we have 4 apples and we want to divide them but we don't have anybody, then we won't divide at all , the process won't occur , in such case the number of apples we divide is zero.The result is zero apples have been divided.
0/0=0, we have nothing to divide and we have no people to divide between , then again the process won't occur at all and while we have nothing , we will divide nothing, and while we have no people then we have noone to give them anything.The result is zero apples have been divided.
4/(infinity)=0
The same idea we have 4 apples but the number of people is not fixed they change and increases without bound then we can't do the process of dividing, and we can't give any apple to anyone.The process didn't occur and the result is 0 apples have been divided.
Infinity here is not a point or a very large number, infinity is increment in the value of some variable continuously and without bound.
According to these definition of simple arithmetic expression, we can redfine some equations so that they have meaning:
v=X/t=0/0=0, and that is at the origin when t=0 and x=0
F=GMm/r^2=GMm/(infinity)=0
Relativistic kinetic energy:
K.E=mc²/√(1v²/c²)mc²
At the speed v=c for a mass, the expression √(1v²/c²) will equal zero:
K.E=mc²/0  mc² , K.E = 0mc²
K.E=mc²
But energyof mass m equals:
E=mc²
K.E=E
The kinetic energy of a mass m moving at the speed of light c equals the energy of the mass, and that what is the kinetic energy of a photon is, the kinetic energy of a photon is its contained energy, the kinetic energy of a mass m moving at v=c is its contained energy E , i.e the energy of its mass
The article in vixra:
http://vixra.org/abs/1908.0119

we can redfine some equations so that they have meaning:
They already had meaning.
However, division by zero does not.

we can redfine some equations so that they have meaning:
They already had meaning.
They don't, speed equation is a function of the graph x/t that represent all points including t=zero
However, division by zero does not.
It should.

The result is zero apples have been divided.
Which means that you have not done any division.
You can't define "division" to mean "not doing division".
It should.
No.
It should not, because , if you do, you get contradictions.
You really are not going to win this argument after hundreds of years.

speed equation is a function of the graph x/t
Speed has nothing to do with graphs.

You can't define "division" to mean "not doing division".
I'm defining " the result of division", division is givining the 4 apples to the two people half for each one"4/2" the result is each one will take two
It should.
No.
It should not, because , if you do, you get contradictions.
What are these contradictions?
You really are not going to win this argument after hundreds of years.
It is like saying Einstein won't win any arguments against Newton.
Speed has nothing to do with graphs.
really? it is only to mention that a graph is a visual represntation of speed equation

Einstein won't win any arguments against Newton.
Einstein had evidence.
I'm defining " the result of division", division is givining the 4 apples to the two people half for each one"4/2" the result is each one will take two
And, your "definition" does not work for division by zero.
What are these contradictions?

speed equation is a function of the graph x/t
Speed has nothing to do with graphs.
This only part of what I said, It is meaningless to qoute a part of a full sentence.
Einstein had evidence.o/o=o for speed,and relatistic kinetic energy
I presented evidence too.
It is all about dividing by zero is not possible, I agree
If you think the video has contradictions give me an example that contradict my post to discuss it here

Let's define other expressions with the same logic, these expressions are the subject of speaking:
4/0=0 , we have 4 apples and we want to divide them but we don't have anybody, then we won't divide at all , the process won't occur , in such case the number of apples we divide is zero.The result is zero apples have been divided.
Another way of stating division is by making reference to multiplication. Asking "what is four divided by two?" is the mathematical equivalent of asking "what multiplied by two equals four?" In this case, the answer is two. When you try this with zero, it doesn't work. "What is four divided by zero?" is equivalent to "what multiplied by zero equals four?" There is no answer to this question. There is no number you can multiply by zero and get four. So the answer is undefined.
0/0=0, we have nothing to divide and we have no people to divide between , then again the process won't occur at all and while we have nothing , we will divide nothing, and while we have no people then we have noone to give them anything.The result is zero apples have been divided.
This is undefined as well, but for a different reason. "What is zero divided by zero?" is equal to "what multiplied by zero equals zero?" The answer can be any number at all. Any number multiplied by zero is zero, so you don't have any one answer. That's why this is also undefined.
4/(infinity)=0
The same idea we have 4 apples but the number of people is not fixed they change and increases without bound then we can't do the process of dividing, and we can't give any apple to anyone.The process didn't occur and the result is 0 apples have been divided.
Infinity here is not a point or a very large number, infinity is increment in the value of some variable continuously and without bound.
What multiplied by infinity equals four? That question also has no answer, so it's undefined.

Let's define other expressions with the same logic, these expressions are the subject of speaking:
4/0=0 , we have 4 apples and we want to divide them but we don't have anybody, then we won't divide at all , the process won't occur , in such case the number of apples we divide is zero.The result is zero apples have been divided.
Another way of stating division is by making reference to multiplication. Asking "what is four divided by two?" is the mathematical equivalent of asking "what multiplied by two equals four?" In this case, the answer is two. When you try this with zero, it doesn't work. "What is four divided by zero?" is equivalent to "what multiplied by zero equals four?" There is no answer to this question. There is no number you can multiply by zero and get four. So the answer is undefined.
0/0=0, we have nothing to divide and we have no people to divide between , then again the process won't occur at all and while we have nothing , we will divide nothing, and while we have no people then we have noone to give them anything.The result is zero apples have been divided.
This is undefined as well, but for a different reason. "What is zero divided by zero?" is equal to "what multiplied by zero equals zero?" The answer can be any number at all. Any number multiplied by zero is zero, so you don't have any one answer. That's why this is also undefined.
4/(infinity)=0
The same idea we have 4 apples but the number of people is not fixed they change and increases without bound then we can't do the process of dividing, and we can't give any apple to anyone.The process didn't occur and the result is 0 apples have been divided.
Infinity here is not a point or a very large number, infinity is increment in the value of some variable continuously and without bound.
What multiplied by infinity equals four? That question also has no answer, so it's undefined.
For other numbers division and multiplication occured .For dividing by zero it didn't occur, so we can't compare the two. but both have a result
Division didn't occur and that resulted in zero.
Division and nondivision are not the same, but division resulted in nondivision and nondivision resulted in zero , so division resulted in zero

Einstein had evidence.
I presented evidence too. 0/0=0 for speed ,gravity equation and the relativistic kinetic energy equation

For other numbers division and multiplication occured .For dividing by zero it didn't occur, so we can't compare the two.
Division didn't occur and that resulted in zero.
Division and nondivision are not the same, but division resulted in nondivision and nondivision resulted in zero , so division resulted in zero
If the division never happened, then you never got an answer. Which is the same as saying that it is undefined.

