# How to prove x^a >= 1?

1. Feb 11, 2012

### julypraise

(Sorry for the wrong title!)

Let a>1 (a in R). Then how do you prove for all x in R, a^x > 0 ?

And also that a^y>1 for all y>0, how do you prove?

And also how do you prove that 0^b = 0 when b is a real number?

(For me this is so difficult. So please just help me. Enough hints will suffice, no full solution.)

Last edited by a moderator: Feb 22, 2012
2. Feb 11, 2012

### micromass

Staff Emeritus
This depends on what you can use. I would prove it by the following scheme:

1) First prove it for all natural numbers x (by induction).

2) Use $a^{-x}=(a^x)^{-1}$ to prove it for all integers.

3) Use $a^{m/n}=\sqrt[n]{a^m}$ to prove it for all rational numbers.

4) Use continuity of $a^x$ to prove it for all real numbers.

3. Feb 11, 2012

### julypraise

Could you help me bit more on proving the part 3)?

So by the theorem in Rudin's Mathematical Analysis in page 10 thm 1.21, I can prove this right?

Last edited by a moderator: Feb 22, 2012
4. Feb 12, 2012

### checkitagain

b must be restricted, as in b > 0, so that 0^b is not

undefined and/or is not indeterminate.

Last edited by a moderator: Feb 22, 2012