Thread: Range of x,y for which x^2 > y^2 holds true

1. Range of x,y for which x^2 > y^2 holds true

The following is the problem on 63rd page of Zakon's "Basic Mathematics"

(iii) If x > y ≥ 0, then
x2 > y2 and x3 > y3 ≥ 0 (where x3 = x2x);
Similarly,
x4 > y4 ≥ 0 (where x4 = x3x).
Which (if any) of these propositions remain valid also if x or y is negative? Give proof.

I am able to solve the first part of the problem. Got stuck in the second. If somebody can give me a hint of how to determine whether the proposition is valid if x or y is negative.

Regards...............D

2. Originally Posted by dawoodvora
The following is the problem on 63rd page of Zakon's "Basic Mathematics"

(iii) If x > y ≥ 0, then
x2 > y2 and x3 > y3 ≥ 0 (where x3 = x2x);
Similarly,
x4 > y4 ≥ 0 (where x4 = x3x).
Which (if any) of these propositions remain valid also if x or y is negative? Give proof.

I am able to solve the first part of the problem. Got stuck in the second. If somebody can give me a hint of how to determine whether the proposition is valid if x or y is negative.

Regards...............D

For example,

$x^2>y^2\Longrightarrow (x-y)(x+y)>0\Longrightarrow (1)\,x-y\,\,or\,\, (2)\,x>y\,\,and\,\,x<-y$

So if $x<0$ say, then (1) above is true for any y s.t. $|x|<|y|$ ,and

(2) is true for any $y.

Tonio

3. My understanding of Tonio's solution

Hello everybody,
Please you all have a look at it and let me know if I understood what Tonio tried to explain me. Please let me know if my understanding of Tonio's solution is Okay or Not.

x2 > y2 => (x-y)(x+y) > 0

=> (x-y) > 0 & (x+y) > 0 OR (x-y) < 0 & (x+y) < 0
(Both the terms +ve) OR (Both the terms –ve)
=> x > y & x > -y OR => x < y & x < -y
=> x > |y| OR => x < |y|
=> x > 0 OR => x < 0

Thus x2 > y2 holds true only when:
1. x > |y| if x > 0
2. x < |y| if x < 0