Answer c,d, AND f. Please don\'t post incomplete, I will give thumbs down. Thank

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Answer c,d, AND f. Please don't post incomplete, I will give thumbs down. Thank you.

Each of the following is a proposed"proof" of a "theorem." However the "theorem" may not be a true statement, and even if it is, the "proof" may not really be a proof. You should read each "theorem" and "proof" carefully and decide whether or not the "theorem" is true. Then:
•If the "theorem" is false, find where the "proof" fails
•If the "theorem" is true, decide and state whether or not the "proof" is correct. If it is not correct, find where it fails.

c) “THEOREM": For any sets A and B, if g (A) n g (B) , then A n B . "Proof": We will prove this by contradiction. Assume A and B are sets, and to get a contradiction, assume A and B are not disjoint. Then there is some element x in A nB. So () E P(A) n P(B), which means that P(A) n P(B) is not empty, and we are done d) “THEOREM": For a, b, c E N, if alb and alc, then a l (bc) 99. "Proof": Assume that the hypothesis is true. Then, by definition of divisibility, there

Explanation / Answer

c)The proof for C is correct

Let us assume that sets are not disjoint, then A (int) B is not equal to phi, which means there will be an element x that will belong to A int B

Since x belongs to A and B, hence the {x} will belong to P(A) int P(B) which means the set P(A) int P(B) will not be empty, hence by contradication we can say that A int B must not be phi

d)The proof for D is correct  

Since a divides b, then we can write it as

b = k1a, where k1 is an natural number----- (i)

Since a divides c, hence we can write it as

c = k2a, where k2 is an natural number ----- (ii)

Now from (i) and (ii) we can write

bc = (k1a)(k2a) = k1k2a^2 = (k1k2a)(a)

Since k1,k2 and a are all natural numbers, hence there product k1k2a will also be a natural number

Therefore we can write bc = k3a, where k3=k1k2a

Hence we can say that a divides bc

Therefore the statement is TRUE

f)

The proof is FALSE

Since ac divides bc, hence we can write

bc = k1ca, since a,b and c belongs to natural number, cancelling the value c from both sides we can say that

b = k1a

Therefore, a divides b or we can say that a|b

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