Vrftjkry

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?

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 $p,qin mathbb{Z}, q>0, 2q<p<3q$. By simple algebra the square root is also equal to $6q/p$, thus

$p/q=6q/ptext{.....Eq. 1}$.

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.

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.

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).