Let $A =${$ m \in \mathbb{Z} | m = 4a + 3 $ for $a \in \mathbb{Z}$ } and let $B =$ {$n\in \mathbb{Z} | n = 12b + 7$ for $b \in \mathbb{Z}$}. Prove that $B \subset A$.

this is what i have gathered so far, i think these are meaningful to the proof but i need some help bringing it all together.

so $A$ contains all integers where $m \equiv 3(mod 4)$

and $B$ contains all integers where $n \equiv 7(mod 12)$

if something is divisible by 12 then it is also divisible by 4 because $12 = 4\cdot 3$

i know that $4(a) +3$ is equal to 7


1 Answer 1


This follows directly from the definitions of $A$ and $B$ and some simple manipulation.


Consider $b \in B$.

$b = 12x + 7$ for some $x \in \Bbb Z$.

$b = 12x + 7 = 4(3x + 1) + 3$.

Hence, $b \in A$.

Since $b$ was arbitrary, $B \subseteq A$

  • $\begingroup$ that was so much simpler and more obvious than i expected, i need to do more of these... thanks $\endgroup$
    – kr1s
    Nov 9, 2017 at 2:13

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