The Answer to the Question
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Here's the Solution to this Question
1) ab is odd, ab = 2k+1. Let's suppose that a(or b) is even, then a(or b) is equal to 2m. But a(or b) is a divisor of ab, so 2m is a divisor of 2k+1, it's impossible, so a(or b) cannot be even.
2) First of all, we have Cauchy's inequality:
(Equality, if )
3) It's easy to check that function is correct for n=1 and n=0, let's suppose that it works for n=N (factorial(N) = N!), now we have to prove that it works for n=N+1 (factorial(N+1) = (N+1)!).
Function factorial (N+1) will perform this line:
return (factorial (N))* (N+1)
So it'll return N! * (N+1) = (because factorial (N) is equal to N! - as we supposed) = (N+1)!
Function's correctness is proven using the induction.