1. ## irrational exponents

hi,

i am trying to prove that the property x<y implies a^x<a^y for any real numbers x and y. i am having a little trouble using two sequences of rational numbers converging to x and y respectively. using some analysis and the theory of convergent sequences i can show that a^x<=a^y but i cannot show STRICT inequality. could someone help me with this? thanks!

2. Originally Posted by kkoutsothodoros
hi,

i am trying to prove that the property x<y implies a^x<a^y for any real numbers x and y. i am having a little trouble using two sequences of rational numbers converging to x and y respectively. using some analysis and the theory of convergent sequences i can show that a^x<=a^y but i cannot show STRICT inequality. could someone help me with this? thanks!
I have an easy proof but I just do not know how you defined exponents. The way I define exponents (let $a\in \mathbb{R}^+$) is by $a^x = e^{x\log a}$. (The exponential function is defined to be the inverse of the logarithm function, and the logarithm function is defined through an integral). Now if $a>1$ then $\log a >0$. Therefore, if $x>y \implies x\log a > y\log a$ which means $e^{x\log a} > e^{y\log a}$ since the regular exponential is an increasing function and so by definition $a^x > a^y$.

3. hi, thanks! i am familiar with that definition of exponents and it does make things much easier to deal with.

the definition i am using for a^x is the limit of a^x_n where x_n is a sequence of rational numbers. i am using the fact that every real number is the limit of a sequence of rational numbers.

thanks!

4. Originally Posted by kkoutsothodoros
the definition i am using for a^x is the limit of a^x_n where x_n is a sequence of rational numbers. i am using the fact that every real number is the limit of a sequence of rational numbers.
To prove the strict inequality using that (horrible!) definition, let s and t be rational numbers with x<s<t<y. You then know that $a^s from properties of rational numbers.

If $(x_n),\ (y_n)$ are sequences of rational numbers with limits x, y respectively, then for all sufficiently large n it will be true that $x_n and $t. Assuming you can prove that this implies $a^x\leqslant a^s$ and $a^t\leqslant a^y$, it then follows that $a^x\leqslant a^s.

5. thanks!

and yes it is a horrible way to define exponents!