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.

Let $n$ be an integer value, then $n+\frac{1}{\sqrt2}$ is irrational. 
Since the set of integer is not bounded from above, the set of irrational number is not bounded from above since $n+\frac{1}{\sqrt2}> n$.
Remark: there is nothing special about the number $\frac1{\sqrt2}$, it can be replaced by any positive irrational number.
