Thread: 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?

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 and in the domain of , we have , provided

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 and in the domain of , we have , provided

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?

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 and

then, you are asked to prove: (we are not considering negative numbers here, of course)

thus, assume to the contrary that but

then in much the same way as you did the last problem, you can show that that would mean , but that is contrary to our assumption

Originally Posted by Jhevon

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

let and

then, you are asked to prove: (we are not considering negative numbers here, of course)

thus, assume to the contrary that but

then in much the same way as you did the last problem, you can show that that would mean , 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.