For every integer $a$, if $32 \nmid ((a^2 + 3)(a^2 + 7))$ then $a$ is even Statement 1: for every integer $a$, if $32 \nmid  ((a^2 + 3)(a^2 + 7))$ then $a$ is even
I just want to double check my work because I'm not sure of my answers but the question asks,
a) Rewrite the given statement in symbolic form.
$\forall a \in \Bbb Z , 32  \nmid  ((a^2 + 3)(a^2 + 7)) \implies |a|$
b) State hypothesis of Statement 1.
$\forall a\in\mathbb{Z} , 32  \nmid  ((a^2 + 3)(a^2 + 7))$
c) State conclusion of Statement 1.
$|a|$
d) State the converse of Statement 1.
$|a| \implies \forall a\in\mathbb Z , 32  \nmid  ((a^2 + 3)(a^2 + 7))$
e) State the negation of Statement 1.
$\exists a \in \mathbb Z , 32  \nmid  ((a^2 + 3)(a^2 + 7)) \implies -|a|$
f) State the contrapositive of Statement 1.
$-|a| \implies \exists a\in \mathbb Z , 32  \nmid  ((a^2 + 3)(a^2 + 7))$
g) Prove or disprove Statement 1.
I'm a bit stuck on this question.
Can someone verify my answers? I'm not too sure of it…. If I got something wrong can you explain what I did wrong?
 A: I'm assuming you are denoting the statement "$a$ is even" by $|a|$. (Also: As a commenter mentioned, a more appropriate notation would be $(2 \mid a)$.)
(a) Use parentheses for more clarity:
$$ \forall a \in \mathbb{Z}\ (32 \nmid (a^2 + 3)(a^2 + 7) \Rightarrow |a| ) $$
This is to signify that the quantifier $\forall a$ applies to the entire implication.
(b) Now, this is tricky. "$\forall a$" is actually a quantifier. You can say that if it is understood that we are only working with integers, the hypothesis is $32 \nmid (a^2 + 3)(a^2 + 7)$. If not, the hypothesis should be $(a \in \mathbb{Z}) \wedge (32 \nmid (a^2 + 3)(a^2 + 7))$.
(c) Ok.
(d) Again, two versions here. If $a$ is understood to be an integer, the converse is
$$|a| \Rightarrow (32 \nmid (a^2 + 3)(a^2 + 7)  ).$$
The full converse is
$$|a| \Rightarrow ((a \in \mathbb{Z}) \wedge (32 \nmid (a^2 + 3)(a^2 + 7))  ).$$
(e) The negation of an implication $p \to q$ is $p \wedge \neg q$. The full negation is
$$((a \in \mathbb{Z}) \wedge (32 \nmid (a^2 + 3)(a^2 + 7))  ) \wedge -|a|.$$
You can remove $a \in \mathbb{Z}$ if it is understood.
(f) This should be
$$ -|a| \Rightarrow ((a \notin \mathbb{Z}) \vee (32 \mid (a^2 + 3)(a^2 + 7))  ).$$
Again, you can remove $a \notin \mathbb{Z}$ if $a$ is understood to be an integer.
(g) The statement is true. Prove it using the contrapositive in (f). If $a$ is odd, write it as $2k + 1$ and expand $(a^2 + 3)(a^2 + 7)$. It should be clear that this is divisible by 32.
