Question

Prove that, for all real numbers x and y, if x < y, then x^3 < y^3. It does not suffice to merely prove this for positive x and y; your proof must handle all cases.

Answer 1

x^3 < y^3

x^3 - y^3 < 0

(x-y)*(x^2 + xy + y^2) < 0 ... difference of cubes factoring rule

if x < y, then x-y < 0. So x-y is negative while x^2 + xy + y^2 is always positive

so if x < y, then (x-y)*(x^2 + xy + y^2) < 0 is always true

Answer 2

how do you know x-y is negative and x^2+xy+y^2 is always positive?

Answer 3

x < y leads to x-y < 0 which means x-y is negative

Answer 4

x^2 is positive
y^2 is positive
since x < y, this means |x*y| < y^2

so even if x*y was negative, the positive x^2 and y^2 terms counterbalance and make x^2+xy+y^2 always positive

Answer 5

but if I were to say x=5 and y=4, then don't I get positive 1?

Answer 6

keep the inequality x < y in mind

Answer 7

so like, 2 < 3 2-3=-1?

Answer 8

yeah that's one example

Answer 9

how does it fulfill all the cases?

Answer 10

you use an algebraic proof like I did above to show it works in every case (as long as x < y)

Answer 11

how did you know that you wanted the equation to equal 0? (in the beginning)

Answer 12

you mean in my step 2? I subtracted y^3 from both sides to make the right side 0

Answer 13

btw it's not an equation. It's an inequality

Answer 14

I mean this step: x^3 - y^3 < 0

Answer 15

I subtracted y^3 from both sides of x^3 < y^3

Answer 16

I guess my proof in my first post is me thinking backwards

Answer 17

but it's still valid in a way

Answer 18

No, I mean...so I guess you are not supposed to go from x < y to x^3 < y^3?

Answer 19

Here's an attempt at doing the proof in the right order (notice how it's in reverse order of my first proof)


if x < y, then x-y < 0


multiply (x-y) with (x^2+xy+y^2). This result is always negative because x-y < 0 and x^2+xy+y^2 > 0


(x-y)*(x^2+xy+y^2) < 0
x*(x^2+xy+y^2)-y*(x^2+xy+y^2) < 0
x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3 < 0
x^3 - y^3 < 0
x^3 < y^3

Answer 20

OK. I see.

Answer 21

Thanks for the help.

Answer 22

no problem

Answer 23

here's a page showing why x^2 + xy + y^2 is always positive (assuming x and y are nonzero)

http://math.stackexchange.com/questions/349940/if-x-neq-0-y-neq-0-then-x2xyy2-is