How can x^n < y^n be proven using Spivak's 12 axioms and mathematical induction?

  • Thread starter alfred_Tarski
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In summary, the problem at hand is to prove that for x < y and n is odd, x^n < y^n. This can be done using Spivak's 12 axioms, theorems derived from said axioms, and mathematical induction. After considering three cases, it is recommended to use mathematical induction starting with the base case n=1 and then using the properties of even powers and inequalities to show that -y < -x, leading to the conclusion that x^n < y^n. It is also possible to prove this for x and y both negative by using the fact that |x| = -x and |y| = -y.
  • #1
alfred_Tarski
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Homework Statement



Prove that x < y and n is odd implies x^n < y^n

This is from chapter 1 problem 6 (b), so I am sure a lot of you are familiar with this. One can only use Spivak's 12 axioms, theorems derived from said axioms, and mathematical induction.

Homework Equations



x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + ... + xy^{n-2} + y^{n-1})

Sorry I do not know how the scripting language works here.

The Attempt at a Solution



I have tried doing this problem for about five hours, *shame*, and I have looked up hints on the internet about it but have been still unsuccessful. I think my problem with this exercise is my accounting.

Here is one of my attempts:

If x < y [itex]\wedge[/itex] n is odd [itex]\Rightarrow[/itex] x^n < y^n

There are three cases to consider:

(1) 0 [itex]\leq[/itex] x < y
(2) x < 0 [itex]\leq[/itex] y
(3) x < y [itex]\leq[/itex] 0

Case (1) is a special case of the previous theorem we proved in problem 6 (a), so it's trivial.
Case (2) is a negative number to an odd power is less than a positive number to an odd power, so again trivial. But since I am so new to proofs, it would be nice if one of you nice gentlemen or ladies would help me show this. (I am about 1/4 the way done with the book How To Prove It and my school has never taught me anything about proofs in my 1 and 1/2 years of attendance)
Case (3) is the case which I have been stuck on.
x < y [itex]\leq[/itex] 0
x - y < 0 < -y
∴ x - y < 0
Notice all three cases can produce this inequality so there must be one logical way to prove them all at once WLOG, but I do not see it?
I've also tried to prove this other ways, namely induction and contrapositive, but they did not get me any closer. Thanks for your time, and I really appreciate any and all responses.
 
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  • #2
I know you've said that you tried induction, but I think you should give it another try. For case 3, this is my recommended outline/hint guide:

1. The base case is n=1. This is trivially true.
2. Let [itex] m\geq 1[/itex] be odd. Then m=k+1 for some k even.
3. [itex] x<y \Leftrightarrow -y<-x [/itex]. What do you know about even powers and inequalities? (note that -y and -x are both positive).
4. Since [itex] -x>0 [/itex], you know that for any [itex] 0<a<b [/itex], the following holds: [itex]-xa<-xb [/itex].
5. But then, you know from 3. that [itex] -y<-x [/itex]...

That's the gist of it.
 
  • #3
You can prove that xn<yn for x<y and n odd if both x and y are positive.

If case both x and y negative, x<y means |x|>|y|. Both are positive numbers, so |y|n<|x|n. And you know that |y|=-y, |x|=-x... Can you proceed?


ehild
 

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