Bee_see
  • Bee_see
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.
Discrete Math
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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jim_thompson5910
  • jim_thompson5910
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
Bee_see
  • Bee_see
how do you know x-y is negative and x^2+xy+y^2 is always positive?
jim_thompson5910
  • jim_thompson5910
x < y leads to x-y < 0 which means x-y is negative

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jim_thompson5910
  • jim_thompson5910
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
Bee_see
  • Bee_see
but if I were to say x=5 and y=4, then don't I get positive 1?
jim_thompson5910
  • jim_thompson5910
keep the inequality x < y in mind
Bee_see
  • Bee_see
so like, 2 < 3 2-3=-1?
jim_thompson5910
  • jim_thompson5910
yeah that's one example
Bee_see
  • Bee_see
how does it fulfill all the cases?
jim_thompson5910
  • jim_thompson5910
you use an algebraic proof like I did above to show it works in every case (as long as x < y)
Bee_see
  • Bee_see
how did you know that you wanted the equation to equal 0? (in the beginning)
jim_thompson5910
  • jim_thompson5910
you mean in my step 2? I subtracted y^3 from both sides to make the right side 0
jim_thompson5910
  • jim_thompson5910
btw it's not an equation. It's an inequality
Bee_see
  • Bee_see
I mean this step: x^3 - y^3 < 0
jim_thompson5910
  • jim_thompson5910
I subtracted y^3 from both sides of x^3 < y^3
jim_thompson5910
  • jim_thompson5910
I guess my proof in my first post is me thinking backwards
jim_thompson5910
  • jim_thompson5910
but it's still valid in a way
Bee_see
  • Bee_see
No, I mean...so I guess you are not supposed to go from x < y to x^3 < y^3?
jim_thompson5910
  • jim_thompson5910
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
Bee_see
  • Bee_see
OK. I see.
Bee_see
  • Bee_see
Thanks for the help.
jim_thompson5910
  • jim_thompson5910
no problem
jim_thompson5910
  • jim_thompson5910
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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