# 0<a^(1/2)<b^(1/2)

• October 29th 2008, 05:43 PM
kathrynmath
0<a^(1/2)<b^(1/2)
If 0<a<b, prove 0<a^(1/2)<b^(1/2).

I'm not sure where to even start except that to assume 0<a<b.
Where do i go from here?
• October 29th 2008, 05:59 PM
Jhevon
Quote:

Originally Posted by kathrynmath
If 0<a<b, prove 0<a^(1/2)<b^(1/2).

I'm not sure where to even start except that to assume 0<a<b.
Where do i go from here?

yes, i am not sure if your professor expects you to use special axioms here or what. but i would just note that the square root function is strictly increasing. we can see this since the derivative is always positive (where defined, it is not defined at zero, but we are not considering zero, so that's ok). thus, for any two nonzero numbers $a$ and $b$ in the domain of $\sqrt{x}$, we have $\sqrt{b} > \sqrt{a}$, provided $b > a$
• October 29th 2008, 06:18 PM
kathrynmath
Quote:

Originally Posted by Jhevon
yes, i am not sure if your professor expects you to use special axioms here or what. but i would just note that the square root function is strictly increasing. we can see this since the derivative is always positive (where defined, it is not defined at zero, but we are not considering zero, so that's ok). thus, for any two nonzero numbers $a$ and $b$ in the domain of $\sqrt{x}$, we have $\sqrt{b} > \sqrt{a}$, provided $b > a$

Ok, that makes sense. Is there another way to do it kinda like I did with a similar problem?

here is my problem and what I did:
If 0<a<b, prove that 0<a^2<b^2
0*a<a*a<a*b
and 0*b<a*b<b*b
0<a^2<ab and 0<ab<b^2
0<a^2<ab<b^2
0<a^2<b^2

Or would doing something your way be better?
• October 29th 2008, 07:03 PM
Jhevon
Quote:

Originally Posted by kathrynmath
Ok, that makes sense. Is there another way to do it kinda like I did with a similar problem?

here is my problem and what I did:
If 0<a<b, prove that 0<a^2<b^2
0*a<a*a<a*b
and 0*b<a*b<b*b
0<a^2<ab and 0<ab<b^2
0<a^2<ab<b^2
0<a^2<b^2

Or would doing something your way be better?

you can use the method here. but the only way i see it working is to use contradiction

let $x^2 = a$ and $y^2 = b$

then, you are asked to prove: $0 < x^2 < y^2 \implies 0 < x < y$ (we are not considering negative numbers here, of course)

thus, assume to the contrary that $0 < x^2 < y^2$ but $0 < y < x$

then in much the same way as you did the last problem, you can show that that would mean $0 < y^2 < x^2$, but that is contrary to our assumption
• October 29th 2008, 07:10 PM
kathrynmath
Quote:

Originally Posted by Jhevon
you can use the method here. but the only way i see it working is to use contradiction

let $x^2 = a$ and $y^2 = b$

then, you are asked to prove: $0 < x^2 < y^2 \implies 0 < x < y$ (we are not considering negative numbers here, of course)

thus, assume to the contrary that $0 < x^2 < y^2$ but $0 < y < x$

then in much the same way as you did the last problem, you can show that that would mean $0 < y^2 < x^2$, but that is contrary to our assumption

Yeah, that does make sense. i think what you originally said amkes more sense to me, but thanks for showing the alternate way, also.