# Follow-up Question: Proof of Irrationality of $\sqrt{3}$

As a follow-up to this question, I noticed that the proof used the fact that $p$ and $q$ were "even". Clearly, when replacing factors of $2$ with factors of $3$ everything does not simply come down to being "even" or "odd", so how could I go about proving that $\sqrt{3}$ is irrational?

• The square of a multiple of $3$ is a multiple of $3$. The square of something that is either $1$ more or $1$ less than a multiple of $3$ is $1$ more than a multiple of $3$. ${}\qquad{}$ Aug 10, 2015 at 0:15
• Aug 10, 2015 at 0:16

It's very simple, actually. Assume that $\sqrt{3}$ = $\frac{p}{q}$, with $p,q$ coprime integers.

Then, $p = \sqrt{3}q$ and $p^2 = 3q^2$. If $3\mid p^2$, then $3\mid p$. So actually, $9\mid p^2$. Then, by similar logic, $3\mid q^2$, meaning $3\mid q$. Since $3$ divides both $p$ and $q$, the two numbers are not coprime. This is a contradiction, since we assumed that they $\textbf{were}$ coprime. Therefore, $\sqrt{3}$ cannot be written as a ratio of coprime integers and must be irrational.

$\textbf{NOTE:}$ The word "even" in the original proof was just a substitution for "divisible by $2$". This same idea of divisibility was used in this proof to show that $p$ and $q$ were divisible by $3$. It really is the same idea. There just isn't a nice concise word like "even" that was used to describe a multiple of $3$ in this proof.

• But it seems like this argument only works for prime numbers, really. Like if we had $\sqrt 4$ we can't say $4|p^2\implies 4|p$. so do we just prove that the square root of every prime is irrational and that the square root of any other integer is the product of roots of primes? Aug 10, 2015 at 0:20
• @ElliotG The proof can be modified to show that $\sqrt{n}$ is irrational unless $n=k^2$ is a square. If $n$ is not a perfect square then it has a prime factor $r^m$ with $m$ odd so if $\sqrt{n} = \frac{p}{q}$ with $(p,q)=1$ then $q^2n = p^2$. The right hand side must be divisible by an even power of $r$ while the left hand side is divisible by an odd power of $r$ contradiction. Aug 10, 2015 at 0:30
• This argument works for all numbers that are not perfect squares. For example, $6$, which definitely is not prime could still work. That is why the proof works: it shows irrationality for the square roots of all numbers that are not perfect squares themselves.
– user134593
Aug 10, 2015 at 0:31
• wow that argument is almost simpler than the specific case $n=2$. also not sure why i momentarily thought $\sqrt 4$ was irrational. thanks. Aug 10, 2015 at 0:39
• How about 'threeven'? =P Aug 10, 2015 at 0:57

Alternatively, a contradiction can be derived as follows: \begin{align} \sqrt3 &=\frac ab \\ a^2&=3b^2 \\ a^2+b^2 &= 4b^2=(2b)^2 \\ \end{align}The contradiction is due to the fact that the integer length of the hypothenuse of a primitive right triangle is odd.

• You need to include no co-primes - as the triangle $[6,8,10]$ has an even hypotenuse. But nice anyway... +1 Aug 10, 2015 at 0:46
• @Downvoter: What's wrong? Aug 13, 2015 at 10:19
• I voted up - but I noticed that last week a lot of down votes are given without any reason. Yesterday I had 5 down votes on several posts, without posting any reason. Aug 13, 2015 at 11:54
• @johannesvalks I see, that's very unpleasant. Do you happen to know if the moderators are aware of this? Aug 13, 2015 at 12:55
• @johannes: If I'm not mistaken, there should be a chat room precisely for this. Aug 14, 2015 at 12:07

Here is a proof that if $n$ is a positive integer that is not a square of an integer, then $\sqrt{n}$ is irrational. This proof does not use any divisibility properties.

Let $k$ be such that $k^2 < n < (k+1)^2$. Suppose $\sqrt{n}$ is rational. Then there is a smallest positive integer $q$ such that $\sqrt{n} = p/q$.

