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


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    If $n \in \mathbb{N}$, then $n + \pi$ is irrational. Since the natural numbers are unbounded, the irrationals are unbounded. – Xander Henderson Feb 14 '18 at 01:29
  • The proof can be similar, or you could pass directly from the first theorem (that the integers have no upper bound) to disproving an upper bound for *irrational numbers*. The point is that while the real numbers (Thm. 2) include the integers, we can't say the same about the irrational numbers. However consider that the sum of an integer and a positive irrational such as $\sqrt 2$ is an irrational number (and larger than the integer used). – hardmath Feb 14 '18 at 01:30
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    The set of irrational numbers is closed with respect to $x\to 2x$, hence it cannot be bounded. – Jack D'Aurizio Feb 14 '18 at 13:02

2 Answers2


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.

Siong Thye Goh
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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.

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