# How to prove that $a^n * b^n = (ab)^n$ [closed]

How do you prove that $$a^n * b^n = (ab)^n$$ This is a basic law of exponents used in calculation, but I am unable to prove it.

• How $*$ is defined ? Jun 25, 2019 at 10:59
• It is true iff the operation is commutative. E.g. note that in that case $(ab)^2=abab=aabb=a^2b^2$ Jun 25, 2019 at 11:04
• For positives $a$ and $b$ and real $n$ it's not so easy. Jun 25, 2019 at 12:17

Let G be an abelian group. Suppose $$a,b \in G$$. Then for an arbitrary natural number n, $$(ab)^n$$ = $$(ab)(ab)(ab).......(ab)$$ =$$(aa.......a)(bb......b)$$ =$$a^nb^n$$

For an integer $$n<0$$, you will require induction.

However, the issue with your question is that it requires context. Are you working in the real numbers under multiplication? Then it isn’t a group, but the property still holds. If not, and you’re working in a group, then the group needn’t be an abelian group for the property to hold. The property still holds if $$a,b$$ commute.

• What happens if $n$ is a real number? If you'll see don-voting it's not mine. Jun 25, 2019 at 14:40
• Doesn’t necessarily hold when a,b are complex. More information is required.
– user643073
Jun 25, 2019 at 14:48
• It's exactly, which I say. What happens if $n$ is a real number and $a$ and $b$ are positives? I think in this case there is a very big problem with your reasoning. See please my comment two hours ago. Jun 25, 2019 at 14:50
• If a is real and n is real then it really depends on how you define $a^n$
– user643073
Jun 25, 2019 at 16:28
• I think all these are your work. But a big problem, I think, it's a proof. Jun 25, 2019 at 17:36

Hints: use induction and the commutativity of the multiplication.

• I agree that this is an answer, but I suppose you could have just commented on the original post... Jun 25, 2019 at 11:15
• Thank you, this helped me solve this myself. Jun 25, 2019 at 11:15
• @PratikApshinge you don't get $50$k points by commenting Jun 25, 2019 at 11:38