Respuesta :
The square root of two is irrational. It means that you can't write it as a fraction:
[tex] \sqrt{2} \neq \cfrac{p}{q}\quad\forall p \in \mathbb{Z},\ q \in \mathbb{Z}\setminus\{0\} [/tex]
Here's a proof. Assume that you could write
[tex] \sqrt{2} = \cfrac{p}{q} [/tex]
for some integers p and q, and that p and q have no common divisors. If you square both sides you have
[tex] 2 = \cfrac{p^2}{q^2} \implies p^2 = 2q^2 [/tex]
So, [tex] p^2 [/tex] is even, which means that also p is even, and thus there exists some number k such that [tex] p = 2k [/tex]. The expression becomes
[tex] p^2 = 2q^2 \implies 4k^2 = 2q^2 \iff q^2 = 2k^2 [/tex]
So, also [tex] q^2 [/tex] is even, and thus q is also even.
But then, p and q are both even, whereas we assumed that they had no common divisors. Contraddiction.
As for the value, any calculator will give you an approximation that starts with
[tex] \sqrt{2} = 1.4142\ldots [/tex]
so, it is best approximated by 1.4.
That answer is amazing! Please give it Brainliest it or make it a verified answer
