1 year ago
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• a year ago

0=0+0+0+...
0=(1-1)+(1-1)+(1-1)+...
0=1-1+1-1+1-1+...
0=1+(-1+1)+(-1+1)+(-1+1)+...
0=1+0+0+0+...
0=1

• a year ago

@mani_bharathy Not that this is structured in any way that follows, but if these are taken as premises then the argument begs the question in the 4th premise by assuming the conclusion.

• a year ago

@vladimir_susic_jna 4th step follows from 3rd step. It's not begging the question

• a year ago

Have you seen the famous proof for sum of all natural numbers is -1/12? They do these kinds of transformation at lot of places

• a year ago

@mani_bharathy The issue with this “proof” is that in math, we can’t do algebraic manipulations with an infinite number of things.

Yes, I’ve done calculus, stick with me here.

When we use the concept of infinity in rigorous math, we always describe it as a limiting process. That is to say, we start with a finite value and ask what happens as that value becomes infinitely large (as in this case) or infinitely small.

If we can find no discernible pattern that emerges as the value tends to infinity, we don’t draw conclusions from that. And we certainly never start from an infinite number of things and try to do our math from there.

Take the proof of the sum of an infinite geometric series. Some math teachers would present it in the following way:

Let S = 1 + x + x^2 + x^3 + x^4 + ...
-( xS = x + x^2 + x^3 + x^4 + ... )
S(1-x) = 1
S = 1/(1-x)

But this is an incorrect proof. The proper proof is the following:

Let S(n) = 1 + x + x^2 + x^3 + x^4 + ... + x^n
-( xS(n) = x + x^2 + x^3 + x^4 + ... + x^(n+1) )
S(n)(1-x) = 1 - x^(n+1)
S(n) = (1-x^(n+1))/(1-x)

Then, using our S(n) formula, we can take the limit as n —> infinity, and if |x| < 1, then S = 1/(1-x)

The first proof is incorrect because it starts with the infinite and tries to get an answer. The second proof is correct because it starts with the finite, gets an answer, then extends that answer to the infinite.

The same issues with the incorrect proof apply to your proof. You can’t start with the infinite.

• a year ago

@kyrothehero There is a lot to say. But I'll start with this one thing.

That is Ramanujan's handwritten proof for the sum of all natural numbers. Based on your above issues with my proof, you should have problems with Ramanujan's proof also.

If you have problem with that proof, mind you, you are messing with Ramanujan

• a year ago

@mani_bharathy It is completely true that mathematicians such as Ramanujan have worked on assigning values to divergent series.

https://brilliant.org/wiki/sums-of-divergent-series/

But to say that the sum of the divergent series “equals” the value which is assigned to it, in any conventional/colloquial sense of the word, is simply wrong.

Your proof is different from Ramanujan’s proof in an important way, which is that you attempt to telescope a divergent series. By trying to group terms together, the steps of your proof effectively assume, at different steps, that infinity is either even or odd; that’s the only way that all of the 1’s and -1’s can cancel out. Telescoping, as with many other algebraic tricks for evaluating series, only works if the value of the terms tends to zero. That is why I said that you need to start with finite values and then extend to infinity.

While Ramanujan’s proofs are questionable in some ways, they do not suffer from the same problems that your proof does. Ramanujan does not require infinity to be even or odd in his proof. Instead, all that he assumes is that the infinite series converge. From that assumption he is able to algebraically manipulate the series and produce a finite result.

And yes, I do have a problem with anyone who says that the sum of all natural numbers is -1/12. To make such a statement without a qualifier as to what we mean by “sum” is ill-advised.

• a year ago

@kyrothehero So your problem with my proof is, it's assuming infinite as either odd or even without any reason. Let me give one other thing which does not have that(odd or even) problem and from there I can again prove my result.

Proof :

sum of all even num
= 2+4+6+...
= 2(1+2+3+...)
= 2 * Sum of all natural numbers

Do you have any issues with this proof? If yes then is it any different from the issues you have with Ramanujan's proof ?

• a year ago

@mani_bharathy I just looked at Ramanujan’s proof and I’ve changed my mind. I don’t think that it’s valid to do this algebraic manipulation of infinite sums at all; I think Ramanujan’s proof is invalid.

Please don’t give me this “you are messing with Ramanujan” stuff if you want to have a productive conversation. If you think Ramanujan is correct and I am incorrect, then provide a reason that I am incorrect rather than citing an authority.

