Question

Prove that if a positive integer is of the form $ 6q+5 $ , then it is of the form $ 3q+2 $ for some integer q, but not conversely.

Answer 1

Hint: We first find a random positive integer of the form $ 6q+5 $ and try to show that the integer is of the form $ 3q+2 $ . Then we try to show that the converse isn’t true as we find an example where we have an integer being of the form $ 3q+2 $ but not of the form $ 6q+5 $ .

Complete step by step solution:
We have to prove that if a positive integer is of the form $ 6q+5 $ , then it is of the form $ 3q+2 $ for some integer q, but the converse isn’t true.
Let us take an integer as $ x $ , where $ x=6q+5 $ .
We now break the 6 as $ 6=3\times 2 $ and take $ x=3\left( 2q \right)+3+2=3\left( 2q+1 \right)+2 $ .
As q is integer therefore, $ \left( 2q+1 \right) $ is also integer.
It proves that if a positive integer is of the form $ 6q+5 $ , then it is of the form $ 3q+2 $ where $ q\in \mathbb{Z} $ .
Now we have to prove that the inverse is not always true. We can imply it just by showing an example.
We take 8 as it is of the form $ 3q+2 $ for some integer q.
We have $ 8=3\times 2+2 $ but 8 is not of the form $ 6q+5 $ as $ 8=6\times 1+2 $ .
Therefore, the converse isn’t always true.

Note: We need to remember that to show a converse not being a theorem, we just need to show an exception where it doesn’t abide by the rule. For a theorem to be true it has to be applied on all the points in its domain.