If x>0 and y>0 and x^2 > y^2, prove that x > y?

Answer 1

Another way of looking at this (and avoiding square roots)
x > 0 and y > 0 => x + y > 0 (adding positives results in a positive)
x^2 > y^2 => x^2 - y^2 > 0
=> (x - y)(x + y) > 0 (factoring)
=> x - y > 0 (dividing by the positive x + y leaves the inequality unchanged)
=> x > y

Answer 2

            x² > y²
      x² - y² > 0
(x+y)(x-y) > 0
this gives us
x+y>0    &     x-y>0
    x>-y   &        x>y
whereby, x>y because x>0 & y>0, |y| < x.

Answer 3

x^2>y^2 -->x^2 - y^2>0 -->(x-y)(x+y)>0
(inequality)(1)
x>0,y>0--> x+y>0 (2)
(1),(2) --> x - y>o --> x>y

Answer 4

x^2>y^2 =>(imply) (x-y)(x+y)>0 (1)
x>0, y>0 => x+y >0
so from (1) x-y must by also >0
=> x-y>0
=> x>y

ps: good luck at school
:)

Answer 5

x^2>y^2
=>x^2-y^2>0
=>(x+y)(x-y)>0
=>x>-y, x>y
x>0,y>0,x>-y=>x>0 is what we already have
so we r left with x>y

Answer 6

x^2 > y^2
so
|x^2| > |y^2|
so
|x| > |y|
so
x > y

Note: Absolute values were used in the equation as x > 0 and y > 0

Answer 7

x>0 y>0
x^2>y^2
so x^2-y^2>0
(x+y)(x-y)>0
if this hads to be positive x>y
as x+y
in any case is positive (given)

Answer 8

verify no remember if the equation is suited: enable M(x, y) = x + y enable N(x, y) = x - y ?M(x, y)/?y = a million ?N(x, y)/?x = a million The equation is suited. hence, there exists a function, F(x, y) the partial spinoff with admire to x is M(x, y) and the partial spinoff with admire to y is N(x, y). we start up with the aid of integrating M(x, y) with admire to x: ?(x + y)dx = (a million/2)x² + yx + h(y) the place h(y) is a function that would disappear as quickly as we differentiate with admire to x. We differentiate this with admire to y: x + h'(y) This might desire to equivalent N(x, y); x + h'(y) = x - y h'(y) = -y h(y) = (-a million/2)y² + C the answer is: F(x, y) = (a million/2)(x² - y²) + xy + C

Answer 9

Take the square root of both terms giving x > y

Answer 10

What that other guy said....