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Proofs must assume that $0^0$ is undefined. Alas, I was able to find proofs only for integer exponents and real bases (by the way, considering that this is already proven, it can enough to prove the rules only for positive bases and fractional exponents, including exponents that are improper fractions). I specifically restricted bases to positive reals in order to avoid complex numbers and implicit divisions by zero.

I want to prove following rules:

1.$x^{a}x^{b}=x^{a+b}\space$

2.$\frac{x^{a}}{x^{b}}=x^{a-b}\space$

3.$(x^{a})^b=x^{ab}\space$

4.$(xy)^a=x^ay^a \space$

5.$(\frac{x}{y})^{a}=\frac{x^{a}}{y^{a}}\space$

KarmaPeasant
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    For those wondering if they've seen this question before, OP has asked similar ones here and here. They may not be duplicates, but they're worth mentioning for context to inform the reader of what OP has been told in comments and a previous answer. – Blue Mar 25 '22 at 18:41
  • @Blue By the way, I was unable to understand that previous answer by Paul Sinclair. This is why I restricted my appetite here, hoping to get answer that is in my reach – KarmaPeasant Mar 25 '22 at 18:46
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    That's fine by me. You should consider commenting on that answer to request clarification; you should also consider describing here (in the body of the question, not a comment) what you find incomprehensible about that answer, so that people don't waste time (theirs or yours) explaining things in the same way. – Blue Mar 25 '22 at 18:52
  • @Blue I'm afraid that I don't understand too much of that answer. – KarmaPeasant Mar 25 '22 at 18:58
  • @user161005 Welcome to MSE. You're expected to show your work on the problem. Your question should show that you put some effort on the question before asking for help. – jjagmath Mar 25 '22 at 20:33
  • @jjagmath You can see my efforts in linked questions – KarmaPeasant Mar 26 '22 at 02:41
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    @user161005 You mean the links that you didn't provide and other user have to make the effort to post in a comment? – jjagmath Mar 26 '22 at 12:02

1 Answers1

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For positive integer exponents and positive real bases $x$, rule (1) is just the observation that multiplying $1$ by $x$ $n$ times and then $m$ more times is the same as multiplying $1$ by $x$ $(m+n)$ times.

For positive integer exponents the other four rules follow from rule (1).

For example, (1) implies (2) because
$$ x^{a-b}x^b = x^{a-b+b} = x^a. $$

Then the definition of $x^a$ for other rational numbers is determined by the requirement that (1) stay true. That requirement forces $$ x^0 = 1 $$ because $$ x = x^1 = x^{1+0} = x^1x^0 = xx^0 $$ and you can cancel the nonzero $x$,

$$ x^{-a} = 1/x^a $$ because $$ 1 = x^0 = x^{-a+a} = x^{-a}x^a $$ and $$ x^{1/a} = \sqrt[a]{x} $$ because $$ (x^{1/a})^a = x^{a \times (1/a)} = x^1 = x. $$

Then the rest of the rules follow from rule (1) as before for all rational exponents. The arguments there depend only on the associative, commutative and distributive laws that are as true for rationals as they are for integers.

Ethan Bolker
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