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Bee_see
 one year ago
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.
Bee_see
 one year ago
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.

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0x^3 < y^3 x^3  y^3 < 0 (xy)*(x^2 + xy + y^2) < 0 ... difference of cubes factoring rule if x < y, then xy < 0. So xy is negative while x^2 + xy + y^2 is always positive so if x < y, then (xy)*(x^2 + xy + y^2) < 0 is always true

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0how do you know xy is negative and x^2+xy+y^2 is always positive?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0x < y leads to xy < 0 which means xy is negative

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0x^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

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0but if I were to say x=5 and y=4, then don't I get positive 1?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0keep the inequality x < y in mind

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0so like, 2 < 3 23=1?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0yeah that's one example

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0how does it fulfill all the cases?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0you use an algebraic proof like I did above to show it works in every case (as long as x < y)

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0how did you know that you wanted the equation to equal 0? (in the beginning)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0you mean in my step 2? I subtracted y^3 from both sides to make the right side 0

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0btw it's not an equation. It's an inequality

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0I mean this step: x^3  y^3 < 0

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0I subtracted y^3 from both sides of x^3 < y^3

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0I guess my proof in my first post is me thinking backwards

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0but it's still valid in a way

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0No, I mean...so I guess you are not supposed to go from x < y to x^3 < y^3?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0Here'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 xy < 0 multiply (xy) with (x^2+xy+y^2). This result is always negative because xy < 0 and x^2+xy+y^2 > 0 (xy)*(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

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0here'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/ifxneq0yneq0thenx2xyy2is
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