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