Then $\sqrt{n} = \sqrt{n}\frac{\sqrt{n}-k}{\sqrt{n}-k} = \frac{n-k\sqrt{n}}{\sqrt{n}-k} = \frac{n-kp/q}{p/q-k} = \frac{nq-kp}{p-kq}$.

Since $k < \sqrt{n} < k+1$, $k < p/q < k+1$, or $kq < p < (k+1)q$, so $0 < p-kq < q$. We have thus found a representation of $\sqrt{n}$ with a smaller denominator, which contradicts the specification of $q$.

Note: This is certainly not original - but I had fun working it out based on the proof I know that $\sqrt{2}$ is irrational.

• But $k^2 < n$ does not imply $k < n$. Aug 10, 2015 at 18:24
• @johannesvalks: I don't say that $k < n$; I say $k < \sqrt{n}$. Aug 16, 2016 at 5:05
• It is misleading to say the proof does not use divisibility. It essentially uses the division algorithm to achieve descent on denominators. I explain this further in this May 20, 2009 sci.math post, where I highlight the beautiful view of irrationality proofs in terms of Dedekind's conductor ideal. Jan 26, 2017 at 21:18

Here's another way: suppose $\sqrt3 = \frac pq$ for some coprime $p,q$, so $$3q^2=p^2.$$ Now reduce this modulo $4$, noting that the quadratic residues modulo $4$ are $0$ and $1$. The only solution is $p^2 \equiv q^2 \equiv 0 \pmod 4$, but then $p$ and $q$ are both even, a contradiction.

Assume that $\sqrt{3}$ is rational . Then this means that $\sqrt{3} = \frac{p}{q}$ where $p$ and $q \neq 0$ are integers in their lowest terms. That is to say that $\gcd(p,q) =1$

Now square both sides to get $3 = \frac{p^2}{q^2} \implies 3q^2 = p^2$

Now clearly, $3 \mid p^2$ but since $3$ is prime then $3 \mid p$.

This is because of euclid's lemma that states that if $p$ is prime and $p$ divides $ab$ then $p$ divides $a$ or $p$ divides $b$. And so here we have $3 \mid pp = p^2$ and so $3 \mid p$

Now $3k = p$ for some integer $k$ right and so $3q^2 = (3k)^2$ and so $3q^2 = 3(3k^2)$ and so $q^2 = 3k^2$ and so $3 \mid q^2$, however $3$ is prime so $3 \mid q$ and so now we have $3 \mid q$ and $3 \mid p$ which contradicts the fact that $\gcd(p,q)=1$ and so our assumption that $\sqrt{3}$ is rational is false and hence it must be irrational

The general case : If $a,b,c,d$ are positive integers with $GCD(a.b)=1=GCD(c,d)$ then $(a/b)^{c/d}$ is irrational unless $a$ and $b$ are $d$-th powers of positive integers.PROOF: Suppose $(a/b)^{c/d}=e/f$ where $f,g$ are positive integers and $GCD(f,g)=1$. Then $a^cf^d=b^ce^d$. We now apply a result (which one book called the "fundamental theorem of arithmetic"), that if $x,y,z$ are non-zero integers where $x$ divides $yz$ and $GCD(x.y)=1$ then $x$ divides $z$. Firstly, this implies that $GCD(a^c,b^c)=1=GCD(f^d,e^d)$. Secondly, applying the theorem with $x=a^c,y=b^c,z=e^d$, we have: $a^c$ divides $e^d$,but applying it with $x=e^d,y=f^d,z=a^c$, we have: $e^d$ divides $a^c$. So $a^c=e^d$.Now since $a^c.f^d=b^c.e^d$ and $a^c=e^d$, we also have $f^d=b^c$.Finally another consequence of the above theorem is that if $x$ is both a $c$-th power and a $d$-th power with $.GCD(c.d)=1$, then $x$ is a $cd$-th power. Applying this with $x=a^c$,and also with $x=b^c$, we know there are positive integers $u,v$ where $a^c=u^{cd}$ and $.b^c=v^{cd}$. So $a=u^d$ and $b=v^d$. QED.