A friend and I are completely stumped on this prompt, and are even having trouble seeing how its statement is true. Any help will be appreciated!

Prove that if $a \equiv b \pmod{3}$, then $2a \equiv 2b \pmod{3}$.

  • $\begingroup$ Hint: $a = 3n + b$ $\endgroup$ – JACKY Li Oct 24 '12 at 5:08
  • $\begingroup$ If $a-b$ is divisible by 3, then $2a-2b=2\cdot(a-b)$ ... ? $\endgroup$ – Hagen von Eitzen Oct 24 '12 at 6:27

$a\equiv b\pmod 3 ⇔ 3\mid (a-b)\implies 3\mid n(a-b) ⇔ na\equiv nb\pmod 3$ where $n$ is any integer.

Also, $3\mid n(a-b)\implies 3\mid(a-b)$ if $(n,3)=1$

So, $3\mid (a-b) ⇔ 3\mid n(a-b)$ if $(n,3)=1$

Here $n=2,(2,3)=1,$ so, $3\mid (a-b) ⇔ 3\mid 2(a-b)$

  • $\begingroup$ Do we not have an equivalence $3|(a-b)\iff 3|2(a-b)$ rather than just an implication? For, $gcd(3,2)=1$, so $3$ must divide $a-b$ when it divides $2(a-b)$. $\endgroup$ – yearning4pi Oct 24 '12 at 5:09
  • $\begingroup$ @peoplepower, ya, we do. But here, implication is sufficient, we don't need to prove the reverse. $\endgroup$ – lab bhattacharjee Oct 24 '12 at 5:11
  • $\begingroup$ Just like the two other equivalences. $\endgroup$ – yearning4pi Oct 24 '12 at 5:11
  • $\begingroup$ @peoplepower, I've generalized the answer. $\endgroup$ – lab bhattacharjee Oct 24 '12 at 5:20

Hint $\rm\ \ n\in\Bbb Z\:\Rightarrow\:2n\in\Bbb Z,\ $ i.e. $\rm\ \dfrac{a-b}3\in\Bbb Z \ \Rightarrow\ \dfrac{2a-2b}3\, =\, 2\,\left(\dfrac{a-b}3\right)\in2\,\Bbb Z\subset \Bbb Z$


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