Discrete Math: Writing Proofs 1) How do I prove the following: 
Let A = {6a + 4b ∈ Z : a, b ∈ Z} and B = {2a ∈ Z : a ∈ Z}. Show that A = B.
Thank you all for the help!
 A: You prove it by proveing that every number that can be written as $6a+4b$ where $a,b$ are integers can be written as $2m$ where $m$ is an integer and vice versa.
If $x \in A$ then $x = 6a + 4b$ for some $a,b\in \mathbb Z$.
so..... you prove that there is an $m\in \mathbb Z$ so that..... $x= 2m$.
So $x \in B$.  So $A \subset  B$.
Then if $y \in B$ then $x = 2n$ for some $n \in \mathbb Z$.
so .... you prove that there are $a,b \in Z$ so that ... $y = 6a + 4b$.
So $y \in A$.  So $B\subset A$.
So $A\subset B$ and $B \subset A$ so $A = B$.
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Hint 1:

 $6a+4b = 2(3a+4b)$.

Hint 2:

 $6n +4(-n) = 2n$.

.
A: You need to prove that any element in $A$ is also an element in $B$. Likewise that any element in $B$ is also in $A$. Once this has been done, you have shown that the sets are equal. 

First direction: show that $A \subseteq B$
Let $x \in A$, then $x = 6a + 4b$ for some $a, b \in \mathbb{Z}$.
Since $x= 2(3a + 2b)$, it is a multiple of two and in particular, $x \in B$. 

Second direction: show that $B \subseteq A$
Conversely, let $x \in B $, then $x = 2a$ for some $a \in \mathbb{Z}$. 
Then $x = 6a + 4(-a)$, so $x \in A$. (You can pick anything in the brackets)

Hence $A = B$
A: For $2)$, the contrapositive of $4\nmid n^2\implies 2\nmid n$ would be $2\mid n\implies 4\mid n^2$.
Well, if $n=2k$, then $n^2=4k^2=4K$.  QED.
A: Just for one of the inclusions, you need to prove that $6a+4b$ can be written as $2z$ where a,b,z are integers. That is, $6a+4b=2(3a+2b)$ and let $z=3a+2b$.
What about the converse?
