0=1

[Aranfan]

I think this goes here.

So I recently came across a proof that 0=1 and I can’t find the mistake I’m doing this from memory but I think I have all the important stuff, can anyone tell me what’s wrong in the proof?

(1) 0=0+0+0+…
(2) 0=[1+(-1)]

Substituting (2) into (1) gets:

(3) 0=[1+(-1)]+ [1+(-1)]+ [1+(-1)]+…

By the Associative Property of Addition:

(4) 0=1+[(-1)+1]+ [(-1)+1]+ [(-1)+1]+…

By the Commutative Property of Addition:

(5) 0=1+[1+(-1)]+ [1+(-1)]+ [1+(-1)]+…

Substituting (2) back into (5) gets:

(6) 0=1+0+0+0+…

Substituting (1) into (6) gets:

(7) 0=1+0

From the definition of Additive Identity:

(8) 0=1+0=1

Applying Transitivity of Equality:

(9) 0=1



What was done wrong?

[Jaepheth]

The alternating series Sum(i=0..inf) -1^(i) does not converge.

[Jaepheth]

Ghetto edit.

It's basically saying if a series S = z1+z2+z3+z4.... converges for zn = c for all n then S also converges if zn is a series... which in no way follows.

[Aranfan]

The alternating series Sum(i=0..inf) -1^(i) does not converge.

Why not? Does the substitution not work?

[Formless]

A very simple theorum exists to prove 1 =/= 0

1 + 1 = 2

0 + 1 = 1

Therefore, 0 =/= 1

[Karza]

(3) 0=[1+(-1)]+ [1+(-1)]+ [1+(-1)]+…

By the Associative Property of Addition:

(4) 0=1+[(-1)+1]+ [(-1)+1]+ [(-1)+1]+…

Moving the brackets around like that would still leave a lone (-1) as the last term there, after the [(-1)+1] terms, so...

By the Commutative Property of Addition:

(5) 0=1+[1+(-1)]+ [1+(-1)]+ [1+(-1)]+…

...the result result here is actually 0=1+0+0+...+0+(-1), which is ye olde 0=0.

[Surlethe]

I thought this would be the old "divide by zero" proof. Let's see here ...

By the Associative Property of Addition:

(4) 0=1+[(-1)+1]+ [(-1)+1]+ [(-1)+1]+…

This is not justified. Writing an infinite sereis is shorthand for writing the limit of partial sums. In this shorthand, the negative 1 that's paired with the 1 at the front "disappears" because it's pushed off to infinity, but in actuality, the sum is still zero. Here's what's really going on behind the scenes: the sum a_0+a_1+a_2+a_3+... is defined as the limit of the sums a_0, a_0+a_1, a_0+a_1+a_2, a_0+a_1+a_2+a_3, and so forth. This means that, in the case of zero, 0 = 0+0+0+... = lim_{n\rightarrow \infty} \sum_{i=0}^n 0 = lim_{n\rightarrow \infty} \sum_{i=0}^n [(1)+(-1)] = [(1)+(-1)] + [(1)+(-1)] + [(1)+(-1)] + ... [(1)+(-1)]. Rearranging the terms in this sequence doesn't alter its value.

Another way to think about it is, as Jaepheth pointed out, you're rearranging terms in the infinite series 1-1+1-1+1-1+1-1+..., which does not converge. You can actually get any integer out of this series by rearrangement, not just 1.

[Jaepheth]

Since the terms remain constant, the partial sums do not converge and if you start on -1 then the partial sums simply alternate between -1 and 0. And if you start with 1, the partial sums alternate between 1 and 0.

If a series X(n) to converges to some L then the sequence of its partial sums also converges. More formally, for every epsilon > 0 there exists an N such that for all n >= N , |Sn - L| <= epsilon.

Proof: Let X(n) be the series SUM(n=1..inf) (-1)^(n) We will assume this series converges to 0 as conjectured in the OP's proof. Let 0<epsilon<1, so... let's say epsilon = 1/3. then there must exist an N such that |Sn-0| <=1/3 for ALL n>=N. Clearly though, the only possible values of |Sn| are 1 and 0 and for any N there is an n> N such that Sn = 1 which is > 1/3. Therefore, the series cannot converge.

[Aranfan]

I thought this would be the old "divide by zero" proof. Let's see here ...

