Proving the converse: if x^n<y^n then x<y for positive x and y

[garyljc]

Homework Statement

1st part , using induction to prvoe that if both x and y are positive then x<y implies x^n<y^n
2nd part, prove the converse, that if both x and y are positive then x^n<y^n implies x<y

Homework Equations

my question is more on the second part. I understand that I have to use the indirect proof using contrapositive.

The Attempt at a Solution

I started with this
If x and y are not positive then it does not imply the following x^n<y^n implies x<y

 

[HallsofIvy]

Homework Statement

1st part , using induction to prvoe that if both x and y are positive then x<y implies x^n<y^n
2nd part, prove the converse, that if both x and y are positive then x^n<y^n implies x<y

Homework Equations

my question is more on the second part. I understand that I have to use the indirect proof using contrapositive.

The Attempt at a Solution

I started with this
If x and y are not positive then it does not imply the following x^n<y^n implies x<y

First, what you have written here is irrelevant. The theorem (and therefore its contrapositive) talks only about positive numbers. "x and y both positive" is not part of the hypothesis, it is a "preliminary requirement" What is true if x and y are not positive does not matter. The theorem itself is simply "x^n< y^n implies x< y".

Are you clear on what the contrapositive is here? The contrapositive requires that the hypothesis and conclusion be negated and reversed. "x and y both positive" is not part of the hypothesis, it is a "preliminary requirement" that stays the same.

If x and y are positive numbers, then $x^n\ge y^n$ implies $x\ge y$.

Also, while you can prove it using induction, you don't have to. $x^n<br /> \ge y^n$ implies that $x^n- y^n\ge 0$ and $x^n- y^n= (x- y)(x^{n-1}+ x^{n-2}y+ \cdot\cdot\cdot+ xy^{n-2}+ y^{n-2}$

 

[garyljc]

i do understand contrapositive
but I'm new to it
does it mean that I'm suppose to prove if x^n < y^n is false therefore it implies that x < y is false ?
am i heading the correct direction ?

 

[garyljc]

Also, while you can prove it using induction, you don't have to. $x^n<br /> ge y^n$ implies that $x^n- y^n> 0$ and $x^n- y^n= (x- y)(x^{n-1}+ x^{n-2}y+ \cdot\cdot\cdot+ xy^{n-2}+ y^{n-2}$

i do not understand how did you arrive at this conclusion

 

[HallsofIvy]

i do understand contrapositive
but I'm new to it
does it mean that I'm suppose to prove if x^n < y^n is false therefore it implies that x < y is false ?
am i heading the correct direction ?

Yes, you are. Don't forget that you are assuming also that x and y are positive and that "x^n< y^n is false" is the same as "$x^n\ge y^n$".

i do not understand how did you arrive at this conclusion

What conclusion do you mean? If you are talking about the factoring (which is not a "conclusion"), what do you get if you multiply (x- y) and (x^n-1+ yx^n-2+ ...+ xy^n-2+ y^n-1? And what would it say if n= 2?

 

[garyljc]

What conclusion do you mean? If you are talking about the factoring (which is not a "conclusion"), what do you get if you multiply (x- y) and (x^n-1+ yx^n-2+ ...+ xy^n-2+ y^n-1? And what would it say if n= 2?[/QUOTE]


sorry i meant like who did you factor that up ?
is there a formula to help me with ?

 

[HallsofIvy]

Did you kow that x²- y²= (x- y)(x+ y)? How about x³- y³= (x- y)(x²+ xy+ y²)?

Yes, there is a formula: it is exactly what I gave: xⁿ- yⁿ= (x- y)(x^n-1+ x^n-2y+ ...+ xy^n-2+ y^n-1. And, again, you can prove that by multiply the right hand side.

 

[garyljc]

Thanks . Got it =)