For other numbers division and multiplication occured .For dividing by zero it didn't occur, so we can't compare the two.
Division didn't occur and that resulted in zero.
Division and nondivision are not the same, but division resulted in nondivision and nondivision resulted in zero , so division resulted in zero
If the division never happened, then you never got an answer. Which is the same as saying that it is undefined.
This y/z=x/0 has no meaning, since there is division in the left which occered and definition in the right which didn't, we can't compare the two with equality sign.
but this has a meaning:
V=x/t=0/0, since we have a vairable in the left we want to know its value , V equals the result of the nondivision

V equals the result of the nondivision
Then V is undefined. Nondivision can't give you an answer, because it isn't a mathematical operation.

V equals the result of the nondivision
Then V is undefined. Nondivision can't give you an answer, because it isn't a mathematical operation.
It was not a mathematical operation before, but I now add it to mathematical operations, which works for undefined cases.
x/0 , x/∞, 0/0, etc are unique cases in which the operation on them is applied to any variable .
I presented evidence of three equation examples, which turn out to satisfy .

It was not a mathematical operation before, but I now add it to mathematical operations, which works for undefined cases.
It doesn't work. It is literally a lack of a mathematical operation.
I presented evidence of three equation examples
They don't give an answer because you didn't actually do anything at all. You can't get an answer by not doing anything. That's like trying to get a result from a computer program by not running the program. By definition, you don't get a result.
which turn out to satisfy .
No it doesn't. It gives the wrong answer. Zero multiplied by zero does not equal four.

They already had meaning. However, division by zero does not.
I beg to differ with some of my esteemed colleagues; I think that the case of 0/0 is very important in many areas of maths and physics.
 But you can't say what the answer is without more information.
 The process of mathematics (mostly) gives you that information.
Differentiation may be used to calculate the velocity of an object falling under gravity.
 Technically, differentiation is a limit of Δy/Δx as Δx & Δy approach 0
 Differentiating (when it works) is effectively the result when Δy=Δx=0
 If you have an equation for this motion (eg s=½gt^{2}), then the equation provides the necessary additional information (provided the equation meets certain criteria such as smoothness and lack of infinities at this location).
 Differentiating, you have v=gt, which is effectively the result when Δs=Δt=0
 Differentiation is a basic function in maths & physics, and it (mostly) works well, provided you are a bit careful
Let's take another example, which is applicable to your cellphone:
 If you generate a signal with a spectrum of sin(f)/f, it has some nice properties about not interfering with other signals at f=±π, ±2π, ±3π, etc
 But what is the amplitude at f=0? sin(f)/f= 0/0, so you need some more information.
 In this case, the additional information can be provided by L'Hopital's rule:
 If you differentiate the top and bottom, the answer remains the same
 At f=0, sin(f)/f = 0/0 = cos(f)/1 = 1/1=1
0/0=0....The result is zero apples have been divided.
The OP gave the wrong answer in this case, because he did not provide (or have access to) the necessary additional information.
If we take another case: sin(2f)/f at f=0, it also looks like 0/0.
However, when you differentiate at f=0, you get sin(2f)/f=0/0=2cos(2f)/1=2/1 = 2
The answer is different because the additional information in the equation showed it was different.
See: https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
Edit: The π was halfbaked...

They already had meaning. However, division by zero does not.
I beg to differ with some of my esteemed colleagues; I think that the case of 0/0 is very important in many areas of maths and physics.
 But you can't say what the answer is without more information.
 The process of mathematics (mostly) gives you that information.
Differentiation may be used to calculate the velocity of an object falling under gravity.
 Technically, differentiation is a limit of Δy/Δx as Δx & Δy approach 0
 Differentiating (when it works) is effectively the result when Δy=Δx=0
 If you have an equation for this motion (eg s=½gt^{2}), then the equation provides the necessary additional information (provided the equation meets certain criteria such as smoothness and lack of infinities at this location).
 Differentiating, you have v=gt, which is effectively the result when Δs=Δt=0
 Differentiation is a basic function in maths & physics, and it (mostly) works well, provided you are a bit careful
Let's take another example, which is applicable to your cellphone:
 If you generate a signal with a spectrum of sin(f)/f, it has some nice properties about not interfering with other signals at f=±1, ±2, ±3, etc
 But what is the amplitude at f=0? sin(f)/f= 0/0, so you need some more information.
 In this case, the additional information can be provided by L'Hopital's rule:
 If you differentiate the top and bottom, the answer remains the same
 At f=0, sin(f)/f = 0/0 = cos(f)/1 = 1/1=1
0/0=0....The result is zero apples have been divided.
The OP gave the wrong answer in this case, because he did not provide (or have access to) the necessary additional information.
If we take another case: sin(2f)/f at f=0, it also looks like 0/0.
However, when you differentiate at f=0, you get sin(2f)/f=0/0=2cos(2f)/1=2/1 = 2
The answer is different because the additional information in the equation showed it was different.
See: https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
Excellent reply.Thanks

I would also like to add that I think that the case of ∞*0 is also very important in many areas of maths and physics.
 But you can't say what the answer is without more information.
 The process of mathematics (mostly) gives you that information.
Integration may be used to calculate the distance that an object has fallen under gravity.
 Technically, integration divides the xaxis into n small segments of width Δx
 Integration is a limit of nΣy(x)*Δx as n approaches ∞
 Integration (when it works) is effectively the result when n=∞ and Δx=0
 If you have an equation for motion (eg v=gt), then the equation provides the necessary additional information (provided the equation meets certain criteria such as smoothness and lack of infinities at this location).
 Integration is a basic function in maths & physics, and it (mostly) works well, provided you are a bit careful

Evan it came to my mind something:
Actually limits are not included in my assumption , Limit of 1/x² as x approaches zero doesn't mean it will actually reach zero in x values , so when I say 1/0=0 I'm not taking about the value of the graph of 1/x² at x=0 ,I'm talking about the value of the arithmetic expression 1/0 .The graph doesn't end up in a a value of y=1/0 "zero is undefined" the graph behaves differently.

