Square root of 6 proof rationality

Question

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?

Answer 1

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.

Answer 2

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).

Answer 3

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.

Answer 4

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 $p,q\in \mathbb{Z}, q>0, 2q<p<3q$. 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 $2q<p<3q$ and thus $0<3q-p<q$. 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.