My professor recently assigned us a homework question but he did not cover any similar questions to it, so I do not know how to approach the problem at all.

The question is: Use the definition of congruence to prove that if $a \equiv b \pmod{8}$, then $2a \equiv 2b \pmod{16}$.

If any tips could be provided as to how to begin the problem, thanks!

  • $\begingroup$ You yourself specified that this is your Homework Question, this may lead to downvotes. BTW Salute to your honesty. $\endgroup$ Apr 14, 2017 at 18:50
  • $\begingroup$ Hint. Start with what the definition of congruence says about $a$, $b$ and $8$. I think that hint doesn't count as doing your homework for you. $\endgroup$ Apr 14, 2017 at 18:53
  • $\begingroup$ Sorry, I was not trying to receive the full solution I just wanted some hints as to how to start the problem :( $\endgroup$
    – Matt Y
    Apr 14, 2017 at 18:55
  • $\begingroup$ I would say it's encouraged to say that this is homework, the community here don't like to do someones homework but like helping people do their homework $\endgroup$
    – kingW3
    Apr 14, 2017 at 19:46
  • $\begingroup$ Tip is: definition is $a\equiv b \mod n$ means $a-b = 8k$ for some integer k. SO you need to prove if $a-b = 8k$ for some $k$, that $2a-2b = 16m$ for some m$. Which... isn't hard. $\endgroup$
    – fleablood
    Apr 14, 2017 at 20:59

4 Answers 4


$a \equiv b \pmod 8 \iff a - b = 8k$ for some integer $k$, $\iff 2a - 2b = 16k$ $\iff 2a \equiv 2b \pmod{16}$.



$a \equiv b \pmod 8$ is equivalent to $\exists k \in \Bbb Z: a-b=8k$

$2a \equiv 2b \pmod {16}$ is equivalent to $\exists k \in \Bbb Z: 2a-2b=16k$

  • $\begingroup$ That's a complete solution man and you've written ##HINT ? $\endgroup$ Apr 14, 2017 at 18:51
  • $\begingroup$ @JaideepKhare I'm just restating definitions. $\endgroup$
    – DHMO
    Apr 14, 2017 at 18:51
  • $\begingroup$ To be fair the problem is pretty simple so a hint is basically equivalent to a solution $\endgroup$
    – mrnovice
    Apr 14, 2017 at 19:12
  • $\begingroup$ That's not a complete solution. One still needs to prove $\exists k \in \mathbb Z:a-b = 8k\implies \exists k' \in \mathbb Z: 2a-2b = 16k'$. I'd call it a bit more than a hint but it isn't a solution. Frankly, I prefer to think of it as a dope slap. It says "use the definition" so.... that IS the definition, right? $\endgroup$
    – fleablood
    Apr 14, 2017 at 20:55

Tip: When something says " Use the definition of congruence" then ... use the definition of congruence.

Def: $a \equiv b \mod n \iff n|(a-b) \iff \exists k\in \mathbb Z; kn = a- b$

Can you prove $8|a-b \implies 16| 2a-2b$ or that if $\exists k; 8k = a-b \implies \exists k'; 16k' = 2a - 2b$?


The definition of congruence says that $x \equiv y \pmod k$ if $k$ divides $x-y$.

Suppose $8$ divides $a-b$. Then $a-b = 8 n$ for some integer $n$. Clearly $2(a-b) = 16 n$, and so $16$ divides $2a - 2b$. Hence, $2a \equiv 2b \pmod {16}$.


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