# Thread: help with absolute value

1. ## help with absolute value

Problem : prove that $|x|<|y| <=> x^2 < y^2$ by giving a reason for each step :

My teacher give solution

step 1 : $|x|<|y| => |x||x| =< |x||y|$ and $|x||y| < |y||y|$

step 2 : $=> |x|^2 < |y|^2$

step 3 : $=> x^2

can anybody justify the first step? i can't understand which properties values to use in order to get to first step. tq

problem :

suppose $|x| =< 2$. Use properties of absolute value to show that

$|(2x^2+3x+2)/(x^2+2)| =< 8$

2. Originally Posted by bobey
Problem : prove that $|x|<|y| <=> x^2 < y^2$ by giving a reason for each step :

My teacher give solution

step 1 : $|x|<|y| => |x||x| =< |x||y|$ and $|x||y| < |y||y|$

step 2 : $=> |x|^2 < |y|^2$

step 3 : $=> x^2

can anybody justify the first step? i can't understand which properties values to use in order to get to first step. tq

He's using the basic property of inequalities: $a0$

problem :

suppose $|x| =< 2$. Use properties of absolute value to show that

$|(2x^2+3x+2)/(x^2+2)| =< 8$

Remember (definition/properties of abs. val.) that for $b>0\,,\,\,|a| , so:

$|(2x^2+3x+2)/(x^2+2)| =< 8\Longleftrightarrow -8\leq \frac{2x^2+3x+2}{x^2+2}\leq 8$ ,and now you have an "and" inequality system: both the left AND the right inequalities have to be true, so solve them separately and then take the intersection of the separate solutions. For example:

$-8\leq \frac{2x^2+3x+2}{x^2+2} \Longrightarrow \frac{2x^2+3x+2}{x^2+2}+8\geq 0\Longrightarrow \frac{10x^2+3x+18}{x^2+2}\geq 0$ , and since both the numerator and the denominator are ALWAYS positive (why?)

this inequality is true for all $x$ ;

$\frac{2x^2+3x+2}{x^2+2}\leq 8\Longrightarrow \frac{2x^2+3x+2}{x^2+2}-8\leq 0\Longrightarrow \frac{-6x^2+3x-14}{x^2+2}\leq 0$...and now can you see why this is ALWAYS negative?

Interestingly enough, the given inequality is true for all x, imo, and thus I can't understand why they gave you that $|x|<2$...

Tonio