Let’s assume that the sum of all natural numbers has a traditional mathematical value S.

Using algebraic manipulation, we get
S = 1 + 2 + 3 + 4 + 5 + ...
-( S = 1 + 2 + 3 + 4 + ... )
0 = 1 + 1 + 1 + 1 + 1 + ...

However, using analytic continuation of the Riemann zeta function, we get
1 + 1 + 1 + 1 + ⋯ = ζ(0) = −1/2.
https://en.m.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_⋯

Do you see the problem? When we do these sorts of algebraic manipulations, we are led to absurd conclusions.

• a year ago

@kyrothehero
"When we do these sorts of algebraic manipulations, we are led to absurd conclusions."

So your problem with my proof is "those" sorts of algebraic manipulations. Okay. I won't do those.

Since you seem fine with using analytic continuation of the Riemann zeta function, what if I say
ζ(-1) = −1/12

Or you are including Rieman zeta function analytical continuation also in "those" algebraic manipulations?

• a year ago

@mani_bharathy I’m completely fine with using analytic continuation to assign values to divergent series. My issue doesn’t lie there, because analytic continuation doesn’t lead you to the conclusion that zero equals one.

My problem is with the algebraic manipulations which, as I’ve demonstrated, contradict the results found through analytic continuation.

I also have a problem with people who say “the sum of all natural numbers is -1/12” without an explanation that the word “sum” is being used in a very different way than we normally use it.

So yes, the zeta function of -1 is -1/12. But arriving at that answer using algebraic manipulation is a flawed methodology.

• a year ago

@kyrothehero I can respond in different ways to your statement "algebraic manipulations is a flawed methodology". A simple one would be the following.

You think it's just co-incidence that Ramanujan got the right answer by using wrong method? What are the chances for that? Mathematically the chances are zero to randomly get a particular number from a set of infinite numbers. Instead it makes more sense to believe that his proof is also a valid way.

• a year ago

@kyrothehero Without infinite series algebraic manipulation, some other proofs are

1^0 = 1^1 => 0=1

1! = 0! => 0=1

1^0 = 2^0 => 1=2 => 0=1

0^1 = 0^2 => 1=2 => 0=1

0*0 = 0*1 => 0=1

(1 - (1/2))^2 = (0 - (1/2))^2 => 0=1

• a year ago

@mani_bharathy All of those proofs rely on believing that if f(x) = f(y), then x = y. This is only true for bijective functions, and since none of your functions are bijective, the proofs are invalid.

• a year ago

@kyrothehero
Okay. Let's take a basic bijective quadratic function f(x)=x^2 + x + 1

x² + x + 1 = 0
x+ 1 + (1/x) = 0 (Divide by x)
x² = 1/x (Substitute x+1=-(1/x) in the first equation)
x³ = 1
x = 1

Substitute x=1 in the equation
1² +1 +1 = 0
3=0
1=0

• a year ago

@kyrothehero If you want some calculus,

Integeration by parts :

∫udv = uv−∫vdu

∫(1/x) dx = (1/x)*x - ∫x(-1/x²)dx

∫(1/x) dx = 1 + ∫(1/x) dx

0=1

• a year ago

@mani_bharathy
For the quadratic one (which, by the way, isn't bijective), see this video: https://youtu.be/SGUZ-8u1OxM

For the calculus one, you forgot to add C. Integration by parts doesn't remove the necessity of the arbitrary constant.

• a year ago

@kyrothehero I took it from that video only. He says that the steps should not be creating extraneous solutions. But he did not say why.

Yeah, my bad, it's not bijective. Although I can say that it's bijective if it's defined for domain [0, inf] and range [1, inf].

• a year ago

@kyrothehero if you want a pictorial proof
https://1millionmonkeystyping.files.wordpress.com/2013/11/currys.jpg

say the area of those pieces are x in the first pic and x-a in the second pic

x=x-a
0=a
0=1

• a year ago

@kyrothehero

If you want a proof from probability, there is Bertrand paradox

• a year ago

@kyrothehero Today came across Banach-Tarski paradox which can get me 1=2

• a year ago

You need to address that (1-1) has to come in pairs when continuing to infinite. You cannot include one without the other. That's simply the problem. (Step 4)

• a year ago

@the_logician There are some other proofs I gave in the comments. You can check those out too