What is the proof for xn<yn when x<y and n is odd?

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Homework Help Overview

The discussion revolves around proving the inequality \( x^n < y^n \) given that \( x < y \) and \( n \) is an odd integer. Participants explore various cases, including when \( x \) and \( y \) are both positive, both negative, or when one is negative and the other is positive.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the validity of the proof for different ranges of \( x \) and \( y \), particularly focusing on cases where both are negative or where one is negative and the other is positive. There is also a mention of mathematical induction as a potential method for proof.

Discussion Status

The conversation is active, with participants questioning the completeness of the original proof and exploring different scenarios. Some have offered corrections and clarifications regarding the implications of multiplying by negative numbers and the specific conditions under which the original statement holds.

Contextual Notes

There is a specific focus on the case where \( n \) is odd, and participants note that the statement does not apply to even \( n \). Additionally, the discussion highlights the need to consider the case where \( x < 0 < y \).

calios
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Homework Statement



hy guys ,help me

prove if x<y and n is odd ,then xn<yn
i have solved this . .but a question in below


Homework Equations





The Attempt at a Solution



if 0=or<x<y then xn<yn (i have proved this ) and if x<y<or=0 we can say 0=or<-y<-x it implies -yn<-xn last we can add both side with +xn and +yn we have xn<yn ..
is proved is correct ? and what about if x<y<or=0 ( n is even ) how to prove this ?

thanx before ^_^
 
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calios said:
last we can add both side with +xn and +yn we have xn<yn

This is correct, or you can also use the fact that multiplying by [tex]-1[/tex] reverses inequalities, i.e., if [tex]A < B[/tex] then [tex]-B < -A[/tex].

calios said:
is proved is correct ?

You are missing the case where [tex]x < 0 < y[/tex].

calios said:
and what about if x<y<or=0 ( n is even ) how to prove this ?

In this case, the correct statement is different; try some values for [tex]x[/tex] and [tex]y[/tex], and you should see what it is.
 
calios said:
and what about if x<y<or=0 ( n is even ) how to prove this ?
You don't need to prove the statement for even n. The statement you're trying to prove explicitly says that n is odd.
 
ystael said:
This is correct, or you can also use the fact that multiplying by [tex]-1[/tex] reverses inequalities, i.e., if [tex]A < B[/tex] then [tex]-B < -A[/tex].



You are missing the case where [tex]x < 0 < y[/tex].



In this case, the correct statement is different; try some values for [tex]x[/tex] and [tex]y[/tex], and you should see what it is.

yes ,thanx for correcting :-p
 
Couldn't you prove these by the process of mathematical induction as well :3 ?
 

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