Proving “No matter how large a real number $x$ is given, there is always a natural number $n$ larger”

I know the statement above is true because of the Archimedian Theory. Would the following proof make sense to prove it?

This is a proof by contradiction.

If the set of natural numbers does have an upper bound, then it has a least upper bound. Let $x = \sup \mathbb{N}$, supposing that each does exist as a finite real number. Then there is not a natural number $n > x$. Therefore $n \leq x$. Then, $n \leq x - 1$ cannot be true for all natural numbers.

There is some natural number $m$ such that $m > x - 1$. Since $m$ is a natural number, $m + 1$ must also exist, so $m + 1 > x$. But this cannot be so since we defined $x$ as the largest number.

Therefore, by contradiction, the above statement stands true.

• Why can't $n \le x - 1$ be true for all natural numbers? – Lucas Mann Sep 10 '14 at 19:13
• It can't be true for all natural numbers since it is less than or equal to x. If n was equal to x to begin with, then it would be greater than x - 1 which isn't true since we supposed that there is no number n > x. – knerd Sep 10 '14 at 19:16
• Ok, suppose all natural numbers are $\le x$. Now take $x' := x + 1$. Then obviously all natural numbers are $\le x'$. If we apply your proof to $x'$, it claims that $n \le x' - 1$ cannot be true for all natural numbers. But $x' - 1 = x$. So aren't you just saying that the assumption is true? – Lucas Mann Sep 10 '14 at 19:21
• Suppose there is an $x$ such that $x\ge n$ for every natural number. Then the set of natural numbers is bounded above, so has a least upper bound $b$. Now you can produce an argument roughly along the lines described. But as it stands what you have is not a proof. – André Nicolas Sep 10 '14 at 19:22
• The statement you're trying to prove is usually taken as an axiom, by the way. – Greg Martin Sep 10 '14 at 19:35

It remains to prove the Archimedean property of the rationals. Assuming the construction as equals class of fractions $p/q$ with $p$ positive, we know that $p/q \leq p$.