At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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
how do you know x-y is negative and x^2+xy+y^2 is always positive?
x < y leads to x-y < 0 which means x-y is negative
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
but if I were to say x=5 and y=4, then don't I get positive 1?
keep the inequality x < y in mind
so like, 2 < 3 2-3=-1?
yeah that's one example
how does it fulfill all the cases?
you use an algebraic proof like I did above to show it works in every case (as long as x < y)
how did you know that you wanted the equation to equal 0? (in the beginning)
you mean in my step 2? I subtracted y^3 from both sides to make the right side 0
btw it's not an equation. It's an inequality
I mean this step: x^3 - y^3 < 0
I subtracted y^3 from both sides of x^3 < y^3
I guess my proof in my first post is me thinking backwards
but it's still valid in a way
No, I mean...so I guess you are not supposed to go from x < y to x^3 < y^3?
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
OK. I see.
Thanks for the help.
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