Prove the following statement by considering two cases based on the parity principle. Use definitions of the terms but do NOT use any other facts previously proved in class or in the text or in the exercises. For all integers a and b, if a divides b then a² divides b²

Prove the given statement: The sum of an integer and its square is even.

The set S contains some real numbers, according to the following three rules.(i) 1/1 is in S.(ii) If a/b is in S, where a/b is written in lowest terms (that is, a and b have highest common factor 1), then b/2ais in S. (iii) If a/b and c/d are in S, where they are written in lowest terms, then a+c/b+d is in S.These rules are exhaustive: if these rules do not imply that a number is in S, then that number is not in S. Can you describe which numbers are in S? For example, by (i), 1/1 is in S. By (ii), since 1/1 is in S, 1/2·1is in S. Since both 1/1 and 1/2 are in S, (iii) tells us 1+1/1+2 is in S. What I have so far: Claim: Set S in contained in interval [½, 1] for a/b where 0<a≤b≤2a The reason is that 1/1 has this form and transformations preserve the property of being in this interval If a≤b≤2a, then b/2a obeys the requirement, since b≤2a≤2b And if a/b and c/d obey the requirement, then so does (a+c)/(b+d), since a+c≤b+d≤2a+2c=2(a+c) However, I feel there is still more to this…

Prove the following statement: P(A|B) := (P(A ∩ B)) / ( P(B) ) if able please write it step by step with some explanation, thank you in advance.