for all integers $a$ and $b$, if $a \equiv 2 (\text{mod}\,4)$ and $b$ is odd, then $ab \equiv2 (\text{mod}\,4)$

Question

Is False? Since $2 \equiv 2\,(\text{mod}\,4)$ and $2 \times 3 = 6\,$ then $\,6 ≠ 2$.

This is a proof by counter-example?

It seemed too easy.

That's wrong.

So;

a ≡ 2 (mod 4) = 4q + 2 and b = 2q + 1

ab = (4q + 2)(2q + 1) = 8q + 2q + 2q + 2

then ab = 16q + 2 which is 2 mod 16

Answer 1

$6\neq2$ is correct, but the task is not saying that $ab$ will equal $2$, it says that it will have the remainder of $2$ when divided by $4$, and since $6=1\cdot 4 + 2$, you have $$6\equiv 2 \mod 4$$

so no, your counterexample is not correct.

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Hint 1:

The statement is not false

Hint 2:

Fill in the gaps:

• If $a\equiv 2\mod 4$, then it can be written as $\_\cdot \_ + \_$
• If $b$ is odd, then it can be written as $\_\cdot\_ +\_$
• Then $ab$ is eqal to $\_\_\_$