# Prove $\sqrt6$ is irrational

Suppose $\sqrt6 = \frac pq$ where $p$ and $q$ have no common factors.

$$6 = \frac {p^2}{q^2}$$

$$6p^2 = q^2$$

So $q^2$ and therefore $q$ is divisible by $6$.

$$p^2 = \frac {q^2}{6}$$

So $p^2$ and therefore $p$ is divisible by $6$.

S0, $p$ and $q$ have a common factor $6$. Contradiction.

Therefore $\sqrt6$ is irrational.

Does the proof work?

• Yes, it's correct. – Relure Jun 2 '14 at 17:54
• I would say that it is incomplete. Why is it that if $6$ divides $p^2$ then $6$ divides $p$? – David Steinberg Jun 2 '14 at 17:56
• You need to justify the inference $\,6\mid q^2\,\Rightarrow\,6\mid q.\$ That is the key step in the proof. Also you need to be more prcise about how you deduced that $\,6\mid p^2,\,$ since it is not clear form what was written that this was done correctly. – Bill Dubuque Jun 2 '14 at 17:57
• David and Bill aren't just being formal: if $9$ divides $p^2$, it may not be the case that $9$ divides $p$, for example. – Strants Jun 2 '14 at 18:00
• Note also that you could use $2$ instead of $6$, and then the reasoning follows from well-known parity arithmetic, just as in the classical irrationality proof of $\sqrt 2\ \$ – Bill Dubuque Jun 2 '14 at 18:01

As others have pointed out, it is important to justify that $6 \mid q^2 \Rightarrow 6\mid q$. And it isn't totally clear from your proof.

I suggest the following.

First show that $\sqrt{6}$ is not an integer. It's not difficult to do that. Since $4<6<9$, it follows that $2<\sqrt{6}<3$ and that means that $\sqrt{6}$ is not an integer.

Now assume that $\sqrt{6}$ is a rational number, $\frac{p}{q}$ where $p$ and $q$ are co-prime positive integers and $q>1$.

Now you can write

$$6=\frac{p^2}{q^2}$$

$$\Rightarrow 6q=\frac{p^2}{q}$$

It is clear that the left hand side is an integer. But the right hand side isn't since $p^2$ and $q$ share no common factors.

So this equality can not hold. And $\sqrt{6}$ can not equal $\frac{p}{q}$.

So it has to be an irrational number.

There's an incredibly short proof of this if you know the rational root theorem. Just notice that $\sqrt{6}$ is a root of the monic polynomial $x^2-6$. The proof is almost immediate.

EDIT: Here's a messy justification of why $q$ does not divide $p^2$. Let $p=\prod {p_i}^{x_i}$ and $q=\prod {p_j}^{y_j}$ such that $x_i$ and $y_j$ are positive integers. This notation is incredibly informal but it gets the message across.

Now since $p$ and $q$ are co-prime, $p_i\neq p_j$ for any $i$ & $j$. Now $p^2=\prod {p_i}^{2{x_i}}$. Notice that $p^2$ has the same prime divisors as $p$. Since $p$ and $q$ share no common prime factors, it follows that $p^2$ and $q$ share no common prime factors.

That means $q\nmid p^2$.

• Alas, this has a bigger gap than the OP's proof. You need to justify the claim: $\,p,q\,$ coprime $\,\Rightarrow\,q\nmid p^2.\ \$ – Bill Dubuque Jun 3 '14 at 20:09
• @BillDubuque, I thought that was fairly obvious. I added an edit to the answer. Personally I'm not a fan of the edit since it's really messy. Any thing else that needs justifying? – mursalin Jun 4 '14 at 13:51
• Well, since you ask So this inequality can not hold. - it's not an inequality. But a very neat proof. Valid for any n which is not a perfect square, and a short cut to "the square root of an integer is either an integer or irrational". – Tom Collinge Jun 4 '14 at 13:58
• @TomCollinge, thanks for pointing that out! It was a typo. – mursalin Jun 4 '14 at 14:02
• @Tom This is a standard proof, i.e. if $\, n = (a/b)^2,\ (a,b)= 1\,$ then by Euclid's Lemma (or Bezout, or unique factorization) also $(a^2,b)=1,\,$ so $\,b\mid a^2\Rightarrow\,b=1,\,$ so $\, n = a^2\,$ is a perfect square. Any such proof needs to explicitly justify why the inference is true in $\Bbb Z$ because it need not be true in other number rings lacking said properties, e.g. quadratic number rings, e.g. $\ 3 = (\sqrt{12}/2)^2$ is the square of a proper fraction over $\,\Bbb Z[\sqrt 12] = \{j + k\sqrt 12\ :\ j,k\in \Bbb Z\}.\ \$ – Bill Dubuque Jun 4 '14 at 14:14

Using the prime factorization theorem (every positive integer has a unique prime factorization), we can write, for any rational number $r$: $$r = {p\over q} = {{2^{a_1}\cdot 3^{a_2}\cdot 5^{a_3} ...} \over {2^{b_1}\cdot 3^{b_2}\cdot 5^{b_3} ...}} = {2^{c_1} \cdot3^{c_2}\cdot 5^{c_3} ...}$$ where $a_i$ and $b_i$ are non-negative integers, and $c_i = a_i - b_i$ are integers (possibly negative). Since the factorizations of $p$ and $q$ are unique, the factorization of $r$ must also be unique.

Thus: $$r^2 = 2^{2 c_1}\cdot 3^{2 c_2} \cdot5^{2 c_3}...$$

If $r^2 = 6 = 2^1\cdot3^1$, then we have $2c_1 = 1$ and $2c_2 = 1$. This is impossible.

This technique generalizes to any number that has any prime factor an odd number of times. It also generalizes to other roots. For example: $^3\sqrt 4$.

If you don't mind using some stronger weapons, then Eisenstein's criterion can help.

Choose $p=2$ and take the polynomial $1x^2+0x-6$. Since $$2\not\mid1 ,\quad 2\mid0 ,\quad 2\mid -6 ,\quad 2^2\not\mid-6 ,$$ we have that the polynomial is irreducible. It has $\sqrt6$ as a root. Therefore $\sqrt6$ is quadratic algebraic and not linear (=rational).

• The rational root test is much simpler. And your polynomials doesn't have $\sqrt{6}$ as zero either. – vonbrand Jun 5 '14 at 0:58
• sorry, corrected. And well, I say that I use stronger weapons, which can be used in a large framework, I know that simpler solutions exist. – yo' Jun 5 '14 at 9:02