Is it permissible to use infinity, which is not defined in physics, to assume the impossibility of traveling at the speed of light?

[Genady]

6 minutes ago, Z.10.46 said:

I haven't done it. The purpose of this is to let the divergence appear in order to apply regularization using the zeta function.

We are talking mathematics here.

If you set c=v on line 5, you divide by 0 on line 6, and everything after that is meaningless.

If you don't set c=v on line 5, then there is no c+c on line 6, and everything after that is wrong.

You can decide, if you have it done or you haven't. The derivation is wrong anyway.

[studiot]

32 minutes ago, Z.10.46 said:

no it's correct you can show it to any mathematician, I'm not going to give a theory either where there is a calculation error and without verification by mathematicians

You have shown it to this Mathematician.

And I thought there were too many misunderstndings in it to be worth detailed consideration.

So +1 to Genady for the patience to do this.

+1 also to PhiforAll who got in before me when I was about to complain.

 

A word of advice.

You can expect ridicule as answers when you make such wildly and obviously incorrect statements.

 

Such as claiming the non existence of zero.

 

Consider this sequence of integers

 

...-4,  -3,  -2,  -1,  1,  2, 3,  4...

...Even, odd, even, odd, odd, even, odd, even...

Do you notice anything missing ?

It implies that there are more odd numbers than even numbers.

 

By the way do you understand the ... symbol?

It is called an ellipsis.

 

You have also been arguing wrongly about the meaning of the equals sign.

It stands for two different properties  -  Identity, and simple equality.

Sometimes the identity symbol with three bars not two is used instead for this.

 

 

[Z.10.46]

20 minutes ago, studiot said:

You have shown it to this Mathematician.

And I thought there were too many misunderstndings in it to be worth detailed consideration.

So +1 to Genady for the patience to do this.

+1 also to PhiforAll who got in before me when I was about to complain.

 

A word of advice.

You can expect ridicule as answers when you make such wildly and obviously incorrect statements.

 

Such as claiming the non existence of zero.

 

Consider this sequence of integers

 

...-4,  -3,  -2,  -1,  1,  2, 3,  4...

...Even, odd, even, odd, odd, even, odd, even...

Do you notice anything missing ?

It implies that there are more odd numbers than even numbers.

 

By the way do you understand the ... symbol?

It is called an ellipsis.

 

You have also been arguing wrongly about the meaning of the equals sign.

It stands for two different properties  -  Identity, and simple equality.

Sometimes the identity symbol with three bars not two is used instead for this.

 

 

It seems like you are searching for errors where there is no ,mathematical demonstration that is 100% correct.

In this discussion, those who understand mathematics and physics very well have highlighted the only flaw in this proof, which is to assume x+2=1/c-v without considering the unit of c.

However, the counterproof was very strong because the relationship remains valid regardless of the unit of c, just like the equation c/c=1, which is always true regardless of the unit of c.

The discussion about the non-existence of 0 in physics is just off-topic. Here, you are talking about mathematical zero, while I am talking about the non-existence of a physical zero.

Edited July 24, 2023 by Z.10.46

[MJ kihara]

Clarification pliz line 3 to line 4 where is denominator c^2 going?

[Z.10.46]

6 minutes ago, MJ kihara said:

Clarification pliz line 3 to line 4 where is denominator c^2 going?

1. M = m0 / √(1 - v^2/c^2) =

2. = m0 / √(c^2 - v^2)/c^2 =

3. = m0 / √(c + v)√(c - v)/c^2 =

Here 1/√1/c^2=c 

4. = m0c / √(c + v)√(c - v) =

here c+v=c+c To isolate the divergence (c - v).

6. = m0c / √(c + c)√(c - v) =

7. = m0c / √2c√(c - v) =

8. = m0√c / √2√(c - v) =

9. = m0√c / √2√n

Edited July 24, 2023 by Z.10.46

[Genady]

1 hour ago, Z.10.46 said:

1. M = m0 / √(1 - v^2/c^2) =

2. = m0 / √(c^2 - v^2)/c^2 =

3. = m0 / √(c + v)√(c - v)/c^2 =

Here 1/√1/c^2=c 

4. = m0c / √(c + v)√(c - v) =

here c+v=c+c To isolate the divergence (c - v).

