Prove only using of the definition of congruence:

if $a \equiv b \pmod 5$ then $2a \equiv 2b \pmod{5}$?

I have thought about the solution as follows:

$2a \equiv 10k+2r \equiv 5 (2k) +2r$ ......(1)

$2b \equiv 10l+2r \equiv 5 (2l) + 2r$ .....(2)

both of (1) and (2) have the same remainder, therefore $2a \equiv 2b \pmod{5}$?


According to your definition of congruence, I believe the following will suffice.


By definition of congruence we know that: if $a \equiv b \pmod{n}$, then $ac \equiv bc \pmod{n}$.

Since it is given that $n=5$, and $a \equiv b \pmod{5}$ then by letting $c=2$ and applying definition of congruence we have:

if $a \equiv b \pmod{n}$, then $2a \equiv 2b \pmod{5}$.

which is true.


What is your definition of congruence? I might assume $a\equiv b \pmod 5$ is defined as $5|(a-b)$, in which case $a-b=5k$ for some $k$, and $2a-2b=5(2k)$. The result follows.

  • $\begingroup$ The definition of congruent is : Let a, b, c ∈ Z and n ∈ N. If a ≡ b (mod n), then ac ≡ bc (mod n). $\endgroup$ – Coheen Dec 6 '15 at 22:45
  • $\begingroup$ In that case your proof is trivial — let $c=2$. $\endgroup$ – Elliot G Dec 6 '15 at 22:46
  • $\begingroup$ Yes, but I need the formal proof for like this question, how can I write the proof formally? $\endgroup$ – Coheen Dec 6 '15 at 22:49
  • $\begingroup$ It would truly be one line. We know $a\equiv b$, therefore $2a\equiv 2b$ by the definition of congruency. $\endgroup$ – Elliot G Dec 6 '15 at 23:04

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