Prove that √p, where p is a prime number, is not a rational number. In other words, there are no whole numbers a and b such that √p = a/b and a/b is reduced.

Question

tkhunny · Answer

Square both sides of $\sqrt{p} = \dfrac{a}{b}$

tkhunny · Answer

Keep in mind that "a" and "b" have NO common factor.

michele_laino · Answer

if sqrt(p) is a rational number, namely, exist two integer, such that sqrt(p)=m/n, as you wrote,, then, we can write:
$$p=\frac{ a ^{2} }{ b ^{2} }$$
so a^2 is divisible by a, also a is divisible by a: namely a=p*k. Inserting the last relationship in our fraction, we have:
$$p=\frac{ p ^{2} k ^{2} }{ b ^{2} }$$
from which also b is divisible by k, then a and b are not relatively prime numbers, as required from our initial hypothesis

michele_laino · Answer

sorry before I meant a^2 is divisible by p, also a is divisible by p...

michele_laino · Answer

from last relationship, we have:
$$b ^{2}=k ^{2}*p$$
so b^2 is divisible by p, and also b is divisible by p.

michele_laino · Answer

more precisely, a^2 is divisible by p, since a is divisible by p, whereas we got an absurd. Similarly for b^2, which is divisible by p, sinc b is divisible by p, otherwise we got an absurd