6. = m0c / √(c + c)√(c - v) =

7. = m0c / √2c√(c - v) =

8. = m0√c / √2√(c - v) =

9. = m0√c / √2√n

Another question for you. In the definition,

,

the sqrt(n) is in the numerator. But in the "derivation" above, on the step 9, it is in the denominator. How come?

 

[Bufofrog]

1 hour ago, Z.10.46 said:

which is always true regardless of the unit of c.

I think there is a language issue.  Please tell me what the 'unit' if c is.

[joigus]

2 hours ago, Z.10.46 said:

It seems like you are searching for errors where there is no ,mathematical demonstration that is 100% correct.

In this discussion, those who understand mathematics and physics very well have highlighted the only flaw in this proof, which is to assume x+2=1/c-v without considering the unit of c.

However, the counterproof was very strong because the relationship remains valid regardless of the unit of c, just like the equation c/c=1, which is always true regardless of the unit of c.

The discussion about the non-existence of 0 in physics is just off-topic. Here, you are talking about mathematical zero, while I am talking about the non-existence of a physical zero.

My Dunning-Kruger effect sensor went off the scale here.  

Your handling of units and zero/infinity is appaling.

[swansont]

37 minutes ago, joigus said:

My Dunning-Kruger effect sensor went off the scale here.  

Your handling of units and zero/infinity is appaling.

Which isn’t a language issue; the math is just flat-out wrong.

[Z.10.46]

54 minutes ago, joigus said:

My Dunning-Kruger effect sensor went off the scale here.  

Your handling of units and zero/infinity is appaling.

Yes, but it's a very solid mathematical proof to justify that x+2 has no unit because the equation x+2=1/c-v is true for any unit of measurement chosen for c, due to the property that infinity=a*infinity, similar to the '1' in the equation 1=c/c which has no unit of measurement as this equation is true for any unit of measurement chosen for c.😁

16 minutes ago, swansont said:

Which isn’t a language issue; the math is just flat-out wrong.

Could you point out a possible error in this proof?

Edited July 24, 2023 by Z.10.46

[swansont]

18 minutes ago, Z.10.46 said:

Could you point out a possible error in this proof?

We’ve been discussing it for some time now. You refuse to acknowledge it, but you can’t arbitrarily add or subtract numbers that do not have the same units. “2” is unitless

Quote

Yes, but it's a very solid mathematical proof to justify that x+2 has no unit because the equation x+2=1/c-v is true for any unit of measurement chosen for c, due to the property that infinity=a*infinity, similar to the '1' in the equation 1=c/c which has no unit of measurement as this equation is true for any unit of measurement chosen for c

1/c-v is not infinite; you can graph it and see that it tends to infinity as v approaches c, but you wrote is as a function. x+2 is not infinite either. You can’t arbitrarily set them equal to each other, and can’t use them as equations where x and v are variables, because you aren’t treating them as variables.

It also looks like you are saying that two functions that tend to infinity in some limit are equal, which is just wrong.

[Z.10.46]

29 minutes ago, swansont said:

1/c-v is not infinite; you can graph it and see that it tends to infinity as v approaches c, but you wrote is as a function. x+2 is not infinite either. You can’t arbitrarily set them equal to each other, and can’t use them as equations where x and v are variables, because you aren’t treating them as variables.

It also looks like you are saying that two functions that tend to infinity in some limit are equal, which is just wrong.

Here, I am not just saying that v tends to c, I am saying that v=c, so 1/c-v egal infinity not just tends to infinity , if I set x+2=1/c-v, it's like having the equality infinity+2=infinity

I agree with you that it is impossible to do this for a finite quantity but 1/c-v is not finite at v=c and in this case setting x+2=1/c-v is valid without a fixed unit as I demonstrated.

