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Some time ago I read Euclid's proof of the irrationality of $\sqrt 2$. The proof I leave below for any $n$ is quite similar in procedure to the one for $n=2$, and I think it is a good generalisation.

Theorem: $$ \not\exists k \in \mathbb N : k^2 = n \implies \sqrt n \not\in \mathbb Q $$

Proof (by contradiction):

For $n \ge 2$, assume $\sqrt n = \frac p q \land \textrm{gcd}(p, q) = 1$. Then $$ \begin{align*} n &= \frac{p^2}{q^2} \\ n q^2 &= p^2 \\ &\implies n | p^2 \iff n | p \\ &\implies \exists r : p = r \cdot n \\ nq^2 &= (rn)^2 \\ nq^2 &= r^2 n^2 \\ q^2 &= r^2 n \\ &\implies n | q^2 \iff n | q \\ &\implies \textrm{gcd}(p, q) \ge n \\ & \implies \bot \\ &\because \textrm{gcd}(p, q) = 1 \land \textrm{gcd}(p, q) \ge n \end{align*} $$

Therefore, $\sqrt n$ is irrational.

Is this a valid proof? Have I made a mistake?

Bill Dubuque
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Ronaldo Gligan
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    How do you get $n | q^2 \iff n | q $ – geetha290krm Sep 24 '23 at 09:18
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    Try running through your proof with $n = 4$ (and $p = 2$ and $q = 1$). Somewhere, the proof must break down. – Theo Bendit Sep 24 '23 at 09:19
  • Observe that $4|2^2$ but not $4|2$. So $n|q^2\Rightarrow n|q$ is not generally true. – Jam Sep 24 '23 at 09:34
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    For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Sep 24 '23 at 17:02

2 Answers2

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The claim is true but your proof attempt does not work. The assertion that $n\mid p^2 \Leftrightarrow n\mid p$ is false. E.g $9\mid 9$ does not imply $9\mid 3$.

In general, it suffices to prove that if $\sqrt[n]{m}$ is rational, then $\sqrt[n]{m}$ is an integer, where $m,n\in\mathbb N$ (see here, for instance). Then it follows that if $n$ is not a square, its square root cannot be rational.

AlvinL
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Let $n$ be an natural number with atleast a prime factor $k$.

Assume, $\exists p,q \in \mathbb Z \ \sqrt n = \frac{p}{q} \land \textrm{gcd}(p, q) = 1\\ \Rightarrow n=\frac{p^2}{q^2} \Rightarrow k\cdot n'=\frac{p^2}{q^2}$

where, $n' \in \mathbb N\\ \therefore \gcd(p^2,q^2)\ne 1 \Rightarrow \gcd(p,q) \ne 1 \text{ as } p,q \in \mathbb Z$

Hence; our assumption cannot be correct. $ \therefore \sqrt n \not\in \mathbb Q \quad \square$

Jam
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O M
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    I believe the aim of this question is to help and edit OP's proof rather than state your own. While your alternate answer is not incorrect, it does not quite "answer" OP's question. It will be unlikely for OP to understand where you are trying to correct him, unless you explain your proof with comments. Good effort in the short and sweet proof though! – Dstarred Sep 24 '23 at 10:30
  • Hey @Dstarred, I firmly believe that Stack Exchange is a valuable platform for sharing knowledge, and it's essential to encourage multiple high-quality solutions to each question. This benefits not only the original poster (OP) but also future users who may have similar questions but worded differently. Marking questions as duplicates can sometimes limit access to alternative solutions, so it's crucial to keep this in mind for the sake of the community's growth and learning. – O M Sep 24 '23 at 10:42
  • How do you get from $k \cdot n' = \frac{p^2}{q^2}$ to $\textrm{gcd}(p^2, q^2) \neq 1$? – Ronaldo Gligan Sep 24 '23 at 11:12
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    @RonaldoGligan full form of gcd is greatest common divisor. If dividing two integer gives a prime factor (or divisor), gcd of the that two integer will not equal to 1. – O M Sep 24 '23 at 11:25
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    Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Sep 24 '23 at 17:03