Proving that $\frac{n+1}{2n+3}$ and $\frac{3n-5}{4n-7}$ are irreducible for all $n$

I am trying to solve the following problem:

Prove that the following fractions are irreducible for any n (n is a natural number and it cannot be null).

1. $\frac{n}{n+1}$
2. $\frac{n+1}{2n+3}$
3. $\frac{3n-5}{4n-7}$

I don't know if my logic is right!

Now for the first one, i used the property which states that $n$ and $n+1$ are always coprime. For the second one i relied on the following property,

1. if $n|a$ and $n|b$ then $n|(a*k - b*p)$.

So lets suppose $\frac{n+1}{2n+3}$ is reducible, then there is a whole number $d$ which divides both the numerator and denominator. By property 1, $d|[2n+3 - 2(n+1)]$ this means $d|1$ so $d=1\ldots$.

So the fraction is not reducible since every number is divisible by 1.

Is my logic right? how would someone go about solving this problem?

Yes: since $(2n+3)-2\cdot(n+1)=1$, every common divisor of $2n+3$ and $n+1$ is also a divisor of $1$. Thus, $2n+3$ and $n+1$ are co-prime.

Ditto for $3n-5$ and $4n-7$, to concude it suffices to find some integers $\color{red}{a}$ and $\color{green}{b}$ such that $\color{red}{a}\cdot(3n-5)+\color{green}{b}\cdot(4n-7)=\pm1$.

Both cases are examples of the more general result that $un+v$ and $u'n+v'$ are co-prime as soon as $uv'=u'v\pm1$.