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$\forall n \in \mathbb{N}^+, r\in \mathbb{N}^+, s\in \mathbb{N}^+, r\cdot s \leq n \implies r \leq \sqrt{n} $ or $s \leq \sqrt{n}$

Assuming that $\mathbb{N}^+$ refers to all positive natural numbers starting at $1$. Can someone pls give me a hint as to how to start this proof? I'm not too sure on how I should approach this.

Larry B.
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2 Answers2

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Not a problem. If $r \le s $ then $r^2 \le rs \le n$. So $r^2 \le n$ and $r \le \sqrt{n}$. And similarly if $s \le r $ then $s \le \sqrt{n}$.

DeepSea
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  • How did you get r≤s if r = s? –  Oct 28 '17 at 04:50
  • The case $r = s$ is included in the argument as well with the $ \le $ symbol....Most folks would use contradiction proof as commented above, and I tried hard...to be slightly new.... – DeepSea Oct 28 '17 at 04:51
  • Thanks for your contribution, but I don't really understand your method. It's a bit too confusing for me. So far, I have tried to use contradiction where rs≤n implies n≤r or n≤s, then I mutliply the right side for n≤rs, which contradicts the left side... Could you verify if this is right? –  Oct 28 '17 at 05:02
  • I also tried contraposition where I rearranged it to be $ r > \sqrt{n}$ or $ s > \sqrt{n} \implies rs > n$, then rs > n $\implies$ rs>n –  Oct 28 '17 at 05:13
  • Yours must be both of inequalities lead to $rs > n$ which is a contradict to given. – DeepSea Oct 28 '17 at 05:18
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Correct me if wrong.

Assume the statement is false:

(The negation, hopefully correct).

There exists an $n\in \mathbb{N^+},$ $r,s \in \mathbb{N^+},$

such that $rs \le n$ implies

$r \gt √n$ and $s\gt √n.$

Then: $ n \ge rs \gt n,$ a contradiction .

Peter Szilas
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