# Square root of 6 proof rationality

I was proving $$\sqrt 6 \notin \Bbb Q$$, by assuming its negation and stating that: $$\exists (p,q) \in \Bbb Z \times \Bbb Z^*/ \gcd(p,q) = 1$$, and $$\sqrt 6 = (p/q)$$.

$$\implies p^2 = 2 \times 3q^2 \implies \exists k \in \Bbb Z; p = 2k \implies 2k^2 = 3q^2$$ and found two possible cases, either $$q$$ is even or odd, if even we get contradiction that $$\gcd(p, q) \neq 1$$, if odd we get contradiction that $$2k^2 = 3q^2$$.

Is it a right path for reasoning it?

• Is the $k$ a typo? What is $p, k$ and $q$? Why is $q$ being even a condradiction with $\gcd(p,q) = 1$? And why is $2k^2 = 3q^2$ a contradiction? You just said $2k^2 = 3q^2$ so that isn't a contradiction? You are on the right track but you haven't explained any of your arguments. – fleablood Nov 15 '18 at 2:13
• Why is $q$ even a contradiction? What if $k$ or $p$ (whats the difference) is odd? – fleablood Nov 15 '18 at 2:20
• I rearranged it! – Papa Nov 15 '18 at 3:51

How about a proof by descent?

First show that $$2^2<6<3^2$$. Then if $$\sqrt{6}$$ is to be rational it must have a form $$p/q$$ where $$p,q\in \mathbb{Z}, q>0, 2q. By simple algebra the square root is also equal to $$6q/p$$, thus

$$p/q=6q/p\text{.....Eq. 1}$$.

Now if $$a/b=c/d$$ then also

$$a/b=(ma+nc)/(mb+nd)$$

for any coefficients $$m,n$$ where the denominator is nonzero. In particular, Eq. 1 implies

$$p/q=(3p-6q)/(3q-p)\text{.....Eq. 2}$$

where we already have $$2q and thus $$0<3q-p. So the proposed rational fraction $$p/q$$ must be equal to an alternative rational fraction with a smaller positive denominator. This causes an infinite descent contradiction forcing the assumption of a rational value to be false.

We can form a similar proof for the square root of any natural number that is not a squared integer. "Not a squared integer" is needed because the square root must be strictly between two adjacent integers to obtain a descent of positive denominators.

Assume $$\sqrt{6} = {a \over b}$$ with $$a, b \in \mathbb{Z}$$ . This implies

$$2 \cdot 3 b^2 = a^2$$

Use the fundamental theorem of algebra to decompose both sides into a unique prime factorization. There are an even number of factors of $$2$$ on the rhs and an odd number of factors of $$2$$ on the lhs... a contradition. Same for factors of $$3$$.

Thus $$\sqrt{6} \neq {a \over b}$$, i.e., is irrational.

• Yeah, I tried to reach it without prime factorisation. A second look? – Papa Nov 15 '18 at 2:42
• You are considering the value of the rhs and lhs of the equation. I am instead looking at the number of prime factors. Frankly, I think that is so much more elegant... but I suppose this is a matter of taste. Mine applies to $\sqrt{5/3}$ whereas yours doesn't... at least not as directly. – David G. Stork Nov 15 '18 at 2:47
• Our professor used the value way, so I am following his steps for this, though, thanks! – Papa Nov 15 '18 at 2:50
• Show him/her the better way! And stand out as a student! – David G. Stork Nov 15 '18 at 2:51
• Well, that doesn t apply in Morocco. Pray for me I win visa lottery so I can join a better faculty in the US :) – Papa Nov 15 '18 at 3:49

I'll play more generally and see what happens.

Suppose $$\sqrt{m} =\dfrac{u}{v}$$ where $$m = \prod_{p \in P} p^{m_i}, u = \prod_{p \in P} p^{u_i}, v = \prod_{p \in P} p^{v_i}$$.

Then $$\prod_{p \in P} p^{m_i} =\dfrac{u^2}{v^2} =\dfrac{ \prod_{p \in P} p^{2u_i}}{\prod_{p \in P} p^{2v_i}}$$ so $$\prod_{p \in P} p^{m_i}\prod_{p \in P} p^{2v_i} =\prod_{p \in P} p^{2u_i}$$ or $$\prod_{p \in P} p^{m_i+2v_i} =\prod_{p \in P} p^{2u_i}$$.

By unique factorization, $$m_i+2v_i =2u_i$$, so $$m_i =2u_i-2v_i =2(u_i-v_i)$$.

Therefore $$m =\prod_{p \in P} p^{m_i} =\prod_{p \in P} p^{2(u_i-v_i)} =\left(\prod_{p \in P} p^{u_i-v_i}\right)^2$$ so $$m$$ is a perfect square.

Therefore the square root of an integer is rational only if it is a square.

I'll now try to generalize this to $$k$$-th roots, with as much cut-and-paste as possible.

Suppose $$\sqrt[k]{m} =\dfrac{u}{v}$$ where $$m = \prod_{p \in P} p^{m_i}, u = \prod_{p \in P} p^{u_i}, v = \prod_{p \in P} p^{v_i}$$.

Then $$\prod_{p \in P} p^{m_i} =\dfrac{u^k}{v^k} =\dfrac{ \prod_{p \in P} p^{ku_i}}{\prod_{p \in P} p^{kv_i}}$$ so $$\prod_{p \in P} p^{m_i}\prod_{p \in P} p^{kv_i} =\prod_{p \in P} p^{ku_i}$$ or $$\prod_{p \in P} p^{m_i+kv_i} =\prod_{p \in P} p^{ku_i}$$.

By unique factorization, $$m_i+kv_i =ku_i$$, so $$m_i =ku_i-kv_i =k(u_i-v_i)$$.

Therefore $$m =\prod_{p \in P} p^{m_i} =\prod_{p \in P} p^{k(u_i-v_i)} =\left(\prod_{p \in P} p^{u_i-v_i}\right)^k$$ so $$m$$ is a perfect $$k$$-th power.

Therefore the $$k$$-th root of an integer is rational only if it is a $$k$$-th power.

When you get $$2p^2 = 3q^2$$ that means that $$2|q^2$$ so $$2|q$$ and that $$3|p^2$$ so $$3|p$$.

Then if you replace $$p = 3p'$$ for some integer $$p'$$ and $$q=2q'$$ for some integer $$q'$$ you get

$$2(3p')^2 = 3(2q')^2$$

$$18p'^2 = 12q'^2$$

$$3p'^3 = 2q'^2$$.

Can you finish from there?

This assumes you know Euclid's lemma that 1) prime numbers exist and 2) if $$p$$ is prime and $$p|a*b$$ then either $$p|a$$ or $$p|b$$ (or both).

• I tried to reach it without prime factorisation, can you have a second look? – Papa Nov 15 '18 at 2:42