It was not a mathematical operation before, but I now add it to mathematical operations, which works for undefined cases.
It doesn't work. It is literally a lack of a mathematical operation.
Lack of mathematical operation is not dividing from the first place, when a moving object with constant speed hit another object it stops, that doesn't mean it didn't move at the beginning., in fact we could know the kinetic energy of the other object from the kinetic energy of the first one.
No it doesn't. It gives the wrong answer. Zero multiplied by zero does not equal four.
You can't reverse the operation , These expressions are a special case , the reverse operation doesn't work for them. This is a rule for my assumption, it can't be proven wrong it can't be proven right, but I presented proofs for my original assumption in the original post, if you follow my OP you can follow this rule since it can't be proven wrong

it can't be proven wrong it can't be proven right,
Can it be proven to be useful?

Relativistic kinetic energy:
K.E=mc²/√(1v²/c²)mc²
At the speed v=c for a mass, the expression √(1v²/c²) will equal zero:
K.E=mc²/0  mc² , K.E = 0mc²
K.E=mc²
But energyof mass m equals:
E=mc²
K.E=E
The kinetic energy of a mass m moving at the speed of light c equals the energy of the mass, and that what is the kinetic energy of a photon is, the kinetic energy of a photon is its contained energy, the kinetic energy of a mass m moving at v=c is its contained energy E , i.e the energy of its mass
Let me add another proof:
m=m0/(1 v²/c²)
if v=c then m=0, a photon moving at the speed of light is massles.

if I can't do the operation then...
Then that operation isn't defined.
Let me add another proof:
m=m0/(1 v²/c²)
if v=0 then m=0, a photon moving at the speed of light is massles.
No
If v=0 then v^2/c^2 = 0
m=m0/(1 0)
m=m0/(1)
m=m0
As expected, if you measure the mass of an object at rest (i.e. v=0) then the mass you measure is the rest mass.

No
If v=0 then v^2/c^2 = 0
m=m0/(1 0)
m=m0/(1)
m=m0
I did a writing mistake the post is about a photon v=c

Actually limits are not included in my assumption
Then you are ignoring useful information, and that will only end badly for you.
If you ignore the fact that sin(x)/x = 1, no matter how small you make x, and then just assume that sin(0)/0 = 4, you are making an unwarranted assumption.
Willful ignorance does not work in mathematics any more than it does in other fields.

Lack of mathematical operation is not dividing from the first place
And, indeed, you didn't divide in the first place.
when a moving object with constant speed hit another object it stops, that doesn't mean it didn't move at the beginning., in fact we could know the kinetic energy of the other object from the kinetic energy of the first one.
That has nothing to do with dividing by zero.
You can't reverse the operation , These expressions are a special case , the reverse operation doesn't work for them. This is a rule for my assumption, it can't be proven wrong it can't be proven right
It can be proven wrong. In math, it is possible to rearrange equations to produce equivalent expressions. The general form x/y = z can be rearranged by multiplying both sides by y. When you do this, you get y(x/y) = yz. In the part y(x/y), the two y's cancel to give you just an x. The result is x = yz. This proves that x/y = z is equivalent to x = yz. Now, replace x with 4 and y and z with 0. You get 4 = (0)(0). This is wrong, so the equivalent expression 4/0 = 0 is also wrong.
I presented proofs for my original assumption in the original post
Except that you presented no such proof. All you did was say that you didn't do division in the first place.

when you differentiate at f=0, you get sin(2f)/f=0/0=2cos(2f)/1=2/1 = 2
How it is possible that 0/0=2? The information I ignored is related to the graph but not the exact value of 0/0
I don't care if the graph was sin(f)/f and changed to 2cos(2f) at f=0 but he graph will give a value of 2 for 2co(2f)/1 but not for sin(f)/f, so the information you are talking about has nothing to do with exact value of 0/0. For 1/x at x=0 , I'm not talking about small numbers closer to 0 I'm talking about exactly x=0. The graph is not defined at x=0 I agree , but the graph has nothing to do with 1/0 it will give the value of numbers closer to zero but not exactly x=0, so the graph can't help determining 1/0

Lack of mathematical operation is not dividing from the first place
If 4/0 is not defined it means that 4/0 can't be done, and we result in nondivision
I add nondivision to mathematical operations.
we have 3 apples and we have 0 apples 3 is a number zero is also a number, zero is nonexistence of 3 apples
You can't say because no apples are there.0 is not a number
Nondivision occurs for undefined mathematical operations."undefined division"
Dividing by zero =Not dividing= nondivision
Nondivision is a separate operation , dividing by zero operation results in nondivision operation
something like multiplication results in addition, 4*2 results in 2+2+2+2, both are equal, both multiplication and addition gives the same result.
If not dividing = nondivision they will give the same result.
so dividing by zero results in nodivision by zero"similar to 4*2 results in 2+2+2+2" and nondivision by zero results in zero, so (any number /0)=0

If 4/0 is not defined it means that 4/0 can't be done
Exactly. If it can't be done, you can't get an answer.
I add nondivision to mathematical operations.
But no calculation is being performed whatsoever, so it can't be a mathematical operation.
If not dividing = nondivision they will give the same result.
Yes, and that result is that you don't get a result because you never did anything.

You can't reverse the operation , These expressions are a special case , the reverse operation doesn't work for them. This is a rule for my assumption, it can't be proven wrong it can't be proven right
It can be proven wrong. In math, it is possible to rearrange equations to produce equivalent expressions. The general form x/y = z can be rearranged by multiplying both sides by y. When you do this, you get y(x/y) = yz. In the part y(x/y), the two y's cancel to give you just an x. The result is x = yz. This proves that x/y = z is equivalent to x = yz. Now, replace x with 4 and y and z with 0. You get 4 = (0)(0). This is wrong, so the equivalent expression 4/0 = 0 is also wrong.
Your proof is simply wrong:
if yz is zero whether y is zero z is zero or both are zero:
you started with:
x/y=z
in case y=0 and z is nonzero:
4/0 is undefined
in case z=0 and y is nonzero
4/y cant equals 0
in case both are zero:
x/y=x/0 also undefined
You can't substitute x=4

If 4/0 is not defined it means that 4/0 can't be done
Exactly. If it can't be done, you can't get an answer.
wrong, this can't be done but gives an answer:
√ x when x is positive can't be done , but √ x could equal a number I in which √ x=I and I²=  x

