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For part a)
1 First, we need to prove that (ab)^n=(a^n)(b^n) for any n>0
2 Do this by induction
3 You can pick any base case>0, so let us pick n=1
4 LHS: (ab)^1=ab
5 RHS: (a^1)(b^1)=ab
6 RHS=LHS so base case is true
7 We now need to show providing, that P(k) is true, P(k+1) is also true
8 P(k): (ab)^k=(a^k)(b^k)
9 P(k+1):(ab)^(k+1)=(a^(k+1))(b^(k+1))
10 We assume P(k) is true
11 LHS: ((ab)^k)*(ab)
12 (a^k)(b^k)(ab)
13 (a^(k+1))(b^(k+1))
14 RHS: (a^(k+1))(b^(k+1))
15 Since LHS=RHS, P(k+1) is true, therefore P(k) is true
16 (ab)^n=(a^n)(b^n) is true for all n>0
17 sqrt(a)*sqrt(b)=sqrt(ab) can also be written as
18 (a^1/2)*(b^1/2)=(ab)^1/2 and we have proved the general case for this
19 by induction
20 Therefore for all positive a and b this identity holds