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Here's my proof:

Let us suppose for sake of contradiction, $\sqrt2$ is rational. Then, we can write $$\sqrt2=\frac{a}{b}$$ for non-zero coprime integers $a$ and $b$. Squaring our equation, $$2=\frac{a^2}{b^2}$$ Here, LHS is an integer, but since $a^2$ and $b^2$ are coprime, RHS is not an integer. This is a contradiction, and so our original assumption was wrong. Therefore, $\sqrt2$ is irrational.

Is this valid? Thanks in advance.

ultralegend5385
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    Of course it is valid. But it is arguably incomplete: you should explain how $a$ and $b$ being coprime implies $a^2/b^2$ is not an integer. – balddraz Feb 14 '21 at 07:09
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    It is possible for the quotient of two coprime integers to be an integer—for example, $\frac21=2$. So something needs to be justified here. (Otherwise the same proof would show that $\sqrt4$ was irrational...!) – Greg Martin Feb 14 '21 at 07:10
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    Why are $a^2$ and $b^2$ coprime? – fleablood Feb 14 '21 at 07:31
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    It's not valid nor complete. You are implying that $a$ and $b$ being coprime will imply that $a^2$ and $b^2$ are coprime. That's not a given and it is not trivial (although it is true). – fleablood Feb 14 '21 at 07:34
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    "It is possible for the quotient of two coprime integers to be an integer" True. But it's easily argued than if $\frac ab\in \mathbb Z$ then $b|a$ so if $a,b$ are coprime and the only factors they have in common is $1$ but then have $b$ in common then $b = 1$. But what isn't explained is that if $a,b$ are coprime, and $c$ and $d$ are coprime it does not follow that $ac$ and $bd$ are coprime. [In fact that isn't true.] (Also, do you know every integer has a unique prime factorization?) – fleablood Feb 14 '21 at 07:39
  • @fleablood: I had that from here. – ultralegend5385 Feb 14 '21 at 07:41
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    Well, then your proof is valid and complete. But I do think that must be included in your proof. – fleablood Feb 14 '21 at 07:45
  • Well, then that poses a problem @fleablood. – ultralegend5385 Feb 14 '21 at 10:06
  • As @GregMartin points out, this proof doesn't distinguish between the cases when the argument under the radical is a perfect square and when it is not. So, there is no flaw in using the same proof for $\sqrt4$. – ultralegend5385 Feb 14 '21 at 10:07
  • "Well, then that poses a problem" Why? Just include a statement that if $a,b$ are relatively prime and $b \ne 1$ then explain why that would imply $a^2, b^2$ would also be relatively prime. (Although I'd completely avoid the methods done in the others). A lot depends an what basics of prime factorizations you are allowed to assume. – fleablood Feb 14 '21 at 19:03

2 Answers2

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  • You need to point out that $\sqrt 2 =\frac ab$ where $a,b$ are relatively prime that $b \ne 1$.

That's easy enough. You just have to show there is no integer so that $n^2 = 2$ and there obviously aren't as $|n| \ge 2 \implies n^2 \ge 4$ and $|n| < 2\implies |n| \le 1\implies n^2 \le 1$ so there are no integer square roots of $2$ or $3$.

  • You need to point out that $a,b$ are relatively prime (and $b\ne 1$) that would imply $a^2$ and $b^2$ are relatively prime (and $b^2 \ne 1$).

If we have the fundamental theorem of arithmetice (all integers have unique prime factorizations) and/or euclid's lemma (if $p$ is prime and $p|nm$ then $p|n$ or $p|m$) that's short work.

$b\ne 1$ so $b^2 \ne 1$ and $b^2$ must have a prime factor and that prime factor, call it $p$ must, by Euclid's, lemma divide $b$. By transitivity that prime factor must divide $a^2$ and must therefore divide $a$. But that contradicts $a,b$ are relatively prime.

  • Then your proof is complete. As $a^2,b^2$ are relatively prime and $b^2 \ne 1$ then $b^2$ is not itself a factor of $a^2$ and $\frac {a^2}{b^2}$ can not be the integer $2$.

If you don't have the FTA or EL then you have your work cut out for you. Better to just do the classic.

(That is proving that if $m$ is even/odd then $m^2$ is even/odd and therefore $2 =\frac {a^2}{b^2}$ must imply $a,b$ are both even and thus not relatively prime.)

(This assumes that all rationals can be written as a pair of relatively prime integers. Although that is not a basic definition and should be proven, I'll allow that one to go assumed.)

fleablood
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  • Thanks! Now it is clear why the proof is valid only for square roots of primes! The key point was the implication $$\gcd(a,b)=1\implies\gcd(a^2,b^2)=1$$ requires something, but I took it to be trivial. – ultralegend5385 Feb 14 '21 at 23:40
  • It's valid for all integers that are not perfect squares. But for perfect square $\sqrt{k^2} = \frac ab; a = k; b = 1$. If $n$ is not a perfect square then $\sqrt{n} =\frac ab$ can't have $b=0$. Then $n = \frac {a^2}{b^2}$. If $a,b$ are rel prime then then $a^2,b^2$ are relatively prime (BUT WE HAVE TO PROVE THAT) and so $\frac {a^2}{b^2}$ can't be an integer. So $\sqrt n$ is not rathional if $n$ is not a perfect square. – fleablood Feb 14 '21 at 23:54
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Suppose that $\sqrt{2}$ is rational number then exists $p$, $q$ coprime natural numbers such that $\sqrt{2} = \dfrac{p}{q}$. Implies $2q^2 = p^2$, so $p^2$ is divisible by 2. So on $p\vdots 2$, assume that $p=2p_1$ then $q^2=2p_1^2$, implies $q\vdots 2$, So $\gcd(p, q)\vdots 2$ is contradict $p$, $q$ are coprime.

Bazar Tumurkhan
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  • Sorry sir, but I (and probably other people) already know this. I am willing to check if the proof I offered is correct or not. – ultralegend5385 Feb 14 '21 at 23:41
  • I mean, your proof is incomplete. You explain why $\dfrac{a^2}{b^2}$ is not integer. – Bazar Tumurkhan Feb 15 '21 at 01:57
  • Every textbook seems to use this $\frac{p}{q} $ idea where $p$ and $q$ are in their irreducible form and then testing $p$ and $q$ for a common factor. Yet people looking at the problem for the first time would surely initially be more likely thinking of the basic operations of the integers, $p$ and $q$, e.g. addition, subtraction, multiplication and division, and seeing the implications of these for the initial assertion. Can anyone suggest why the $\frac{p}{q}$ proof is more "popular" ? – Trunk Jun 13 '22 at 20:27
  • Is it because professional mathematicians find disproofs are generally easier in the stricter constraint, i.e. the irreducibility of the ratio of the integers, rather than in the looser constraint, i.e. that $p$ and $q$ are simply integers ? If so, I would proffer this problem to be an exception to that rule of thumb ! https://math.stackexchange.com/questions/3758045/prove-that-13-sqrt2-is-irrational/4471688#comment9372974_4471688 – Trunk Jun 13 '22 at 20:35