0
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Alternatively, I could have said to subtract $n+1$ from both sides $\endgroup$ – J. W. Tanner Feb 7 '19 at 23:31
1
$\begingroup$

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 $\frac n {n+1} > \frac n {n+2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.