# State True or False and explain. If 2a ≡ 4b mod 8, then a ≡ 2b mod 8

So I have concluded that the statement is false however i'm having a hard time explaining them, or writing a proof which I need to do for full marks. Could any one help me in explaining this, please be thorough so that I could understand more of proving these types of questions.

edit: I am no longer sure than the statement is false. I have tried my example again and have realized that I made a mistake. Now I need clarification on whether it is false or not.

2nd edit: Okay my progress is that, in order for the first equation to be true a must be an even number and b must be odd, we then get a modulus of 4 for all even integers a and odd integers b. I have tried many examples and they all turn out to be true. Again i'm having a hard time explaining or proving why.

• If it's false, a single counterexample is all you need. – lulu Apr 4 '18 at 23:17
• So assigning a and b integer values would be enough of a proof? – Alick Lazare Apr 4 '18 at 23:50
• One counterexample is all it takes to disprove a result. If you want to add more, you can point out that $2$ hasn't got an inverse $\pmod 8$ so $2m\equiv 2n\pmod 8$ does not imply that $m\equiv n\pmod 8$. But, really, a single counterexample already kills the theorem. – lulu Apr 4 '18 at 23:52
• guys check edits I have come to a new conclusion, I need clarification – Alick Lazare Apr 5 '18 at 0:14
• Your original instinct was right, The statement is false. Take $a=2,b=3$ for example. The condition $2a\equiv 4b \pmod 8$ just implies that $a\equiv 2b \pmod 4$ which is weaker. – lulu Apr 5 '18 at 0:17

The statement is false. Consider the following counterexample. Let $a = 6$ and $b = 9$. Then \begin{align*} 2a & \equiv 12 \equiv 4 \pmod{8}\\ 4b & \equiv 36 \equiv 4 \pmod{8} \end{align*} so $2a \equiv 4b \pmod{8}$. However, \begin{align*} a & \equiv 6 \pmod{8}\\ 2b & \equiv 18 \equiv 2 \pmod{8} \end{align*} so $a \not\equiv 2b \pmod{8}$.

Observe that if $2a \equiv 4b \pmod{8}$, then $2a = 4b + 8k$ for some $k \in \mathbb{Z}$, so $a = 2b + 4k$ for some $k \in \mathbb{Z}$, which implies that $a \equiv 2b \pmod{4}$. Observe that if $a = 6$ and $b = 9$, then \begin{align*} a & \equiv 2 \pmod{4}\\ 2b & \equiv 18 \equiv 2 \pmod{4} \end{align*} so $a \equiv 2b \pmod{4}$, as claimed.

• Thanks so much i understand much better now – Alick Lazare Apr 5 '18 at 3:05

Nothing wrong with going back to definitions if one has a hard time seeing it directly.

$2a \equiv 4b \mod 8 \implies$

$2a = 4b + 8k$ for some $k \implies$

$a = 2b + 4k \implies$

$a \equiv 2b \mod 4$.

Now we lost the $8$ because we divided by $2$.

This implies it doesn't have to be true because although $2a$ and $4b$ must be a multiple of $8$ apart, $a$ and $2b$ only need to be a multiple of $4$ apart.

To find a specific counter example simply let $2b - a = 4$. Example $b= 3$ and $a = 2$.

Than $2a = 4 \equiv 4b = 12 \mod 8$.

But $a = 2\not \equiv 2b = 6 \mod 8$.

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Maybe clearer to see is

$2a \equiv 4b \mod 8 \iff 2a - 4b \equiv 0 \mod 8 \iff 8|2a -4b$

$a \equiv 2b \mod 8 \iff a-2b \equiv 0 \mod 8 \iff 8|a - 2b$.

So the question is... is it ever possible for $8|2a - 4b$ but not have $8|a-2b$. And the answer to that should be easie. $8|2a - 4b \implies 4|a - 2b$ but not that $8|a-2b$.

• Thank you!!!! I got it! – Alick Lazare Apr 5 '18 at 3:05