4/0 is undefined
Which is what has been said all along...
this can't be done but gives an answer:
This is an oxymoron. You can't get an answer for a problem that can't be solved. If you got an answer, then it meant that the problem could indeed be solved.
This whole thing is nonsense anyway because you aren't ever actually dividing by zero. You said yourself that what you are doing is "nondivision" and that it is a new operation. By definition, 4/0 is division. Are you dividing or are you not dividing? If you are dividing, then this "nondivision" stuff is meaningless. If you aren't dividing, then you aren't addressing 4/0, which is specifically a division problem. So which is it?
(1) You are actually dividing, in which case, by definition, you aren't using nondivision, or
(2) You aren't dividing and therefore aren't solving 4/0.
Option 1 and option 2 are mutually exclusive. You can't have it both ways.
√ x when x is positive can't be done , but √ x could equal a number I in which √ x=I and I²=  x
Now you are contradicting yourself. If the square root of a negative number literally cannot be done, then x cannot equal any number. But i is a number, an imaginary number. So the operation can, in fact, be done. And you can rearrange the equations in an internally consistent way (i.e. you can "do it in reverse"). Your "nondivision" nonsense can't even do that, according to your own admission.

It is no use to "Redefining simple Arithmetic Expressions that contradict physics" if you have to use "arithmetic expressions" that contradict themselves.

(1) You are actually dividing, in which case, by definition, you aren't using nondivision, or
(2) You aren't dividing and therefore aren't solving 4/0.
1) If I'm dividing then I'm acting on something which has a result of not dividing or I'm doing nondivision then I will have a result.So I actually have a result of dividing which is nondivision then dividing by zero is defined according to my idea.
2) If I'm not dividing nothing should happen
I add another proof:
F=GMm/r² when r approaches 0 "at the centre for massive sphere" is said to be that F approaches infinity but according to my definition when r=0 F =0 which is a true fact and F approaching infinity is wrong , now I have five equations that satisfy my definition.

So I actually have a result of dividing which is nondivision then dividing by zero is defined according to my idea.
Why would you call something which is not dividing, "dividing"? Are you being deliberately confusing?
Also, what use is the outcome of this process?

So I actually have a result of dividing which is nondivision then dividing by zero is defined according to my idea.
Why would you call something which is not dividing, "dividing"? Are you being deliberately confusing?
OK, l have values for undefined operations in mathematics, whatever the explanation of these values is , and my values satisfy five equations in physics, those equations give wrong results according to limits, do you think this is coincidence ?

and my values satisfy five equations in physics,
What values and what equations?
Do you realise that physics was doing OK with the equations it already had?
What "problem" do you think you are solving?

1) If I'm dividing then I'm acting on something which has a result of not dividing
An oxymoron. If you are dividing the result cannot be "not dividing". That makes about as much since me saying If I made this post then I'm acting on something which has a result of not making this post. It's completely loopy.
2) If I'm not dividing nothing should happen
So are you dividing or are you not dividing? Don't say both because that's complete nonsense.
but according to my definition when r=0 F =0 which is a true fact
When was this "true fact" ever demonstrated?
and my values satisfy five equations in physics
But they don't. They give the wrong answer and in order to defend your wrong answers you have to claim that your process is not reversible. This, in turn, means that your mathematical process is internally inconsistent.

An oxymoron. If you are dividing the result cannot be "not dividing". That makes about as much since me saying If I made this post then I'm acting on something which has a result of not making this post. It's completely loopy.
Forget about the terminology nondivision, dividing that has no answer is different from not dividing
dividing with no answer is like this 4/0=undefined or no answer.
Not dividing is like this:
some operation but not division=something
4/0=no answer I say there is an answer which is zero.And I presented proof for that.
but according to my definition when r=0 F =0 which is a true fact
When was this "true fact" ever demonstrated?
Everywhere.
https://physics.stackexchange.com/questions/165526/ifthegravityatthecenteroftheearthiszerowhyareheavyelementslikei
It is simple, there are equal forces on the object from all directions,And still my values give the true answer and current values give the wrong answer
and my values satisfy five equations in physics
But they don't.
Why do you say for something obvious it is not true?
F=GMm/r^2 , at r=zero or at the centre of earth, according to current view , the limit of r as r approaches zero is infinity , however in the centre of earth F=0 and the truth is only obtained by my definition.
The relativistic mass of a photon is undefined , it equals 0/0: m=m0/(1 v²/c²), m0=0, v=c
But according to my definition 0/0=0, and that is exactly the definition of relativistic mass of a photon, it is always zero.
velocity at the origin=0/0 "x=0 , t=0 "=undefined but it actually has a value which is 0/0=0, and that is exactly the velocity of an object at stationary.
According to what Newton suggested if we assume infinity is a number then at infinity , force F=0, according to my definition, I exactly did what Newton suggested , F=GMm/r^2=GMm/∞=0
By the same logic of Newton , if we assume v=c , the kinetic energy of mass equals its contained energy, just like a photon's moving at the speed of light has kinetic energy equals its contained energy.
Relativistic kinetic energy:
K.E=mc²/√(1v²/c²)mc²
At the speed v=c for a mass, the expression √(1v²/c²) will equal zero:
K.E=mc²/0  mc² , K.E = 0mc²
K.E=mc²
But energy of mass m equals:
E=mc²
K.E=E
So why do you say for something obvious not true ? if you do that then the discussion is meaningless, and it is just time wasting.

And still my values give the true answer and current values give the wrong answer
Prove it.
Show that current physics gives the wrong answer.
Why do you say for something obvious it is not true?
F=GMm/r^2 , at r=zero or at the centre of earth, according to current view , the limit of r as r approaches zero is infinity , however in the centre of earth F=0 and the truth is only obtained by my definition.
You have made an interesting error there.
You have ignored very nearly the whole of the Earth.
(If I am near the middle of the Earth, there is a lot of it above me pulling me upwards, you have not accounted for that).
And, because you have not done the physics properly, you have not got the right answer.
Well, I have good news.
You do not need to invent new mathematics (which is very difficult).
You just have to learn to do physics which is much easier.
Incidentally, the inverse square law only works for "large" distances. Sufficiently large that teh object can be considered as a point.
Misusing formulae will cause problems.
Again, the answer is not "make up new maths", the answer is "learn when to use the formulae".

