Its been a while since I've written a proof and would appreciate some feedback on this one.

Question: Given the set of rational positive values, $\{q | q \in \mathbb{Q} \wedge 0 \lt q \lt \sqrt{2}\}$, show that there is no maximum value for $q \lt \sqrt{2}$


Given $x=\sqrt{2}$, suppose that $x$ can be defined as a positive rational number such that $x$ is composed of positive numbers $p, q$ such that $x=\frac{p}{q}$ where $q \neq 0 $ and that $p,q$ are simplified to the lowest possible terms.

It follows that $2=\frac{p^{2}}{q{^2}}$ or $p^{2} = 2 \cdot q{^2}$. Therefore, $p^{2}$ must be an even number as it is the product of some $n$ and an even number. As a result, $p$ is an even number because otherwiese, $p^{2}$ would be odd.

If $p$ is an even number, then $p=2n$ for some number $n$.

Substituting $p=2n$ into the original equation:

$$2= \frac{(2n)^{2}}{q^{2}}$$ $$2= \frac{(4n^{2}}{q^{2}}$$ $$2q^{2} = 4n^{2}$$ $$q^{2} = 2n^{2}$$

Therefore, $q^{2}$ is an even number, which makes $q$ even as well. This is a contradiction as $p, q$ are defined to be simplified to the lowest possible terms, which would not be possible if $p, q$ were even. Therefore, $\sqrt{2}$ must be an irrational number.

An irrational number is defined as a number with no terminal or repeating decimals. Because $\sqrt{2}$ repeats to infinite decimal places, there is no exact value that can be defined as a maximal value $< \sqrt{2}$

Therefore, there is no positive rational value that can be defined to be less than or equal to the $\sqrt{2}$

My strategy for this proof is to prove that $\sqrt{2}$ is an irrational number and then explain why there is no possible maximal value for some number $\lt \sqrt{2}$. I think I've satisfied the first section, but feel there is rigor lacking in the second section. How can I redefine the proof that there is no maximal value to a number that is $\lt \sqrt{2}$?

  • $\begingroup$ You can prove that $\sqrt {2}$ is irrational in one line: If $\sqrt {2}=m/n$ is in lowest terms then $\sqrt{2}=(2n-m)/(m-n)$ is in lower terms. $\endgroup$ – Shahab Sep 11 '12 at 14:04
  • $\begingroup$ Related question: No maximum(minimum) of rationals whose square is lesser(greater) than 2. $\endgroup$ – Martin Sleziak Sep 11 '12 at 14:10
  • 2
    $\begingroup$ One can show that he set of rational numbers less than any given number does not have a maximal element. $\endgroup$ – Baby Dragon Sep 11 '12 at 14:17
  • $\begingroup$ @BabyDragon - indeed - I was interrupted by a phone call just as I was about to say the same. My example was rational numbers less than 1. However, it is easier to prove directly in this case (ie without a general theorem about the density of the rationals in the reals), because if $q<1$ then $q<\frac{q+1}2<1$. $\endgroup$ – Mark Bennet Sep 11 '12 at 14:39

To show that no maximum is attained you should show that given any positive rational number $0<\frac{a}{b} < \sqrt{2}$ that there is another rational number $\frac{c}{d}$ such that $\frac{a}{b}<\frac{c}{d}$.

This shows that no matter what you might try to use as a maximum, it misses some rationals. Thus no max can exist.

Also, one note: irrational numbers are defined to be real numbers that are not rational. Taking the definition to be reals with non-terminating non-repeating decimal expansions is equivalent but not natural and quite messy to use (formally).

A possible patch to your proof (spoiler):

Suppose $\frac{a}{b}<\sqrt{2}$. Then let $\epsilon=\sqrt{2}-\frac{a}{b}>0$. Since $1<2<\cdots<n<\cdots \rightarrow \infty$, $0 \leftarrow \cdots<\frac{1}{n}<\cdots<\frac{1}{2}<1$. So given any positive real numbers (such as $\epsilon$) we can find a positive integer $N$ such that $\frac{1}{N}<\epsilon$. Then $\frac{a}{b}<\frac{a}{b}+\frac{1}{N}<\frac{a}{b}+\epsilon=\sqrt{2}$. Thus $\frac{a}{b}+\frac{1}{N}$ is a rational number bigger than $\frac{a}{b}$ but still in your set.


The Title asks about "positive rational numbers less than or equal to two". In the body of the question, you ask about the non-existence of a maximal rational number in the interval, $(0,\sqrt{2})$. Let me address the case where we have an open interval, $(a,b)$. Regardless of whether or not $b$ is rational, this interval does not have a maximal rational. The first case is when $b$ is rational. Suppose that $x$ is a maximal rational in $(a,b)$. Then we construct the mean of $x$ and $b$, which is a larger rational number in the interval. Thus no maximal rational can exist. If $b$ is irrational, then we may take it's decimal expansion and produce the sequence given by all of the finite truncations as a strictly increasing sequence of rationals converging to $b$, giving the fact that no maximal rational less than $b$ can exist.


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