Edited July 24, 2023 by Z.10.46

[joigus]

1 hour ago, swansont said:

Which isn’t a language issue; the math is just flat-out wrong.

Agreed. Not a language issue to me. Just plain wrong.

Dear @Z.10.46, there is a game in town that's very similar to what you're trying to play here. It's called asymptotics.

Quote

In mathematical analysis, asymptotic analysis, also known as asymptotics, is a method of describing limiting behavior.

From: https://en.wikipedia.org/wiki/Asymptotic_analysis

In asymptotics we have to be very careful though. We never use the = sign. But we use an equivalence relation \( \approx \). If, eg, \( x \) is "very small", you're allowed to do things like \( \left( 1+x \right)^{2} \approx 1+2x \), but never things like \( x \approx 0 \), which lead to sorry mistakes. There is no such thing as "infinity" in asymptotics; only very big quantities, which must be either possitive or negative. Otherwise --like in your quite absurd, dimensionally inconsistent expression--, the "quantity" \( x \) defined by \( x + 2 = 1/(c-v) \) with c=v is:

1) +infinity? That is, the right limit, or,

2) -infinity? That is, the left limit.

Of course infinity +2 = infinity, but also infinity + 3 = infinity, infinity +17=infinity-1/pi, etc.

And -infinity+2 = -infinity, etc. "Infinity" is not a number. It breaks all the algebraic rules. Example infinity + x = infinity + y does not imply x=y.

I like to think of infinity more like a topological entity (the boundary of the real numbers). I think that's the way in which modern mathematics we tend to look upon it. But that's another story.

And Swansont's objection about units still stands.

[swansont]

1 hour ago, Z.10.46 said:

Here, I am not just saying that v tends to c, I am saying that v=c, so 1/c-v egal infinity not just tends to infinity , if I set x+2=1/c-v, it's like having the equality infinity+2=infinity

I agree with you that it is impossible to do this for a finite quantity but 1/c-v is not finite at v=c and in this case setting x+2=1/c-v is valid without a fixed unit as I demonstrated.

If v=c you do not have a valid equation

1/c-v is a function, valid for v≠c. If you choose v=c (or any fixed value) you do not have a function.

You can do one or the other. Not both.

Functions are not equal simply because they both diverge to the same limit.

x and x+2 both go to infinity, but x ≠ x+2

[Z.10.46]

1 hour ago, joigus said:

Agreed. Not a language issue to me. Just plain wrong.

Dear @Z.10.46, there is a game in town that's very similar to what you're trying to play here. It's called asymptotics.

From: https://en.wikipedia.org/wiki/Asymptotic_analysis

In asymptotics we have to be very careful though. We never use the = sign. But we use an equivalence relation ≈ . If, eg, x is "very small", you're allowed to do things like (1+x)2≈1+2x , but never things like x≈0 , which lead to sorry mistakes. There is no such thing as "infinity" in asymptotics; only very big quantities, which must be either possitive or negative. Otherwise --like in your quite absurd, dimensionally inconsistent expression--, the "quantity" x defined by x+2=1/(c−v) with c=v is:

1) +infinity? That is, the right limit, or,

2) -infinity? That is, the left limit.

Of course infinity +2 = infinity, but also infinity + 3 = infinity, infinity +17=infinity-1/pi, etc.

And -infinity+2 = -infinity, etc. "Infinity" is not a number. It breaks all the algebraic rules. Example infinity + x = infinity + y does not imply x=y.

I like to think of infinity more like a topological entity (the boundary of the real numbers). I think that's the way in which modern mathematics we tend to look upon it. But that's another story.

And Swansont's objection about units still stands.

Yes, I know that. I didn't write x+1~x or x^3+x^2+x+1x^3~x^3. 

I wrote x+2=1/c-v=1/c-c. You notice that I have two different notations, x and c.

Why am I choosing ~? 

It's like saying I have f(x)=g(y)=1/c-c=1/0=a*1/0 with a in R+*.

 

1 hour ago, swansont said:

If v=c you do not have a valid equation

1/c-v is a function, valid for v≠c. If you choose v=c (or any fixed value) you do not have a function.

