1
$\begingroup$

Let $p$ be a prime number. Show that $\sqrt{p}$ is a irrational number.

Attempt: Suppose for contrary that $\sqrt{p}$ is a rational number. Then, $$\sqrt{p}=\frac{m}{n} \qquad (1),$$ for some integers $m$ and $n$ with $n \ne 0$ and $\gcd(m,n)=1$. By squaring both side of $(1)$, we obtain $$p=\frac{m^2}{n^2} \implies m^2 = pn^2 \implies p\mid m^2. \qquad (2)$$ We'll show that if $p\mid m^2$, then $p \mid m$. Suppose for the contrary that $p \nmid m$. Then, $m=pq+r$, for some integer $q$ and $r$ with $0<r<p$. Hence, $$m^2=(pq+r)^2=p^2q^2+2pqr+r^2.$$ Since $p \mid m^2$, then $p \mid p^2q^2+2pqr+r^2$, which means $p \mid r^2$. But, since $0<r<p$, we have $p \nmid r^2$, a contradiction. So, if $p \mid m^2$, then $p \mid m$. Now, on $(2)$, we get $p \mid m$. It means there is an integer $k$ such that $m=pk$. Plugging this in to the relation $m^2=pn^2$, we have $p^2k^2=pn^2$, which implies that $p \mid n$. Since $p \mid m$ and $p \mid n$, then $\gcd(m,n) \ge p>1$, since $p$ is a prime number. Contradiction with the assumption that $\gcd(m,n)=1$. Therefore, $\sqrt{p}$ is a irrational number.

Does the above approach correct, especially in the proof of implication: "If $p\mid m^2$, then $p \mid m$" (without using Euclid's Lemma; that is, by an elementary approach)? Thanks in advanced.

$\endgroup$
4
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Sep 22, 2022 at 14:27
  • 1
    $\begingroup$ I prefer to show that $\gcd(m,n) = 1 \implies \gcd(m^2, n^2) = 1$, therefore if $n \nmid m$, then $n^2 \nmid m^2$. So if $\frac mn$ is not integer, neither is $\frac {m^2}{n^2}$. Therefore if $p$ is any integer other than a perfect square, $\sqrt p$ cannot be rational. $\endgroup$ Sep 22, 2022 at 22:53
  • $\begingroup$ Does this answer your question? How to prove: if $a,b \in \mathbb N$, then $a^{1/b}$ is an integer or an irrational number? $\endgroup$ Sep 24, 2022 at 23:29
  • $\begingroup$ This question is a multi-duplicate in that many similar or identical questions have been closed as duplicates. The above answer seems to be the generic one which all the duplicates are pointed at. $\endgroup$ Sep 24, 2022 at 23:30

2 Answers 2

2
$\begingroup$

The equality $$ m^2=pn^2 $$ contradicts the fundamental theorem of arithmetic because $p$ appears an even number of times on the left hand side and an odd number of times on the right hand side.

$\endgroup$
0
$\begingroup$

Your method seems quite elementary. The one of Sinclair seems (my opinion) more sophisticate because $n\not\mid m$ does not mean $n\wedge m=1$.

$\endgroup$

You must log in to answer this question.