Prove that there does not exist a largest natural number I need a very simple proof for this.
I could say let $x$ be the largest natural number. Then $x + 1 > x$. QED.
But that doesn't really work does it? What's a very simple proof (no number theory or any sort like that)?
 A: Your proof is good. However, I see some places where you could add more detail:


*

*Why are we assuming that $x$ is the largest natural number? (Answer: because we are doing a proof by contradiction.)

*Why is the fact that $x + 1 > x$ relevant? (Answer: because it contradicts the fact that $x$ is the largest natural number.)


If we add these details in, the proof looks like this:

We will show that there is no largest natural number. In order to do this, let's assume that $x$ is the largest natural number. By the definition of "largest", there is no natural number which is greater than $x$. However, $x + 1$ is a natural number and $x + 1 > x$. This is a contradiction. QED.

There's one more question that this proof doesn't answer:


*

*How do we know that $x + 1 > x$?


If you haven't already proved that $x + 1 > x$, then you should think about giving it a shot.
A: I am a philosophy major, not mathematics, all the same, the following will prove what you are looking for:


*

*Assume the existence of a largest natural number (p)

*Being the largest natural number, 'p' cannot be followed by a greater natural number. Meaning:


p + 1 ≤ p


*Now subtract 'p' from both sides. The result is:


1 ≤ 0


*This has shown that the assumption of 'p' leads to a contradiction.

*Therefore, it is not the case that a largest natural number exists.
QED
A: Here is a constructive proof.
Let $P(n)$ be the property that there exists a natural number $n'$ where $n' > n$. We prove this by induction.
Base case - $n=1$: Clearly, $P(1)$, since $484000 > 1$.
Step – assume for $n$, prove for $n+1$: Assume by induction hypothesis that there exists an $n''$ such that $n'' > n$. By the precongruence properties of $>$ we easily see that $n''+484000 > n+1$. So take $n' = n'' + 484000$.
If we want to prove the statement that it is not the case that there exists an $n_0 \in \mathbb{N}$ such that $n \leq n_0$ for all $n \in \mathbb{N}$, then we assume that there exists such an $n_0$. But then we have $n_0 + 484000 \leq n_0$, which is a contradiction. Therefore $n_0$ cannot exist.
A: Maybe that can be shown in a million of ways, but here is one proof just for fun (which surely lacks some foundational rigor). We will use this facts:
1) $1$ is not the largest natural number
2) For every  natural number $n \neq 1$ the number $n^2-n+1$ is a natural number that is not equal to $n$ (that is, we have $n^2-n+1 \neq n$)
3) $(n-1)^2>0$ for every natural $n \neq 1$.
Suppose there exists largest natural number, call it $n_L$, now form a natural number $n_L^2-n_L+1$. Because $n_L$ is the largest and because of 2) we have: $n_L>n_L^2-n_L+1$, which is equivalent to $0>(n_L-1)^2$ but this is not possible because $n_L \neq 1$ and because of 3).