Incidentally, the inverse square law only works for "large" distances. Sufficiently large that teh object can be considered as a point.
You missed something which is r is measured from centre to centre, in this case an object center will be exactly at earth's centre, which the same as saying r=0

gravityatthecenteroftheearthiszero...
It is simple, there are equal forces on the object from all directions,And still my values give the true answer and current values give the wrong answer
The current answer is given by Newton's Shell Theorem (https://en.wikipedia.org/wiki/Shell_theorem).
 This says that the gravitational attraction of a spherical object is proportional to the mass between your centerofmass, and the other object's centerofmass.
 As you approach zero distance from the center of the Earth, the mass closer to the center decreases as r^{3}
 So Newton's attraction becomes:
F=kGr^{3}m/r^{2}= kGrm
where...
 r is your distance from the center of the Earth
 k is a constant that accounts for the density of matter at the center of the Earth (probably mostly nickel/iron)
 m is your test mass
 G is the universal gravitational constant
So, according to Newton's physics, when r=0, F=0.
 Without anyone going there, we believe this is the correct answer
Now you have to explain how your method manages to compress your mass into a sphere of radius 0 (ie a black hole).
The relativistic mass of a photon is undefined , it equals 0/0: m=m0/(1 v²/c²), m0=0, v=c
But according to my definition 0/0=0, and that is exactly the definition of relativistic mass of a photon, it is always zero.
Perhaps you are thinking of E=mc^{2} ?
 A mass of zero would imply that the energy of a photon is zero  which we know to be false.
This familiar quotation of "E=mc^{2}" is only half of Einstein's equation:
E^{2}=(pc)^{2} + (m_{0}c^{2})^{2}
 where p is the momentum of the particle
How do you account for this if the photon's mass is zero?
See: https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation
√ x when x is positive can't be done
First year university maths covers "imaginary" numbers.
 High school advanced maths covers "imaginary" numbers (at least, back when I went to high school).
 You really need to get some basic grounding on finite numbers before you start tackling (potentially) infinite numbers.
See: https://en.wikipedia.org/wiki/Imaginary_number
my values satisfy five equations in physics
You really need to tell us what these 5 mysterious and significant equations are!
For all we know they could be:
1) 1=1
2) 2=2
3) 3=3
4) 4=4
5) 5=5

The relativistic mass of photon is undefined , it equals 0/0: m=m0/(1 v²/c²), m0=0, v=c
But according to my definition 0/0=0, and that is exactly the definition of relativistic mass of a photon, it is always zero.
Perhaps you are thinking of E=mc^{2} ?
No, this the relativistic mass of a photon, the photon has rest mass m0=0 , its mass doesn't increase " relativistic mass" and I proved that, however current physics says its relativistic mass is undefined"0/0"
I wrote about kinetic energy of an object in another paragraph.
my values satisfy five equations in physics
You really need to tell us what these 5 mysterious and significant equations are!
For all we know they could be:
1) 1=1
2) 2=2
3) 3=3
4) 4=4
5) 5=5
It appears that if we have 3 are apples , then the other three apples will equal 9 apples , 3 apples for each of them, and this is wrong.
Let's say we have three apples, and another three apples, these three apples will equal those three apples"3=3"
Let's say we have collection of the first three apples and collection of the other three apples, C1 consists of the first 3 apples and C2 consists of the other 3 apples.C1=C2 or 3=3
If C1=C2 then we will have C3=9, but also C2=C1, so C4=9, C4=9=C3
But C1=C2 and C2=C1 in this case we can consider this process as infinite process because the two "C1 and C2" will always equal.
C1=3, C2=3
C1=C2=9
C1=C2=9*3
C1=C2=27*3
The rule is:
C1=C2=3^n , n=1,2,3,4.......
We can start with C1=C2 and end with C2=C1 and this exactly the definition of C1=C2.
As we will have C3=C4 this will be defined by the next C5=C6, and C5=C6 will be defined by C7=C8,and so on, and because the process is infinte C1=C2 is always defined.

Incidentally, the inverse square law only works for "large" distances. Sufficiently large that teh object can be considered as a point.
You missed something which is r is measured from centre to centre, in this case an object center will be exactly at earth's centre, which the same as saying r=0
It's not at all clear that I missed anything.
However, you seem not to understand that zero is not large.
, its mass doesn't increase " relativistic mass" and I proved that,
The fact that they have to bolt the magnets down to big blocks of concrete at places like CERN shows that the relativistic mass increase is real.
If your hypothesis doesn't give the same outcomes as reality then it is not because reality has made a mistake.

It appears that if we have 3 are apples , then the other three apples will equal 9 apples , 3 apples for each of them, and this is wrong.
You seem to have failed to understand as usual.
For all we know they could be:
(Equation number 1): 1=1
(Equation number 2): 2=2
(Equation number 3): 3=3
(Equation number 4): 4=4
(Equation number 5): 5=5
The point is that you talk about "Five equations", but you do not say what the equations are.

Forget about the terminology nondivision, dividing that has no answer is different from not dividing
If they are different then you finally agree that you are no longer dividing and therefore are no longer trying to solve 4/0.
dividing with no answer is like this 4/0=undefined or no answer.
So on one hand you agree that division by zero is undefined, yet at the same time you claim that it is zero. How can zero be undefined?
Not dividing is like this:
some operation but not division=something
Stop pretending that you are solving a division equation by literally not solving a division equation.
4/0=no answer I say there is an answer which is zero.
This question has no answer, but oh, it also does have an answer! How do you no see the blaring contradiction in that logic?
And I presented proof for that.
Except for, you know, the part where you never did.
Everywhere.
https://physics.stackexchange.com/questions/165526/ifthegravityatthecenteroftheearthiszerowhyareheavyelementslikei
It is simple, there are equal forces on the object from all directions,And still my values give the true answer and current values give the wrong answer
Congratulations on completely misusing the equation. It only works if you are above the surface of the Earth because it models gravity as a point source.
F=GMm/r^2 , at r=zero or at the centre of earth, according to current view , the limit of r as r approaches zero is infinity , however in the centre of earth F=0 and the truth is only obtained by my definition.
Gravity does not tend towards infinity as you go deeper into the Earth. It gets weaker. The Earth is not a point source for gravity. For a black hole, on the other hand, the equation would be okay.
relativistic mass of a photon is undefined , it equals 0/0: m=m0/(1 v²/c²), m0=0, v=c
But according to my definition 0/0=0, and that is exactly the definition of relativistic mass of a photon, it is always zero.
Yet another error. The relativistic mass of a photon is equal to its energy content, which is always a finite number.
velocity at the origin=0/0 "x=0 , t=0 "=undefined but it actually has a value which is 0/0=0, and that is exactly the velocity of an object at stationary.
Photons are never stationary.
According to what Newton suggested if we assume infinity is a number then at infinity , force F=0, according to my definition, I exactly did what Newton suggested , F=GMm/r^2=GMm/∞=0
That isn't division by zero.
the same logic of Newton , if we assume v=c , the kinetic energy of mass equals its contained energy, just like a photon's moving at the speed of light has kinetic energy equals its contained energy.
Relativistic kinetic energy:
K.E=mc²/√(1v²/c²)mc²
At the speed v=c for a mass, the expression √(1v²/c²) will equal zero:
K.E=mc²/0  mc² , K.E = 0mc²
K.E=mc²
But energy of mass m equals:
E=mc²
K.E=E
The kinetic energy of a mass traveling at the speed of light isn't zero, so that actually proves that your "dividing by zero equals zero" crap is wrong.
So why do you say for something obvious not true ? if you do that then the discussion is meaningless, and it is just time wasting.
This whole thread is time wasting,

