# Proving that there is no biggest number

Problem Elsewhere, I have seen it proposed by a mathematician that there is a "greatest number", as opposed to infinitely many numbers. This number is supposed to be exceedingly large, larger than Graham's number or other such "big" numbers, yet still finite. I dislike this notion immensely, and so have a rough proof that no such greatest number exists. This is outlined below.

Proof Assume that there is some such greatest number $z$. If such a number $z$ exists, then $\frac{1}{z}$ must also exist, and be the smallest, non-zero positive number. Because this number can be expressed as a fraction, $\frac{1}{z}\in\mathbb{Q}$, we now encounter an issue. If there is a greatest number, then there cannot be any irrational numbers, as there is a smallest non-zero number. There are various proofs that there are irrational numbers, and so that is not included here. Because of this contradiction, irrational numbers must exist, yet cannot exist, there is a contradiction, and so the assumption that there is a biggest number $z$ must be false. Therefore there is no biggest number.

Request Is this proof correct? Is there some logical error or unknown construct which invalidates the proof? If not, is there a better proof?

The closest proof I could find is that $\mathbb{N}$ has infinite elements on proof-wiki, but this is not quite the same issue.

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Why not simply note that $z+1>z$? – Alex Becker Aug 7 '12 at 8:41
It is claimed that $z+1=0$ due to something that I can only understand as being similar to integer overflow. This seeks to avoid that oddity. – 512 Aug 7 '12 at 8:46
Well if you're working with modular arithmetic, then "biggest" doesn't even make sense b/c there is no order on the set of integers mod n. – Alex Becker Aug 7 '12 at 8:48
Indeed you would get a contradiction: $0<1<2<\cdots<n<n+1=0\implies 0<0$. – anon Aug 7 '12 at 8:49
I think if you're willing to accept a "largest number", then abandoning transitivity of < is a reasonable next step. – Niel de Beaudrap Aug 7 '12 at 14:11

I suppose that you are referring to Zeilberger's appearance on that BBC Horizon - To Infinity and Beyond.

There is a problem with what you are trying to do. When you try to prove something, one should specify the axioms being used.

If you are assuming the commonly used Peano's axioms then it is easy to prove that there is no largest number. If you assume that there are only finitely many numbers (e.g. you are working in $\mathbb Z/(3)$) then there is no way of proving that.

Zeilberger's belief is that Peano's axioms are wrong (or maybe even inconsistent) and he chooses to work in a different setting where a largest number exists.

Why is that okay? Recall that we cannot prove the consistency of Peano's axioms from the same theory, we need to assume a stronger theory in the background (e.g. set theory like ZFC), but we cannot prove that stronger theory is consistent either. At some point we need to rely on belief, and sure there is a good evidence that ZFC is consistent, but we proved that we cannot prove it unless it's inconsistent to begin with.

So working with a setting which seems unusual is not the end of the world. Much like assuming the axiom of choice was a controversial settings for mathematics.

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