im working on an simple question as: Using induction, prove that for all complex numbers a and b and for all natural numbers $m$ and $n$, we have $(ab)^n = a^n \cdot b^n $and also $(a^m )^n = a^{mn}$ . If $a$ and $b$ are nonzero, prove that these equations hold even if $m = 0$ or $n = 0$.
and for the first part, prove that for all complex numbers $a$ and $b$ and for all natural numbers $m$ and $n$, we have $(ab)^n = a^n \cdot b^n$ and also $(a^m )^n = a^{mn}$ . i believe i have no problem to prove that for all natural numbers $m$ and $n$ by induction but im very not sure does the induction axiom work on complex number(ie: should i some how prove $M = \{a,b∈ C | ...\} = C$ or something else, like, can i just do $M =\{m,n ∈ N | \forall a,b ∈ C ....\} = N)$ and just to make the question a little more valuable, can we write induction for complex number? if so, how should we do it?
by the remind from @peek-a-boo , here is the solution i come up with:
Let $M =\{n ∈ N | \forall (a,b) ∈ C , (ab)^n = a^n \cdot b^n\}$
base case:
let $ n = 1, so (ab)^1 = ab $ which satisfy $M$
induction step:
Assume $ (ab)^n = a^n \cdot b^n$
prove $(ab)^{n + 1} = a^{n + 1} \cdot b^{n +1}$:
$ (ab)^{n+1} = (ab)^n \cdot ab $ and from assumption,
$(ab)^(n+1)=(ab)^n⋅ab=a^nb^n⋅ab$
which is the same as: $(ab)^n \cdot ab = a^n \cdot a \cdot b^n \cdot b $
Which shows $(ab)^{n + 1} = a^{n + 1} \cdot b^{n +1}$
Thus, we concludes that $(ab)^n = a^n \cdot b^n \implies (ab)^{n + 1} = a^{n + 1} \cdot b^{n +1}$
And by induction now $M = N$
For $(a^m )^n = a^{mn}$ : Let $L =\{m,n ∈ N | \forall a ∈ C , (a^m )^n = a^{mn}\}$ base case: Let $ m = 1$ and $ n = 1$ thus we have: $(a^1)^1 = a^1$ which satisfy the base case. induction step: recall that in this case, all natural number must be $ 1>=$, so if $ n> 1, n =1 + q , q ∈ N$ . and if $ m> 1, m = 1 + p, p ∈ N$
Assume $(a^{1 + q})^{1 + p} = a^{(1 +q)(1 + p)}$,
prove $(a^{(1 + q) +1} )^{(1 + p)+1} = a^{((1 +q)+1)((1 + p)+1)}$
i will jump this simple math since im very lazy, thank you :)
Thus we concludes that $ (a^m )^n = a^{mn} \implies (a^{m +1} )^{n +1} = a^{(m+1)(n+1)}$
and from induction, we have$ L = N$ as we desire
Q.E.D
Given suggestion from @ CyclotomicField , i will try the second part again by using the first part:
For $(a^m )^n = a^{mn}$ : Let $L =\{m ∈ N | \forall a ∈ C , \forall n ∈ N a^m )^n = a^{mn}\}$ (which n ∈ Nis given by the first part) base case: Let $ m = 1$ thus we have: $(a^1)^n = a^n$ which satisfy the base case. induction step: Assume $(a^m )^n = a^{mn}$ for $m$ is any given natural number, $U$. prove $(a^(U+1) )^n = a^{(U+1)n}$: $(a^(U+1) )^n = (a^(U)(a) )^n $ which is the same as $(a^(U)^n(a)^n$ #recall M now from assumption, we have: $(a^(U)^n(a)^n = a^{Un}(a)^n $ Thus $(a^(U)^n(a)^n = a^{nU}(a)^n = a^{n(U+1)} = a^{(U+1)n}$ from above we concludes that : $(a^m )^n = a^{mn} => (a^(m+1) )^n = a^{(m+1)n}$ which by induction, shows that $ L = N$ as we desire Q.E.D
Also, recall what @ peek-a-boo has just say that is :induction itself is a principle about natural numbers, not the "property" itself. which my argue is that should induction work on any system that is close under successor or is that in this case all system that is close under successor was being defined as natural number? if the first case holds, than can we not some how defined complex number be a system under successor and drive it though induction? thank you so much for everyone that's been helping!