Proving x<0, y<0, and x<y Implies [y][/2] < [x][/2]

[llanoda]

Hi,

Anybody can help me prove this? Thanks...

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

 

[CompuChip]

Hi, welcome to PF.

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

 

[llanoda]

That is y squared is greater than x squared

 

[Dragonfall]

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

 

[Office_Shredder]

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

 

[HallsofIvy]

Hi,

Anybody can help me prove this? Thanks...

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

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.

 

[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.

 

[CompuChip]

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

 

[poutsos.A]

Hi,

Anybody can help me prove this? Thanks...

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

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