You really need to tell us what these 5 mysterious and significant equations are!
1) 1=1
2) 2=2
3) 3=3
4) 4=4
5) 5=5
These are my 5 equations , I already mentioned them:
F=GMm/r^2 , at r=zero or at the centre of earth, according to current view , the limit of r as r approaches zero is infinity , however in the centre of earth F=0 and the truth is only obtained by my definition.
The relativistic mass of a photon is undefined , it equals 0/0: m=m0/(1 v²/c²), m0=0, v=c
But according to my definition 0/0=0, and that is exactly the definition of relativistic mass of a photon, it is always zero.
velocity at the origin=0/0 "x=0 , t=0 "=undefined but it actually has a value which is 0/0=0, and that is exactly the velocity of an object at stationary.
According to what Newton suggested if we assume infinity is a number then at infinity , force F=0, according to my definition, I exactly did what Newton suggested , F=GMm/r^2=GMm/∞=0
By the same logic of Newton , if we assume v=c , the kinetic energy of mass equals its contained energy, just like a photon's moving at the speed of light has kinetic energy equals its contained energy.
Relativistic kinetic energy:
K.E=mc²/√(1v²/c²)mc²
At the speed v=c for a mass, the expression √(1v²/c²) will equal zero:
K.E=mc²/0  mc² , K.E = 0mc²
K.E=mc²
But energy of mass m equals:
E=mc²
K.E=E

The relativistic mass of a photon is undefined , it equals 0/0: m=m0/(1 v²/c²), m0=0, v=c
That makes perfect sense.
The relativistic mass of a photon depends on the energy.
So it does not have a defined value in terms of v and c because they do not define the energy of a photon.
A gamma ray photon has more energy, and thus more relativistic mass than a visible photon.
If you had some formula that tries to calculate the mass from the velocity, you will get the wrong answer because it's impossible to calculate it without knowing the energy.
So "undefined" is the right answer.
If you don't get "undefined" as the answer to "what is the mass of a photon", then you are doing it wrong.
Do you understand that?

F=GMm/r^2 , at r=zero or at the centre of earth, according to current view , the limit of r as r approaches zero is infinity , however in the centre of earth F=0 and the truth is only obtained by my definition.
We already explained why that's wrong (you are not using the right formula).

velocity at the origin=0/0 "x=0 , t=0 "=undefined but it actually has a value which is 0/0=0, and that is exactly the velocity of an object at stationary.
Graphs are not reality.

I exactly did what Newton suggested
No he didn't

By the same logic of Newton , if we assume v=c , the kinetic energy of mass equals its contained energy, just like a photon's moving at the speed of light has kinetic energy equals its contained energy.
Relativistic kinetic energy:
K.E=mc²/√(1v²/c²)mc²
At the speed v=c for a mass, the expression √(1v²/c²) will equal zero:
K.E=mc²/0  mc² , K.E = 0mc²
K.E=mc²
But energy of mass m equals:
E=mc²
K.E=E
At the very best, you are wrong because KE 1/2 mv^2

Okay, if I want to solve 5 x 7, I can stop doing multiplication and use nonmultiplication to get an answer (0) and then claim that is the answer to the original multiplication equation (5 x 7 = 0). That's no different than what you are doing. It's crazy. It's stupid. It's wrong.

It's crazy. It's stupid. It's wrong.
It is wrong but It is not stupid, discovering a new theory is hard , my threads and posts are deep , I bet if you try so you will post garbage, you are not even able to do it, you do not and never have new ideas in your mind,.

discovering a new theory is hard
And you have not done it. nor have you even got close.
All you have done is made up a nonsensical way of doing something which does not need doing.

It is wrong
Woah, woah, wait a minute. Now you are saying that your idea is wrong?
but It is not stupid
It's stupid because you consistently refuse to accept that you are wrong.
discovering a new theory is hard
I never said otherwise.
my threads and posts are deep
You mean like the one where you claimed that gold atoms are made of actual gold?
I bet if you try so you will post garbage
The difference between you and me is that I am willing to relent on a new idea when others demonstrate that I am wrong. So if I posted garbage and others explained why it was garbage, I would be able to learn from my mistakes.
you are not even able to do it,
I've made threads before about new ideas of mine, so this is a lie.
you do not and never have new ideas in your mind,.
What do you think you accomplish by lying? Almost everyone has new ideas about one thing or another, even if it isn't necessarily sciencerelated.

It is wrong but It is not stupid,
Let's be clear about this.
Claiming that not doing division is the same as doing division is actually objectively stupid.

