Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I’ve been taught that if $a \equiv b$ and $c \equiv d \pmod {m}$, then $a +c \equiv b+d \pmod{m}$ and $ac \equiv bd \pmod {m}$. But I would like to know how one can prove it. Can you give me a hint? Thank you.

share|cite|improve this question
See this answer for proofs of the basic rules of congruence arithmetic. – Bill Dubuque Aug 13 '14 at 19:26

2 Answers 2

up vote 8 down vote accepted

If $a \equiv b$ and $c \equiv d \pmod {m}$, then there exist integers $k$ and $h$ such that $a = b + km$ and $c = d + hm$.

Whence, $$a+c = b+d + (k+h)m;$$ thus, $a +c \equiv b+d \pmod {m}$.

In addition,
$$\begin{align} ac &= bd+dkm +bhm + khm^2 \\ & = bd + (dk + bh + khm)m;\end{align}$$ thus, $ac \equiv bd \pmod {m}$.

share|cite|improve this answer
Thanks a lot. (Also, thank you for using the line to separate the hint from the complete solution. I figured the whole proof out when I looked at the hint and then checked it :)) – user169569 Aug 13 '14 at 19:10
You are welcome. – user161303 Aug 13 '14 at 19:12
  • $$ \\ a \equiv b \pmod m, c \equiv d \pmod m \Rightarrow m \mid a − b , m \mid c − d \Rightarrow m \mid (a-b)+(c-d) \\ \Rightarrow m \mid (a+c)-(b+d) \Rightarrow a+c \equiv (b+d) \pmod m $$

  • $$\\ a \equiv b \pmod m, c \equiv d \pmod m \Rightarrow m \mid a − b, m \mid c − d \\ \Rightarrow m \mid (a − b)c + (c − d)b \Rightarrow m \mid ac − bd \Rightarrow ac \equiv bd \pmod m$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.