Problem Solving

if you know 0< x < y, then is x^(1/n) < y^(1/n) for any number n?

Short answer:

If n > 0, yes.

If n < 0, no.

can you please elaborate why? thanks so much!

el_torero wrote:can you please elaborate why? thanks so much!

Sure sorry, didn't mean to be abrupt - thought you were looking for confirmation rather than explanation

In general, exponent inequalities like this one depend on more than just the value of n; they also depend on the values of x and y. In this case it only depends on n, as you can see in the examples below:

YES and NO are the answers to the question, Is x^(1/n) < y^(1/n)?

CASE 1: 0< x < y < 1; n > 0

eg. x = 0.1; y = 0.2; n = 0.5

=> x^(1/n) = 0.01; y^(1/n) = 0.04 => YES

eg. x = 0.01; y = 0.04; n = 2

=> x^(1/n) = 0.1; y^(1/n) = 0.2 => YES


CASE 2: 0 < x < y < 1; n < 0

eg. x = 0.01, y = 0.04, n = -2

=> x^(1/n) = 10; y^(1/n) = 5 => NO

eg. x = 0.1, y = 0.2, n = -0.5

=> x^(1/n) = 100; y^(1/n) = 50 => NO


CASE 3: 1 < x < y ; n > 0

eg. x = 2, y = 3, n = 0.5

=> x^(1/n) = 4; y^(1/n) = 9 => YES

eg. x = 4, y = 9; n = 2

=> x^(1/n) = 2; y^(1/n) = 3 => YES


CASE 4: 1 < x < y; n < 0

eg. x = 2, y = 3, n = -1

=> x^(1/n) = 1/2 ; y^(1/n) = 1/3 => NO

eg. x = 2, y = 3, n = -0.5

=> x^(1/n) = 1/4; y^(1/n) = 1/9 => NO

Abhay,

I've been reading a lot of your answers in different threads, and they are always very clear and elaborate; I'm learning a ton from reading them, so thanks a lot.

I see that your exam date is coming up. Good luck with that. I think you're aiming a little low relative to your skill level, but you're probably just being modest.

jeremy8 wrote:Abhay,

I've been reading a lot of your answers in different threads, and they are always very clear and elaborate; I'm learning a ton from reading them, so thanks a lot.

I see that your exam date is coming up. Good luck with that. I think you're aiming a little low relative to your skill level, but you're probably just being modest.

Thanks a lot Jeremy!

Lol @ the target score - I always aim way high; that way I might just hit something decent

jeremy8 wrote:Abhay,

I've been reading a lot of your answers in different threads, and they are always very clear and elaborate; I'm learning a ton from reading them, so thanks a lot.

I see that your exam date is coming up. Good luck with that. I think you're aiming a little low relative to your skill level, but you're probably just being modest.

I AGREE

el_torero wrote:if you know 0< x < y, then is x^(1/n) < y^(1/n) for any number n?

If n is positive x < y = YES
If n is negative y < x = NO