MHB Alll Positive Integers proof by contraposition

AI Thread Summary
The discussion revolves around proving that for all positive integers \( n \), \( r \), and \( s \), if \( rs \le n \), then either \( r \le \sqrt{n} \) or \( s \le \sqrt{n} \). A participant expresses confusion about the inequalities derived from assuming \( r > \sqrt{n} \) and \( s > \sqrt{e} \), specifically questioning the transition from \( sr > s\sqrt{n} \) to \( rs > \sqrt{ns} \). Another contributor clarifies that for any positive number, \( s \geq \sqrt{s} \), which helps establish the relationship between the inequalities. The conversation emphasizes understanding the manipulation of inequalities in the context of positive integers. The proof ultimately hinges on correctly applying properties of square roots and inequalities.
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For all positive integers $n$, $r$, and $s$, if $rs \le n$ then $r \le\sqrt{n}$ or $s \le \sqrt{n}$

Proof:

Suppose $r$ , $s$ and $n$, are integers and $r > \sqrt{n}$ and $ s > \sqrt{e}$.

Multiply both sides of the first inequality by $s$.

I get $sr > s\sqrt{n} $, but the book gives $rs > \sqrt{ns}$. How is this possible.

Also, if I multiply the second inequality by $\sqrt{n}$, I get $s \sqrt{n}> n$, but the book gives $\sqrt{ns} > n$ . What am I doing wrong?
 
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tmt said:
For all positive integers $n$, $r$, and $s$, if $rs \le n$ then $r \le\sqrt{n}$ or $s \le \sqrt{n}$

Proof:

Suppose $r$ , $s$ and $n$, are integers and $r > \sqrt{n}$ and $ s > \sqrt{e}$.

Multiply both sides of the first inequality by $s$.

I get $sr > s\sqrt{n} $, but the book gives $rs > \sqrt{ns}$. How is this possible.

Also, if I multiply the second inequality by $\sqrt{n}$, I get $s \sqrt{n}> n$, but the book gives $\sqrt{ns} > n$ . What am I doing wrong?

Hi tmt,

Note that for any positive number, $s\geq\sqrt{s}$. Therefore, $sr > s\sqrt{n}\geq\sqrt{s}\sqrt{n}=\sqrt{sn}$.
 
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