Prove that if that if $p > \sqrt{n} + \sqrt{n+1}$ where $n \in\Bbb N$*; then $p² \ge 4n + 2$

Asked Nov 04 '21 at 17:11

Active Nov 04 '21 at 21:20

Viewed 53 times

Here's what I tried:

$$p > \sqrt{n} + \sqrt{n+1}$$ $$p² > (\sqrt{n} + \sqrt{n+1})²$$ $$p² > 2n + 1 + 2\sqrt{n(n+1)}$$

So we need to prove that

$$2n + 1 + 2\sqrt{n(n+1)} > 4n + 2$$ $$2n + 1 - 2\sqrt{n(n+1)} < 0$$ $$n + \frac{1}{2} < \sqrt{n(n+1)}$$ $$n² + n + \frac{1}{4} < n² + n$$ $$\frac{1}{4} < 0$$

Which is absurd. What am I doing wrong ?

edited Nov 04 '21 at 21:03

asked Nov 04 '21 at 17:11

Amin Guermazi

You mishandled the additive factors of $1$. And you confused the use of $≥$. – lulu Nov 04 '21 at 17:14
A good way to track down errors in these things is to pick particular values for $n$ and go through it line by line using those numbers. – lulu Nov 04 '21 at 17:17
1

You're on the right track: $p^2 > 2n + 1 + 2\sqrt{n(n+1)} > 2n + 1 + 2\sqrt{n\cdot n} = 4n+1$ and so $p^2 \ge 4n+2$. – lhf Nov 04 '21 at 17:34
guys i wrongly copied 1 instead of 2. check my edit pls. its 4n + 2 not 4n + 1 – Amin Guermazi Nov 04 '21 at 21:04
@lhf thanks a lot, that's the solution. You can post an answer and I'll accept it and upvote it. – Amin Guermazi Nov 04 '21 at 21:19
@AminGuermazi The original proposition must be wrong, is or use directly the truthnes condition this is a dead end. – Erdel von Mises Nov 04 '21 at 21:21
It isn't, you can try any example of n and it'll be correct. The only case where it will be wrong is $n \in Z^-$ @ErdelvonMises – Amin Guermazi Nov 04 '21 at 21:32
@lhf, minor quibble: Nowhere is it asserted that $p$ is an integer.... – Barry Cipra Nov 04 '21 at 21:41
@BarryCipra, oops, somehow I assumed that $p$ is an integer. – lhf Nov 05 '21 at 13:26
It's given as an integer. Sorry forgot to state it – Amin Guermazi Nov 05 '21 at 20:09

Prove that if that if $p > \sqrt{n} + \sqrt{n+1}$ where $n \in\Bbb N$*; then $p² \ge 4n + 2$

0 Answers0