The relativistic mass of a photon is undefined , it equals 0/0: m=m0/(1 v²/c²), m0=0, v=c
That makes perfect sense.
The relativistic mass of a photon depends on the energy.
So it does not have a defined value in terms of v and c because they do not define the energy of a photon.
You are wrong ., the relativistic mass doesn't depend on its energy. What equation says that? photon energy is hf, h=planks constant and f= photon frequency. And relativistic mass is always 0 even if light changed its frequency and its energy because photon is massless and its rest mass is always zero.See the equation in this post

At the very best, you are wrong because KE 1/2 mv^2
You are wrong , this is Newton's old equation replaced by Einstein equation.
S0 K.E=  E
where E=mc^2
kinetic energy for an object equals its contained energy supposing v=c

Okay, if I want to solve 5 x 7, I can stop doing multiplication and use nonmultiplication to get an answer (0) and then claim that is the answer to the original multiplication equation (5 x 7 = 0). That's no different than what you are doing. It's crazy. It's stupid. It's wrong.
Stop using multiplication result in several options" that result in 0" , you can't use what you want, however if you stopped using multiplication it just means you are not using it , it doesn't imply you are using something else.
You have to improve yourself in simple logic.

Claiming that not doing division is the same as doing division is actually objectively stupid.
OK forget about nondivision, we are dividing:
4 apples are to be divided between 0 people, the result is zero apples will be given to zero people , since zero is a number.
The expression 0  3 =  3has no meaning in words but it has meaning in mathematics. In fact it appears to be undefined.
What does it mean when we have empty bucket and we want to take three apples out of this bucket ? but because 3 is a number it has meaning

You are wrong ., the relativistic mass doesn't depend on its energy.
The mass of literally everything depends upon its energy content.
What equation says that?
E=mc^{2}.
As far as I can tell, light has a measurable mass regardless of what technique you use to measure it:
(1) Light is pulled towards a gravitational field (detectable as gravitational lensing).
(2) Let's say you have a hollow sphere with a perfectlymirrored internal surface that has light bouncing around inside of it. If this sphere is put into a gravitational field (such as on the surface of the Earth), then those photons that are moving towards the bottom of the sphere will be blueshifted while those moving towards the top will be redshifted (gravitational redshifting). Since those photons hitting the bottom have more energy than those hitting the top, this will result in a net force (radiation pressure) pushing the sphere towards the Earth. The net result is that this sphere is heavier when there is light inside of it than when it is empty.
(3) Conservation of energy requires light to have a mass. We now consider a similar shell to the one described above, except that it contains the entire Earth. The entire Earth's mass is now converted into light energy as per E=mc^{2}. If light had no mass and did not generate its own gravitational field, then objects resting on top of the shell now become weightless and can be moved up away from the sphere's center with no energy input. If you move a weight that was initially resting on top of the shell up to a height of 1 kilometer then convert the light energy back into conventional matter, gravity comes back and the weight will fall, generating brand new kinetic energy out of nowhere. In order to prevent this violation of energy conservation from occurring, it must have been the case that the gravity never actually disappeared when the mass of the Earth was converted into light. So moving the weight away from the lightfilled shell must have actually required energy and the Earth still has a measurable mass even in the form of light.
you can't use what you want
Why not? That's what you're doing.
however if you stopped using multiplication it just means you are not using it , it doesn't imply you are using something else.
Just like not using division doesn't imply you are using something else.
4 apples are to be divided between 0 people, the result is zero apples will be given to zero people , since zero is a number.
Actually, you would be giving four apples to zero people.
In fact it appears to be undefined.
That's not what "undefined" means in mathematics.
What does it mean when we have empty bucket and we want to take three apples out of this bucket ?
Funny, but if you follow your own logic, this should mean that 0  3 = 0, since you end up with zero apples if you literally try to take three apples out of an empty bucket.

4 apples are to be divided between 0 people, the result is zero apples will be given to zero people , since zero is a number.
Actually, you would be giving four apples to zero people.
zero in reality and the natural world means the absence of things , zero people , zero buses, etc, if we have 4 people who left a room , then in this room we have zero people, if we have zero people in the room then we won't give them anything because they are absent , even if you have the the 4 apples
So is it the four apples you want to give, or 0 people will be given? they are the same , you have 2 people you want to give them 4 apples"4/2" or you have 4 apples you want to give to two people.
if you follow your own logic, this should mean that 0  3 = 0, since you end up with zero apples if you literally try to take three apples out of an empty bucket.
You can't use my own logic to prove me wrong, my logic is my logic I'm the only one who is responsible of its results
You have to learn basic logic.

zero in reality and the natural world means the absence of things , zero people , zero buses, etc, if we have 4 people who left a room , then in this room we have zero people, if we have zero people in the room then we won't give them anything because they are absent , even if you have the the 4 apples
So is it the four apples you want to give, or 0 people will be given? they are the same , you have 2 people you want to give them 4 apples"4/2" or you have 4 apples you want to give to two people, if you think this is not the same then you have problem in your language .
I never said that it wasn't the same. It gives the wrong answer because you're not actually doing division.
You can't use my own logic to prove me wrong
Of course I can. If I can demonstrate your logic to be internally inconsistent, then it proves itself wrong. Since you claim that the inability to perform an operation in real life (such as dividing four apples among zero people) automatically results in an answer of zero, then it follows that taking four apples out of an empty bucket also results in an answer of zero because no apples were actually taken out of the bucket. So not only does your logic dictate that 4/0 = 0, but also that 0  4 = 0.
my logic is my logic I'm the only one who is responsible of its results
Okay, so if I invent my own logic, I can make anything true that I want to be true and no one can say otherwise because it's "my" logic. Is that right?
You have to learn basic logic.
Basic logic dictates that you can't make up your own logic, as you claim to have done. Either something is logical or it isn't. If everyone's own personal systems of logic were equally valid, the world would be rife with logical contradictions.

There are not explanation , I knew that x/0=0/0=x/∞=0 and they have nothing to do with division , multiplication.
In words they have no meaning and undefined , but I gave them specific values, that satisfy physics equations

and they have nothing to do with division
If they have nothing to do with division then they can't be the answer to a division problem.
but I gave them specific values, that satisfy physics equations
But they don't. The gravitational attraction between two point sources with a separation of zero is infinite, not zero.

If they have nothing to do with division then they can't be the answer to a division problem.
They are undefined so they are already not solving division problem

The expression 0  3 =  3has no meaning in words
It was freezing (i.e. 0 degrees C) and the temperature dropped by three degrees.
It is now minus three degrees.
Since I have now explained to you that the expression does have a meaning in words, do you accept that you were wrong to base any further ideas on your inaccurate view?
In particular
In fact it appears to be undefined.
In fact it only appears that way to you.
And that is because you don't understand what you are talking about.

If they have nothing to do with division then they can't be the answer to a division problem.
They are undefined so they are already not solving division problem
When, in the real world, is this a problem?
When do you need to divide by zero?

