# Thread: Prove that it is smaller than 6

1. ## Prove that it is smaller than 6

Prove that .

How do i prove this? thank you...

2. ## Re: Prove that it is smaller than 6

A possibility is to take the square of both sides.

3. ## Re: Prove that it is smaller than 6

Originally Posted by MichaelLight
Prove that .

How do i prove this? thank you...
$\displaystyle \sqrt{10}+2\sqrt{2}<6$

$\displaystyle \Leftrightarrow 10+8+8\sqrt{5}<36$ (Squaring both sides)

$\displaystyle \Leftrightarrow 8\sqrt{5}<18$

$\displaystyle \Leftrightarrow 320<324$ (Squaring both sides)

which is true.

4. ## Re: Prove that it is smaller than 6

Originally Posted by MichaelLight
Prove that .

How do i prove this? thank you...
Suppose otherwise, then by assumption:

$\displaystyle \sqrt{10}+2\sqrt{2}\ge 6$

Now square both sides and follow the consequences to get a contradiction.

CB

5. ## Re: Prove that it is smaller than 6

Originally Posted by alexmahone
$\displaystyle \sqrt{10}+2\sqrt{2}<6$

$\displaystyle \Leftrightarrow 10+8+8\sqrt{5}<36$ (Squaring both sides)

$\displaystyle \Leftrightarrow 8\sqrt{5}<18$

$\displaystyle \Leftrightarrow 320<324$ (Squaring both sides)

which is true.
Normally, one would NOT start a proof by assuming what you want to prove, as alexmahone does here. However, this is a perfectly valid "synthetic proof". Every step is invertible. A "standard" proof would be given by going from "320< 324" upward.