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Suppose for a contradiction that $\sqrt{3}$ is rational. Thus $\sqrt{3} = \frac{a}{b}$ where $a, b \in \mathbb{N}$ \ {$0$} and $gcd(a,b) = 1$.

$\sqrt{3} = \frac{a}{b} \Rightarrow 3 = \frac{a^2}{b^2} \Rightarrow a^2 = 3b^2: (1)$

By assumption $gcd(a,b) = 1$, thus $b^2\nmid a^2$ as $gcd(a^2,b^2) = 1$, thus $3 \lvert a^2$ thus $3|a$. Thus $a=3k$ for some $k \in \mathbb{N}$ / {$0$}.

Substituting $a=3k$ into (1):

$3b^2 = (3k)^2 \Rightarrow 9k^2 = 3b^2 \Rightarrow b^2 = 3k^2$.

$k^2 \nmid b^2$ as $k^2 \lvert a^2$ and $gcd(a^2,b^2) = 1$ from $gcd(a,b) = 1$ so if hypothetically $k^2 \lvert a^2$ then $gcd(a^2,b^2) \neq 1$ which is impossible by assumption that $gcd(a,b) = 1$. Thus $3 \lvert b^2$. Thus $3 \lvert b$. $3 \lvert b$ and $3 \lvert a$, but by assumption $gcd(a,b) = 1$, thus this is a contraction. Thus $\sqrt{3}$ is irrational.

Ethan Bolker
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Nostradamus
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    yes, this is a standard proof in the textbook – MathFail Jul 24 '22 at 14:42
  • You need to prove that if $3|xy$ then $3|x$ or $3|y$. (Which you can do by writing $x$ and $y$ in the form $3n+j$, $0\le j<3$.) – David C. Ullrich Jul 24 '22 at 14:43
  • Does this answer your question? Is my proof of the irrationality of $\sqrt{3}$ legitimate? The answers there show the proof you are trying to write. You do not really want "$b^2$ does not divide $a^2$" you want "$3$ divides $a$" and you don't need $a$ and $b$ relatively prime for that. – Ethan Bolker Jul 24 '22 at 14:50
  • From $a^2=3b^2$ (or $b^2=3k^2$), one could immediately conclude that $3 \mid a^2$ (or $3 \mid b^2$) without needing the assumption that $a$ and $b$ are coprime. Also, $b^2 \nmid a^2$ (or $k^2 \nmid b^2$) obviously contradicts $a^2=3b^2$ (or $b^2=3k^2$). – Geoffrey Trang Jul 24 '22 at 14:55
  • @GeoffreyTrang So its right, just some verboseness? I'm new to writing proofs, so not really sure the amount of detail required. – Nostradamus Jul 24 '22 at 14:57
  • How can you do it without the assumption that a and b are coprime? I just can't see it. @EthanBolker – Nostradamus Jul 24 '22 at 14:59
  • @Nostradamus You don't need coprimality to see that $3$ divides $a^2$ and hence divides $a$. You do need it in the next step to reach a contradiction when you show $3$ also divides $b$. – Ethan Bolker Jul 24 '22 at 15:01
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    This is one of the most common proofs, which has been discussed here many times, e.g. here. Please search for answers before posting questions. Also, solution verification quesions are only on-topic if you specify which step of the proof you have doubts about, and why. This site is not meant to be an open-ended proof checking machine. – Bill Dubuque Jul 24 '22 at 15:16
  • @BillDubuque What if there are multiple aspects of my proof that I am unsure about? I just want to know if my proof is correct or not as a whole. – Nostradamus Jul 24 '22 at 15:56
  • "By assumption gcd(a,b)=1, thus b2∤a2 as gcd(a2,b2)=1, thus 3|a2 " Leave all this out. As $a^2 = 3b^2$ (and $b^2$ is an integer) then $3\mid a^2$. It has nothing to do with $\gcd(a,b)=1$ or $b^2\not\mid a^2$ or anything. Just say "$a^2=3b^2$ thus $3|a^2$". Then when you say $3|a$ explain why. Then after you show $b^2=3k^2$, leave out all this "k2∤b2 as k2|a2 and gcd(a2,b2)=1 from gcd(a,b)=1 so if hypothetically k2|a2 then gcd(a2,b2)≠1 which is impossible by assumption that gcd(a,b)=1" which has nothing to do with anything and just say "$b^2=3k^2$ and thus $3|b^2$. – fleablood Jul 24 '22 at 16:27
  • Your proof is valid but be aware it assumes that all rational numbers can be expressed in "lowest terms", and it assumes Euclids lemma (If $p$ is prime and $p|ab$ then $p|a$ or $p|b$). Make sure you are allowed to assume these. – fleablood Jul 24 '22 at 16:30
  • @fleablood Thank you so much for you reply. But I'm just wondering why 3 | a^2 has nothing to do with gcd(a,b)=1 or b2∤a2 or anything? – Nostradamus Jul 24 '22 at 18:24
  • If you have $a^2 = 3b^2$ and $b^2$ is an integer then that means $3|a^2$. That's the definition of $3|a^2$. That's all. It doesn't matter that $\gcd(a,b)$.....Also, if $a^2 = 3b^2$ then $b^2$ DOES divide $a^2$ because $3$ is an integer and $a^2 =3\times (b^2)$ – fleablood Jul 24 '22 at 18:47
  • BTW why do you think $\gcd(a,b)\implies \gcd(a^2, b^2)$? If you DO know that it makes your entire proof trivial..... If $\frac {a^2}{b^2} = 3$ then BY DEFINITION we have $b^2|a^2$ so $b^2$ is a factor of $a^2$. But as $\gcd(a^2,b^2)=1$ we must have $b^2 = 1$. So $b= 1$. So we have $a^2 =3$ and $3$ is a perfect square.... which it isn't... this is actually why after we prove fundiment principal of arithmetic (that every integer has a unique factorization) we usually just prove "The only rational square roots are integers and only perfect squares have rational square roots" in one swoop. – fleablood Jul 24 '22 at 18:49

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