1. Mar 24, 2009

### llanoda

Hi,

Anybody can help me prove this? Thanks...

If x<0 and y<0 and x<y, then [y][/2] < [x][/2]

2. Mar 24, 2009

### CompuChip

Hi, welcome to PF.

What does the brackets notation mean?
What is [y][/2] ?

3. Mar 27, 2009

### llanoda

That is y squared is greater than x squared

4. Mar 27, 2009

### Dragonfall

You seem to have said that y squared is LESS than x squared in your original post.

5. Mar 28, 2009

### Office_Shredder

Staff Emeritus
That's correct. Because x<y<0.

This might be the kind of thing you want to do in parts. For example, prove that x2 = (-x)2 so you can switch this to looking at positive numbers

6. Mar 28, 2009

### HallsofIvy

You mean, "if x< 0, y< 0 and x< y, then y2< x2". If you don't want to use HTML tags or LaTex, the y^2< x^2 is the best way to show an exponent.

It is easy to show that if 0< a< b, then a2< b2. Here, since x< y< 0, 0< -y< -x so (-y)2< (-x)2 which, because (-x)2= x2 and (-y)2= y2, leads to your result.

7. Mar 29, 2009

### derek e

edit:
Use
If a < b, then c * b < c * a if and only if c < 0.
Spoiler :
We have x < 0, y < 0 and x < y. Then x * y < x * x, because x < 0. Similarly, y * y < y * x, because y < 0. So,
y^2 = y * y < y * x = x * y < x * x = x^2,
giving y^2 < x^2.

Last edited: Mar 29, 2009
8. Mar 29, 2009

### CompuChip

See, if you'll wait long enough with replying, we'll solve the entire question for you

9. Apr 6, 2009

### poutsos.A

We have x<y and x<0 and y<0 ,so if we multiply x<y by x<0 we get:

$$x^2>xy$$.................................................................1

and if we multiply x<y by y<0 we get:

$$xy> y^2$$..................................................................2

And from (1) and (2) we have : $$x^2> y^2$$,using the fact .

If A>B AND B>C ,then A>C