# Contrapositive proof

1. Oct 4, 2008

### garyljc

1. The problem statement, all variables and given/known data
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 postive then x^n<y^n implies x<y

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

3. 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

2. Oct 5, 2008

### HallsofIvy

Staff Emeritus
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. $x^n \ge y^n$ implies that $x^n- y^n\ge 0$ and $x^n- y^n= (x- y)(x^{n-1}+ x^{n-2}y+ \cdot\cdot\cdot+ xy^{n-2}+ y^{n-2}$

Last edited: Oct 5, 2008
3. Oct 5, 2008

### 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 ?

4. Oct 5, 2008

### garyljc

i do not understand how did you arrive at this conclusion

5. Oct 5, 2008

### HallsofIvy

Staff Emeritus
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 "$x^n\ge y^n$".

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 (xn-1+ yxn-2+ ...+ xyn-2+ yn-1? And what would it say if n= 2?

6. Oct 6, 2008

### 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 (xn-1+ yxn-2+ ...+ xyn-2+ yn-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 ?

7. Oct 6, 2008

### HallsofIvy

Staff Emeritus
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)(xn-1+ xn-2y+ ...+ xyn-2+ yn-1. And, again, you can prove that by multiply the right hand side.

8. Oct 7, 2008

### garyljc

Thanks . Got it =)