How to prove that if $x>0$, $y>0$, and $x>y$, then $x^2>y^2$?

Question

Prove that if $x>0$, $y>0$, and $x>y$, then $x^2>y^2$.

Here is my attempt:

Since $x>0$, $y>0$, and $x>y$,

may implies that $x+y>0$ and $x-y>0$.

Thus, $(x+y)(x-y)>0$.

Simplifying this

$x^2-y^2>0$

Therefore, $x^2>y^2$.

Does this answer your question? If $a > b$, is $a^2 > b^2$? – Bill Dubuque Dec 03 '21 at 15:24 — Bill Dubuque, Dec 03 '21 at 15:24

score 3 · Answer 1 · answered Jun 19 '20 at 02:05

3

Hint: use the fact that $x^2 - y^2 = (x+y)(x-y)$.

answered Jun 19 '20 at 02:05

angryavian

1

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Dec 06 '21 at 12:39

score -1 · Answer 2 · answered Jun 19 '20 at 02:11

-1

$x>y>0$ multiply on itself.

answered Jun 19 '20 at 02:11

zkutch

1

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Dec 06 '21 at 12:39
@Bill Dubuque. How can your remark be applied to what was written more then year ago? Please allow yourself to comment on recent incidents. – zkutch Dec 06 '21 at 16:53

2 Answers2