How do we prove that the irrational numbers have no upper bound From Calculus to Apostol I know that real numbers do not have upper bound, I also know that irrational numbers belong to real numbers. Would the mathematical proof be different?
I quote the theorems to determine that the real numbers are not upper bounded.

Theorem #1: The set P of positive integers 1,2,3,... is unbounded above.
Proof #1: Assume P is bounded above. We shall show that this leads to a contradiction. Since P is nonempty, P has a least upper bound, say b. The number b−1, being less than b, cannot be an upper bound for P. Hence, there is at least one positive integer n such that n>b−1. For this n we have n+1>b. Since n+1 is in P, this contradicts the fact that b is an upper bound for P.
Theorem #2: For every real x there exists a positive integer n such that n>x.
Proof #2: If this were not so, some x would be an upper bound for P, contradicting Theorem #1.\

Because of my lousy English I also quote the commentary from which I took the quote from Apostol:

frosh (https://math.stackexchange.com/users/211697/frosh), How do we prove that the real numbers have no upper bound, URL (version: 2016-01-06): https://math.stackexchange.com/q/1602018

Thanks.

Suppose irrational numbers have an upper bound, say, $B$. Let $x$ be any irrational number. Without loss of generality, assume $x$ is positive (if it isn't, replace $x$ with $-x$, which is postive and still irrational. 
I say that there is a natural number $n$ such that $nx > B$: indeed, it is equivalent to asking for a number $n$ such that $n > B/x$ (remember $x$ is positive). Such a number exists -- otherwise, $B/x$ would be an upper bound for all natural numbers, which we already know doesn't exist.
But if $x$ is irrational, so is $nx$: if $nx = a/b$, a rational number, then $x = a/nb$, also a rational number. Since $nx$ is greater than $B$, $B$ cannot be an upper bound for all irrational numbers, so irrational numbers have no upper bound.
