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I am trying to solve this Modular arithmetic problem but I have no idea how to verify if the proposition holds or not.

$\forall a,b \in \mathbb{Z}: (a \equiv b \pmod{5} \Rightarrow 2a \equiv 2b \pmod{5} )$

rtybase
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user3602563
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  • see "Properties" in https://en.wikipedia.org/wiki/Modular_arithmetic – Matti P. Feb 26 '19 at 08:20
  • Welcome to MSE. Your questions lacks details and context. Why are you interested in this question and what have you tried so far? You can find some remarks about how to ask a good question here: https://math.stackexchange.com/help/how-to-ask –  Feb 26 '19 at 08:21
  • It's a direct application of Euclid's lemma. You can even conclude $\forall a,b \in \mathbb{Z}: (a \equiv b \pmod{5} \iff 2a \equiv 2b \pmod{5} )$ – rtybase Feb 26 '19 at 08:23

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Try it this way:

$$ a \equiv b \pmod 5 \implies \exists (k_a, k_b, m \in \Bbb N) : \\a = 5k_a+m , \\b = 5k_b+m, \\0 \leq m <5 $$ Okay, now what about comparing $2a$ and $2b$?

One thing to try is to multiply those relations above by 2, getting $$ 2a = 10k_a+2m , \\2b = 10k_b+2m, \\0 \leq 2m <10 $$ If $2m < 5$ it is easy to see that $2a \equiv 2b$ by using multipliers $k'_a = 2k_a$ and $k'_b = 2k_b$. That leaves the headache of the possibility that $2m > 5$.

Can you see how to handle that case and thus complete the proof? (Hint - what happens if you use $k'_a = 2k_a + 1$ ?)

Mark Fischler
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You can prove this directly.

$$a \equiv b \pmod {5} \iff 5|(b-a) \iff b-a = 5k ~(\text{for some } k \in \Bbb Z)\\ \iff 2b-2a =5(2k) \iff 5|(2b-2a) \iff 2a \equiv 2b \pmod {5}.$$

Robert Shore
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    Transition $2b-2a =5(2k) \iff 5|(2b-2a)$ is not justified. You need Euclid's lemma for this. And it only works for all $a,b$ because $\gcd(2,5)=1$. – rtybase Feb 26 '19 at 08:33