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

[calios]

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 ^_^

 

[ystael]

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 $$-1$$ reverses inequalities, i.e., if $$A < B$$ then $$-B < -A$$.

is proved is correct ?

You are missing the case where $$x < 0 < y$$.

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 $$x$$ and $$y$$, and you should see what it is.

 

[Mark44]

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.

 

[calios]

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



You are missing the case where $$x < 0 < y$$.



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

yes ,thanx for correcting

 

[jegues]

Couldn't you prove these by the process of mathematical induction as well :3 ?