Q.a) Write the contrapositive of the given statement.
"If a number is divisible by 9, then it is divisible by 3".

The contrapositive of "if $p$ then $q$" is "if not $q$ then not $p$". For the irrationality proof, assume the opposite (that $\sqrt7$ is rational) and derive a contradiction using divisibility by the prime 7.

a) Contrapositive

Statement: "If a number is divisible by 9, then it is divisible by 3."

This has the form "if $p$ then $q$", where $p$: divisible by 9, $q$: divisible by 3. The contrapositive is "if not $q$ then not $p$":

"If a number is not divisible by 3, then it is not divisible by 9."

b) Proof by contradiction that $\sqrt7$ is irrational

Assume, for contradiction, that $\sqrt7$ is rational. Then it can be written as

$\sqrt7 = \dfrac{p}{q}$

where $p,q$ are integers with no common factor (in lowest terms), $q\neq0$.

Squaring: $7 = \dfrac{p^2}{q^2} \Rightarrow p^2 = 7q^2$

So $7$ divides $p^2$. Since $7$ is prime, $7$ must divide $p$ itself. Write $p = 7k$ for some integer $k$.

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