Prove that if $y \in \mathbb{Z}$, then $\dfrac{2y+3}{4y-2} \notin \mathbb{Z}$.

Question

I have to show that if $y \in \mathbb{Z}$ $\Rightarrow$ $\dfrac{2y+3}{4y-2} \notin \mathbb{Z}$. I tried using proof by contradiction but I got stuck:

Assume $y \in \mathbb{Z}$ and $\dfrac{2y+3}{4y-2} \in \mathbb{Z}$.

We have:

$$\dfrac{2y+3}{4y-2} = \dfrac{2y-1+1+3}{4y-2} = \dfrac{2y-1+4}{2(2y-1)}= \dfrac{2y-1}{2(2y-1)} + \dfrac{4}{2(2y-1)} = \dfrac{1}{2} + \dfrac{2}{2y-1}$$

I don't see any way of reaching an obvious contradiction. Is there any other way of doing this. Or how should I continue, reaching a contradiction?

Answer 1

$2y+3$ is odd, $4y-2$ is even. So the denominator is divisible by 2, while the numerator is not. You can get an integer only if the $\gcd$ is equal in absolute value to the denominator.

Answer 2

Suppose you can write $\frac{2}{2y-1} =\frac{k}{2}$ for odd $k$. Then $4=2ky-k$, and $4+k=2ky$. Now, the LHS is odd while the RHS is a multiple of 2.

Answer 3

To show $\dfrac{1}{2} + \dfrac{2}{2y-1}$ cannot be an integer,

show that it's between $0$ and $1$ if $y>\dfrac52$ or $y<-\dfrac32$,

and it isn't an integer if $y=-1, 0, 1, $ or $2$ either.