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.
$\begingroup$
$\endgroup$
3
-
$\begingroup$ How $*$ is defined ? $\endgroup$– user659895Jun 25, 2019 at 10:59
-
1$\begingroup$ It is true iff the operation is commutative. E.g. note that in that case $(ab)^2=abab=aabb=a^2b^2$ $\endgroup$– drhabJun 25, 2019 at 11:04
-
$\begingroup$ For positives $a$ and $b$ and real $n$ it's not so easy. $\endgroup$– Michael RozenbergJun 25, 2019 at 12:17
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
5
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.
-
$\begingroup$ What happens if $n$ is a real number? If you'll see don-voting it's not mine. $\endgroup$ Jun 25, 2019 at 14:40
-
$\begingroup$ Doesn’t necessarily hold when a,b are complex. More information is required. $\endgroup$– user643073Jun 25, 2019 at 14:48
-
$\begingroup$ 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. $\endgroup$ Jun 25, 2019 at 14:50
-
$\begingroup$ If a is real and n is real then it really depends on how you define $a^n$ $\endgroup$– user643073Jun 25, 2019 at 16:28
-
$\begingroup$ I think all these are your work. But a big problem, I think, it's a proof. $\endgroup$ Jun 25, 2019 at 17:36
$\begingroup$
$\endgroup$
3
Hints: use induction and the commutativity of the multiplication.
-
$\begingroup$ I agree that this is an answer, but I suppose you could have just commented on the original post... $\endgroup$– PCeltideJun 25, 2019 at 11:15
-
$\begingroup$ Thank you, this helped me solve this myself. $\endgroup$ Jun 25, 2019 at 11:15
-
$\begingroup$ @PratikApshinge you don't get $50$k points by commenting $\endgroup$ Jun 25, 2019 at 11:38