By the Associative Property of Addition:

(4) 0=1+[(-1)+1]+ [(-1)+1]+ [(-1)+1]+…

This is not justified. Writing an infinite sereis is shorthand for writing the limit of partial sums. In this shorthand, the negative 1 that's paired with the 1 at the front "disappears" because it's pushed off to infinity, but in actuality, the sum is still zero. Here's what's really going on behind the scenes: the sum a_0+a_1+a_2+a_3+... is defined as the limit of the sums a_0, a_0+a_1, a_0+a_1+a_2, a_0+a_1+a_2+a_3, and so forth. This means that, in the case of zero, 0 = 0+0+0+... = lim_{n\rightarrow \infty} \sum_{i=0}^n 0 = lim_{n\rightarrow \infty} \sum_{i=0}^n [(1)+(-1)] = [(1)+(-1)] + [(1)+(-1)] + [(1)+(-1)] + ... [(1)+(-1)]. Rearranging the terms in this sequence doesn't alter its value.

Another way to think about it is, as Jaepheth pointed out, you're rearranging terms in the infinite series 1-1+1-1+1-1+1-1+..., which does not converge. You can actually get any integer out of this series by rearrangement, not just 1.

Ah, yeah, somebody I showed it to thought that that step wasn't kosher, but she couldn't figure out how exactly.

Since the terms remain constant, the partial sums do not converge and if you start on -1 then the partial sums simply alternate between -1 and 0. And if you start with 1, the partial sums alternate between 1 and 0.

If a series X(n) to converges to some L then the sequence of its partial sums also converges. More formally, for every epsilon > 0 there exists an N such that for all n >= N , |Sn - L| <= epsilon.

Proof: Let X(n) be the series SUM(n=1..inf) (-1)^(n) We will assume this series converges to 0 as conjectured in the OP's proof. Let 0<epsilon<1, so... let's say epsilon = 1/3. then there must exist an N such that |Sn-0| <=1/3 for ALL n>=N. Clearly though, the only possible values of |Sn| are 1 and 0 and for any N there is an n> N such that Sn = 1 which is > 1/3. Therefore, the series cannot converge.

Thank you! My calculus teacher said that the series didn't converge because otherwise you could prove 0=1 as in my original post. I am happy to know that the Associative Property is not in danger.

[Kuroneko]

Well, the question has already been answered (twice, in fact), but Surlethe's comment about changing values can be carried further. The general reason for failure is not the really that the relevant series, here Sum[ (-1)^k ], diverges (although that a sufficient reason, obviously), but that it is not absolutely convergent, i.e., the sum of the absolute values of the terms diverges.

In the reals, a series that is not absolutely convergent will have some permutations of addends giving different values. In other words, it's also possible to do a similar trick with a convergent series, say Sum[ (-1)^k / k ], with no divergent re-arrangement appearing in the problem.

[NoXion]

A very simple theorum exists to prove 1 =/= 0

1 + 1 = 2

0 + 1 = 1

Therefore, 0 =/= 1

Indeed. The OP looks to my untrained eye like a victim of Blinding by Mathematics. If someone tried to prove to me that I have 1 cigarette instead of 0, I would certainly look askance at them. Isn't mathematics supposed to be an abstraction of reality? Therefore it is impossible for 0=1, since cigarettes do not mysteriously appear in empty fag packets.

Much to my chagrin.

[Kuroneko]

Indeed. The OP looks to my untrained eye like a victim of Blinding by Mathematics. If someone tried to prove to me that I have 1 cigarette instead of 0, I would certainly look askance at them.

It's not serious; there was probably never any doubt that 0 or 1 are unequal on the part of the OP. Rather, the objective is to identify the mistake in the purported proof--an exercise to test one's understanding of infinite series.

Isn't mathematics supposed to be an abstraction of reality?

I'd prefer to say just abstraction of structure. Some special relationship to reality makes a piece of mathematics pragmatically useful, and for some people also more interesting, but it's hardly necessary. Otherwise, it'd be physics.

[Steel]

S = 1 - 1 + 1 - 1 + ...

S = 1 - S

=> 2S=1

=> S=0.5

Ta da!

Just remember divergent means the same as does not exist. It is actually impossible to write down S = blah . There is no (real) number whose value this expression takes.

By cleverdickery(tm) you can make them equal whatever you want (easier done in the case of divergent integrals). So of course the best thing to do once you assign a value to something which does not exist is to make a proof that 1 = 0.

[Grog]

There are ways that makes sense and are used to "sum" things like that. For example Cesaro summation where you take the mean value of the first n partial sums. Then kind of "1/2=1-1+1-1+1-..."

[Kuroneko]

Yes, one can also generalize the series of terms into a formal power series with the terms as the coefficients and work with that instead. That's rather common in physics, actually. There are also other methods.