If x^n=y^n and n is odd, then x=y.

[wifi]

This is all seems fairly obvious to me and proving it is a bit awkward. In the previous exercise we proved that if x<y and n is odd, then x^n<y^n.

Spivak argues that, from the previous exercise, x<y would imply that x^n<y^n and y<x would imply that y^n<x^n. That's his whole proof. Is he simply saying that, by analogy, this relationship must hold?

I mean qualitatively it makes sense, because anything raised to an odd exponent retains it's original sign. Thus for x^n=y^n to hold, either x=y or -x=-y (which is obviously the same as x=y).

 

[gerben]

if x is not equal to y then either x<y or y<x which both imply that x^n is not equal to y^n

 

[HallsofIvy]

No. What he is doing is using "trichotomy", that, for any numbers x, y, one and only one must hold:
x= y
x< y
y< x. If x is not equal to y then either x< y or y< x. If x< y then $x^2< y^2$, contradicting "$x^2= y^2$ so that is not the case. If y< x then $y^2< x^2$, contradicting "$x^2= y^2$. I presume that Spivak feels the "y< x" case is so similar to the "x< y" case that the reader could see that for himself.

 

[wifi]

No. What he is doing is using "trichotomy", that, for any numbers x, y, one and only one must hold:
x= y
x< y
y< x. If x is not equal to y then either x< y or y< x. If x< y then $x^2< y^2$, contradicting "$x^2= y^2$ so that is not the case. If y< x then $y^2< x^2$, contradicting "$x^2= y^2$. I presume that Spivak feels the "y< x" case is so similar to the "x< y" case that the reader could see that for himself.

That makes perfect sense. I guess I'm not quite up to the level of the average reader then. My apologies.

Thanks guys.