They are undefined so they are already not solving division problem
Exactly, because the equation cannot be solved.

The expression 0  3 =  3has no meaning in words
It was freezing (i.e. 0 degrees C) and the temperature dropped by three degrees.
It is now minus three degrees.
This example says simply that 03=3, you do not even understand how to define expressions in words, 0 is 0 , dropping is  , and 3 is 3 , so it is just 0  3.
Dropping three units is just like going up 3 three units, we can't say that the direction gives a minus, how do you say that decrement is 0  3 and increament is 0+3 ? how do you say that decrement gives negative sign? we can just say it increased 3 units and it decreased 3 units, we have value of increment " 3 " and we have value of decrement " 3 ", whether it decreased or increased the value is just 3 so you were not able to define it in words and 03=0+3 which is wrong
In fact 03 appears to be undefined.
In fact it only appears that way to you.
How this appears defined to you?
we have a bucket with zero apples inside it, we want to take 3 apples . what is the number of apples we will take out? this is obviously undefined, only if you do not accept obvious things in order to prove you are right , in fact this says that we cant do subtracting , but it has a number, just like my example of 4/0 which appears to be undefined but it has a number

You are wrong ., the relativistic mass doesn't depend on its energy What equation says that?
E=mc^{2}.
This is the contained energy of a mass , photon energy equation is hf, h=plank constant , f=photon frequency
E=mc^2 is not the true equation for photons, photon is massles if m=0 E will equal zero, , but photon has energy

E=mc^2 is not the true equation for photons
Mass and energy are equivalent, so it has to work for photons just as much as any other particle.
photon is massles if m=0 E will equal zero, , but photon has energy
You have it backwards. The fact that a photon has an energy means that it also has a mass. I already posted three different reasons why light must have a mass in post #62.

E=mc^2 is not the true equation for photons
Mass and energy are equivalent, so it has to work for photons just as much as any other particle.
kinetic energy=mc²/1v²/c²  mc²
E=mc² your equation for photon's contained energy"
Kinetic energy of a photon equals its contained energy:
If we use what you are talking about :
K.E=E=mc²
mc²=mc²/1v²/c²  mc²
While v approaches c the left hand side of the equation will approach infinity:
So it will be :
mc²=∞mc² , what you are talking about is wrong,
the alternative equation is E=hf I mentioned in my previous post.

we have a bucket with zero apples inside it, we want to take 3 apples . what is the number of apples we will take out? this is obviously undefined,
We have a bank account with zero dollars inside it, we want to take 3 dollars.
 what is the number of dollars we will take out? this is obviously $3.
 How much is left in the bank account: You owe the bank $3.
 If you now deposit $6 in the bank account, you don't have $6, you only have $3 in the bank account. That's because it previously had a $3 debit.
Bored Chemist gave another valid example, using degrees Celsius.
 However, the answer would have been different if the example used the Kelvin temperature scale.
That's because mathematics can represent many different types of systems
 And you have to do a sanity check to decide whether the mathematical solution makes sense in your realworld environment
 You can't have negative 3 apples in a bucket  but you could be short 3 pallets of apples from a shipment
 You can't have negative 3 degrees Kelvin*, but you can have negative 3 degrees Celsius (or Fahrenheit)
Take another example: 10+4=14 in the natural numbers
 But 10+4=2 on a clock face
 And 10+4=3 in base 11
 These are all valid answers, in different realworld scenarios
So you have to define the system you are talking about before you define the mathematics of that system.
 Or in the case of some nuclear physics problems, they observe the behavior of the subatomic particles, then deduce what mathematical system applies to them (usually some form of symmetry group)
*Negative temperatures in the Kelvin scale can be defined for some exotic systems (not ordinary refrigerators)
https://en.wikipedia.org/wiki/Negative_temperature

E=mc² your equation for photon's contained energy"
It's not "my" equation. It's relativity's equation.
Kinetic energy of a photon equals its contained energy:
If we use what you are talking about :
K.E=E=mc²
mc²=mc²/1v²/c²  mc²
While v approaches c the left hand side of the equation will approach infinity:
So it will be :
mc²=∞mc² , what you are talking about is wrong,
That equation obviously only works for particles that get their kinetic energy from their velocity (photons don't).
the alternative equation is E=hf I mentioned in my previous post.
E =hf and E=mc^{2} are not mutually exclusive. They tell you different things. The first tells you the energy associated with a particular frequency while the other tells you the energy associated with a particular mass. They are both correct.
You seem to keep forgetting that the first law of thermodynamics could be violated if photons didn't have a mass (as I pointed out in post #62). Here is yet another example:
Let's say I have a container that weighs 1 kilogram. Inside this are two magnetized compartments. One compartment contains 0.5 kilograms of electrons and the other 0.5 kilograms of positrons. I put this container on a scale and it weights 2 kilograms total. Now, I open a gate between the compartments that allows the electrons and positrons to reach each other. They annihilate, transforming into photons. Has the mass of the electrons and positrons disappeared when they get turned into photons? If so, then the container should now weigh 1 kilogram: equal to its empty weight.
However, this would allow for the creation of a perpetual motion machine of the first kind. We could raise this container of photons up to some arbitrary height while its contents are in the form of light, have those photons turned back into positrons and electrons, and get the extra weight back. Connect this to an appropriate contraption, and you could get twice as much energy out when the container falls than it requires to raise the container back up again. This is a nono. So we must conclude that the weight of the container did not change when the electronsand positrons annihilated.

How this appears defined to you?
we have a bucket with zero apples inside it, we want to take 3 apples . what is the number of apples we will take out? this is obviously undefined, only if you do not accept obvious things in order to prove you are right , in fact this says that we cant do subtracting , but it has a number, just like my example of 4/0 which appears to be undefined but it has a number
OK, so the problem here is that you don't understand the difference between a temperature which can be negative, and a number of apples which can't (usually).
You should probably work on your understanding of that.
However the simple version is this; different things can be different.

I ran across a Numberphile Youtube video that explains "the problem with zero" in a very accessible way.
There are 3 examples, and I recommend that you watch all three: 1/x (the topic of this thread), 0^{0} and y/x when x=0 (also the topic of this thread): 12 minutes total.
PS: I recently read Matt Parker's book "Humble Pi", which is about mathematical failures in the real world  very instructive if you want some anecdotes for your maths, science or engineering class! Recommended...