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$\sqrt a$ is either an integer or an irrational number.

I have this homework problem that I can't seem to be able to figure out:

Prove: If $n\in\mathbb{N}$ is not the square of some other $m\in\mathbb{N}$, then $\sqrt{n}$ must be irrational.

I know that a number being irrational means that it cannot be written in the form $\displaystyle\frac{a}{b}: a, b\in\mathbb{N}$ $b\neq0$ (in this case, ordinarily it'd be $a\in\mathbb{Z}$, $b\in\mathbb{Z}\setminus\{0\}$) but how would I go about proving this? Would a proof by contradiction work here?



Let $n$ be a positive integer such that there is no $m$ such that $n = m^2$. Suppose $\sqrt{n}$ is rational. Then there exists $p$ and $q$ with no common factor (beside 1) such that

$\sqrt{n} = \frac{p}{q}$


$n = \frac{p^2}{q^2}$.

However, $n$ is an positive integer and $p$ and $q$ have no common factors beside $1$. So $q = 1$. This gives that

$n = p^2$

Contradiction since it was assumed that $n \neq m^2$ for any $m$.

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    $\begingroup$ Oh, I see. We basically use the fact that $n$ must be an integer to draw that the denominator must be 1. Got it, thank you very much! I got up until then and couldn't progress further, but this makes it clear! $\endgroup$ – roboguy12 Aug 31 '12 at 4:57
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    $\begingroup$ It is important to explicitly mention that the proof uses unique factorization (or Euclid's Lemma) to deduce that $q = 1$. This can fail in rings lacking such properties. $\endgroup$ – Bill Dubuque Aug 31 '12 at 5:07
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    $\begingroup$ Here how can I write q=1 $\endgroup$ – codeomnitrix Oct 7 '14 at 15:02
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    $\begingroup$ @BillDubuque Yes. This proof seems to use the property that when elements $p$ and $q$ have no common divisor (other than units), then $p^2$ and $q^2$ have no common divisor (other than units). It sounds like a property that holds only in some rings (which ones?). $\endgroup$ – Jeppe Stig Nielsen Apr 24 '16 at 8:03

Here’s an explanation that I find clearer, and that uses unique factorization explicitly: If a positive number $n$ is not the square of any integer, then when you write it as a product of primes, at least one prime shows up to an odd power. Let one such prime be $p$, and look at the supposed equation $\sqrt{n}=a/b$, with $a$ and $b$ positive integers. This gives $n=a^2/b^2$, hence $nb^2=a^2$. How many times does $p$ show up in the factorization of the left side and of the right? Oddly many times on the left, evenly many on the right. Contradiction to the unique factorization of $nb^2$.


This can also be done with the rational root test: consider the polynomial equation $$x^2 - n = 0$$

and suppose that it has a rational root. Then, this rational root must be an (integer) factor of $n$. So, if $\sqrt{n}$ is rational, then there exists $t\in \mathbb{N}$ (since $x^2 - n$ is an even function of $x$, we may assume, without loss of generality, that $t>0$) with $t \vert n$ such that $$t^2 - n = 0$$ which is to say $$n = t^2$$ and hence $n$ is the square of a natural number.

In fact, this argument generalizes to showing that if $\sqrt[m]{n}$ is rational, then $n$ is an $m^{th}$ power.


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