Proving x<y x^n<y^n if n is Odd

  • Thread starter Ronnin
  • Start date
In summary, the conversation is discussing how to prove the statement x<y x^{n}<y^{n} if n is odd. The participants suggest breaking the proof into different cases or using the technique of induction. One participant also suggests proving the statement for positive numbers first and then using the equality (-x)n=-xn if one of the numbers is negative. Another participant recommends reading "An Introduction to Inequalities" by Bellman for help with proofs in Spivak's chapter 1. They also mention that the trichotomy axiom can provide a hint for the proof.
  • #1
Ronnin
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1

Homework Statement



x<y [tex]x^{n}[/tex]<[tex]y^{n}[/tex] if n is odd

Homework Equations





The Attempt at a Solution


I am trying to prove this. I understand that if n were even it would eliminate any negatives for negative values of x or y, but how would I go about writing the proof for this. Would I break it down into different cases? Could I define what odd number is and just somehow state that because of that there would not be a possible sign change for x or y. I'm new to proofs so translating some of the most ovious stuff into mathmatical terms is tricky for me at times.
 
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  • #2
I would break it to pieces. Consider 0<x<y, x<0<y and x<y<0.

Alternatively, you could also proceed by induction, if you know this technique...
 
  • #3
Have not started using induction yet. I am still muddling along with these basic direct proofs. Thanks for the suggestion.
 
  • #4
This problem has been giving me hell. I have used 2k+1 for n and have tried to factor it out giving this [tex](x^{k}x^{1/2}+y^{k}x^{1/2})(x^{k}x^{1/2}-y^{k}x^{1/2})[/tex] which is just one big circle because this only yeilds a [tex]\sqrt{x^{n}}[/tex]= [tex]\sqrt{y^{n}}[/tex]. I have even tried using the difference of cubes to attempt to factor this out with no luck. Could someone drop a hint as to whether I'm way off beat here or I'm missing a logical detail that would let me make some kind of algebraic magic happen. Thanks!
 
  • #5
You could maybe start by proving: If x and y are positive and x<y, then xn<yn.
You can do this by induction.
This would show the result for positive numbers. If one of the numbers is negative, then you could perhaps use the equality (-x)n=-xn if n is odd...
 
  • #6
Since this is from Spivak chapter 1 I don't think I can use induction. I'm new to proofs and have never used that technique.
 
  • #7
Honestly if you study the chapter deeply enough you'll see that the way to prove this
example is given to you. One thing I was really really confused about with Spivak in
this chapter was his explanation of the Trichotomy Law, P10, & didn't really get it
until I read "An Introduction to Inequalities" by Bellman. I would seriously advise reading
that little book to learn how to deal with some of the proofs in chapter 1 of Spivak first but if
you can't do that then yeah the material is contained in the chapter & the trichotomy
axiom is a big hint as are the responses in this thread :wink:
 

Related to Proving x<y x^n<y^n if n is Odd

1. What is the meaning of "proving x

This phrase refers to a mathematical concept known as the power inequality, which states that if two numbers, x and y, are both positive and x is less than y, then raising them to an odd power will still maintain the inequality, meaning that x^n will be less than y^n.

2. Why is it important to prove this inequality?

Proving this inequality is important because it helps to establish a fundamental mathematical concept that is used in many other areas of mathematics, such as calculus and algebra. It also allows us to make accurate predictions and solve complex equations involving powers.

3. What is the proof for this inequality?

The proof for this inequality can be easily demonstrated using mathematical induction. It involves showing that the inequality holds for n=1 (the base case) and then assuming it holds for n=k and proving that it also holds for n=k+1 (the inductive step).

4. Can this inequality be applied to negative numbers?

No, this inequality only applies to positive numbers. If the numbers are negative, the inequality would be reversed, meaning that x^n would be greater than y^n if x is less than y.

5. How is this inequality related to even powers?

This inequality only applies to odd powers, but it can still be useful when dealing with even powers. For example, if we know that x^n is less than y^n for odd n, we can deduce that x^(n+1) will also be less than y^(n+1), even though the powers are now even. This is because (x^n)^2 is still less than (y^n)^2, and taking the square root of both sides maintains the inequality.

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