Question

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

Answer 1

@Noemi95 @terenzreignz

Answer 2

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.

Answer 3

@terenzreignz 's got it gofor! (:

Answer 4

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.

Answer 5

Thanks

Answer 6

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?