Prove by contradiction that if $n$ is a natural number then $n/(n+1) > n/(n+2)$

Question

Prove by contradiction that if $n$ is a natural number then $n/(n+1) > n/(n+2)$

Since it is a proof by contradiction, I think I start by assuming that $n/(n+1) < n/(n+2)$, but then I don't know how to prove that either. The graphs make it clear but I don't know how to proceed to give a contradiction

Answer 1

If possible, let $\frac n {n+1} \leq \frac n {n+2}$. This gives $n(n+2) \leq (n+1)n$ or $n^{2}+2n\leq n^{2}+n$. This means $2n \leq n$ or $n \leq 0$ which is a contradiction. Hence $\frac n {n+1} > \frac n {n+2}$.

Answer 2

Given that $n>0$, if we assume the contrary, i.e., $$\frac n {n+1} \le \frac n {n+2}, $$ then we could divide both sides by $n$ and cross-multiply to get $n+2 \le n+1.$

Subtract $n$ from both sides, and there is clearly a contradiction.