Alll Positive Integers proof by contraposition

[tmt1]

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?

 

[Sudharaka]

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, $sr > s\sqrt{n}\geq\sqrt{s}\sqrt{n}=\sqrt{sn}$.