You can do one or the other. Not both.

Functions are not equal simply because they both diverge to the same limit.

x and x+2 both go to infinity, but x ≠ x+2

Is it true that 1/(c-v) = 1/(c-c) = 1/0 = a*1/0 with a in R*+ when v=c?

If it is true, then I would set x+2=1/c-c=1/0=a/0 with a>0. Therefore, x+2 would not have any unit because even if I change the value of c, I would still have x+2=1/c-c=1/0=a/0, and thus x+2 would not have any unit.

Edited July 25, 2023 by Z.10.46

[MJ kihara]

32 minutes ago, Z.10.46 said:

Yes, I know that. I didn't write x+1~x or x^3+x^2+x+1x^3~x^3. 

I wrote x+2=1/c-v=1/c-c. You notice that I have two different notations, x and c.

Why am I choosing ~? 

It's like saying I have f(x)=g(y)=1/c-c=1/0=a*1/0 with a in R+*.

 

Is it true that 1/(c-v) = 1/(c-c) = 1/0 = a/1/0 with a in R*+ when v=c?

If it is true, then I would set x+2=1/c-c=1/0=a/0 with a>0. Therefore, x+2 would not have any unit because even if I change the value of c, I would still have x+2=1/c-c=1/0=a/0, and thus x+2 would not have any unit.

Where did +2 come from....did you just fixed it to suit your arguments?

[Z.10.46]

14 minutes ago, MJ kihara said:

Where did +2 come from....did you just fixed it to suit your arguments?

No, I didn't change 2 ,here we are discussing something else, we are talking about an equality and not a limit. And why does x+2 in this equality have no unit...😀

if they agree with my argument, we can move on to calculating the limit.😀

 

Edited July 25, 2023 by Z.10.46

[Z.10.46]

To further support my argument, I give an example with f(x) and g(y), where f(x)=x and g(y)=1/y. I set f(x)=g(y), so I have x=1/y, which is true for any real x and y, even if I set y=0, and I get x=infinity. This equality remains valid even if x is no longer a number and equals infinity because I have infinity=1/0.

Next, with f(x) and g(v), where f(x)=x+2 and g(v)=1/(c-v). I set f(x)=g(v), so I have x+2=1/(c-v), which is true for any real x and v, even if I set v=c, and I get x+2=infinity. This equality remains valid even if x+2 is no longer a number and equals infinity because I have infinity=1/0. Therefore, at v=c, I have x+2=1/(c-v)=1/(c-c)=1/0=a/0, where c can have any unit, so x+2 does not have a unit of measurement.

I believe this argument is clear. If not, could you please point out any errors in this argument as well?

Edited July 25, 2023 by Z.10.46

[Z.10.46]

And here's another argument: at v=c, 1/(c-v)=1/(c-c)=1/0=infinity, and infinity has neither quantity nor unit of measurement. Therefore, I have no problem setting x+2=1/c-v.

 

Edited July 25, 2023 by Z.10.46

[Bufofrog]

6 hours ago, Z.10.46 said:

I believe this argument is clear. If not, could you please point out any errors in this argument as well?

How many times do you need to be told that 1/0 is undefined?  1/0 does not equal infinity.  Your math makes no sense.

As far as infinity goes, it appears this thread is an infinite loop.

Edited July 25, 2023 by Bufofrog

[swansont]

7 hours ago, Z.10.46 said:

I have x+2=1/(c-v), which is true for any real x and v, even if I set v=c, and I get x+2=infinity. This equality remains valid even if x+2 is no longer a number and equals infinity because I have infinity=1/0. Therefore, at v=c, I have x+2=1/(c-v)=1/(c-c)=1/0=a/0, where c can have any unit, so x+2 does not have a unit of measurement.

When you write down a function it needs to be valid for any value of the variables. Not just at one, particularly when it diverges. Units matter.

!

Moderator Note

Since you’re just repeating this nonsense, there’s no point in the thread remaining open.

Do not bring this up again