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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.

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  1. jim_thompson5910
    • one year ago
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    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

  2. Bee_see
    • one year ago
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    how do you know x-y is negative and x^2+xy+y^2 is always positive?

  3. jim_thompson5910
    • one year ago
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    x < y leads to x-y < 0 which means x-y is negative

  4. jim_thompson5910
    • one year ago
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    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

  5. Bee_see
    • one year ago
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    but if I were to say x=5 and y=4, then don't I get positive 1?

  6. jim_thompson5910
    • one year ago
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    keep the inequality x < y in mind

  7. Bee_see
    • one year ago
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    so like, 2 < 3 2-3=-1?

  8. jim_thompson5910
    • one year ago
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    yeah that's one example

  9. Bee_see
    • one year ago
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    how does it fulfill all the cases?

  10. jim_thompson5910
    • one year ago
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    you use an algebraic proof like I did above to show it works in every case (as long as x < y)

  11. Bee_see
    • one year ago
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    how did you know that you wanted the equation to equal 0? (in the beginning)

  12. jim_thompson5910
    • one year ago
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    you mean in my step 2? I subtracted y^3 from both sides to make the right side 0

  13. jim_thompson5910
    • one year ago
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    btw it's not an equation. It's an inequality

  14. Bee_see
    • one year ago
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    I mean this step: x^3 - y^3 < 0

  15. jim_thompson5910
    • one year ago
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    I subtracted y^3 from both sides of x^3 < y^3

  16. jim_thompson5910
    • one year ago
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    I guess my proof in my first post is me thinking backwards

  17. jim_thompson5910
    • one year ago
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    but it's still valid in a way

  18. Bee_see
    • one year ago
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    No, I mean...so I guess you are not supposed to go from x < y to x^3 < y^3?

  19. jim_thompson5910
    • one year ago
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    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

  20. Bee_see
    • one year ago
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    OK. I see.

  21. Bee_see
    • one year ago
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    Thanks for the help.

  22. jim_thompson5910
    • one year ago
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    no problem

  23. jim_thompson5910
    • one year ago
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    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

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