# Is this an axiom or does it have to be proved? a and b are positive integers and x is greater than 1. In Rudin principles of real analysis it is not given as an axiom but proving is seems difficult to me

• I suggest you read Ethan Bloch's Real Numbers and Real Analysis. It will save your live. – Eric Sep 2 '17 at 11:41
• Does he prove it? – roglol Sep 2 '17 at 11:42
• How is $x^a$ defined? Is it $e^{a\ln x}$? – lulu Sep 2 '17 at 11:43
• certain rules only apply in certain domains as well. – user451844 Sep 2 '17 at 11:44
• I think it is so obvious in the case of integers it's almost not worth writing - if you see what each means. E.g. x^(2*3) = x^(6) = xxxxxx = (xx)(xx)(xx) = (x^2)^3. It's pretty apparent that they're the same based on the definition of exponents for positive integers. Though if you want a more formal proof you could use induction – Franz Sep 2 '17 at 11:54

The base case is $x^{1\cdot b}=x^b=(x^1)^b$.
Then, if $x^{a\cdot b}=(x^a)^b$, $x^{(a+1)\cdot b}=x^{ab+b}=x^{ab}x^b=(x^a)^bx^b=(x^ax)^b=(x^{a+1})^b$ where we used $x^{a+b}=x^ax^b$ and $x^by^b=(xy)^b$.
This proves the identity for natural numbers. After you define the exponents of the form $\frac{1}{n}$ (this will be done in Theorem 1.21), you can expand this to the rational numbers and real exponents will be defined in the exercises (ex. 6).
• +1 This is the proof for positive integers $a,b$. Also works for any $x$, not merely $x \ge 1$. It seems the OP is for positive integers, so your last paragraph is just a bonus, not required for this question. – GEdgar Sep 2 '17 at 12:10