Verify by the method of contradiction.
p: root7
is irrational

Question

goformit100 · Answer

@Noemi95 @terenzreignz

terenzreignz · Answer

Yikes.... I think we need a new statement for this, a lemma...
Let a be an integer...
then if $a^2$  is divisible by 7, then $a$ is also divisible by 7.

anonymous · Answer

@terenzreignz 's got it gofor! (:

terenzreignz · Answer

We have to prove that, first...
So, let a be an integer that is NOT divisible by 7 (proof by contrapositive)
Then
$$\large a = 7q + r \quad\quad\quad q,r\in\mathbb{Z}, \quad r<7$$
then
$$\large a^2=(7q+r)^2=49q^2+14qr+r^2=7(\color{blue}{7q^2+2qr})+r^2$$which is also not divisible by 7 since $r^2 $  is not divisible by 7.

goformit100 · Answer

Thanks

terenzreignz · Answer

thus, if $a^2$  is divisible by 7, then a is also divisible by 7.
That's a nice tool for this question... should we move to your next?