Please complete all 14 Claim For all integers p and q if 2p

Please complete all

#14. Claim: For all integers p and q, if 2(p + q) is a multiple of 4, then p and q are even. Consider the following \"proofs\" of the claim.

Proof A: Suppose p and q are any even integers. By definition of even, integer n such that p = 2n and q = 2n. Then 2(p+q) =2 (2n+2n)= 2(4n)= 4(2n). Let k= 2n, which is an integer. Thus, 2(p + q) = 4k for some integer k, and by definition of multiple, 2(p + q) is a multiple of 4.

Proof B: Suppose p and q are any even integers. By definition of even, integer m such that p = 2m and integer n such that q = 2n. Then 2(p+q) = 2(2m+2n)= 4(m+n). Let k= m+n, which is an integer. Thus, 2(p + q) = 4k for some integer k, and by definition of multiple, 2(p + q) is a multiple of 4.

Proof C: (By contraposition; i.e., proving the contrapositive) Suppose p is not an even integer or q is not an even integer. So, suppose p or q is an odd integer. Without loss of generality, suppose p is odd and q is even (otherwise, just switch the roles of p and q). We want to show that 2(p + q) is not a multiple of 4. By definition of odd, integer m such that p= 2m+1 and integer n such that q=2n Then 2(p+q) = 2(2m+1+2n)= 4m + 2 + 4n =4(m+n) +2. Let k= m+n, which is an integer. So, 2(p+q)= 4k+2 for some integer k, but since 2 is not a multiple of 4, the quantity 2(p+q) can not be a multiple of 4.

Proof D: Suppose p and q are integers and 2(p + q) is a multiple of 4. By definition of multiple of 4, integer k such that 2(p + q) = 4k = 2(2k)= 2(2(m + n)) = 2(2m + 2n) for some integers m and n. Then p= 2m and q= 2n for some integers m and n. By definition of even, p and q are even integers.

INSTRUCTIONS:

(a) Critique each proof (A, B, C, D). For each proof, is it a logically valid argument proving the claim? What are the flaws, if any?

(b) Is the Claim true or false? Explain.

Solution

our claim is \" if 2(p+q) is multiple of 4 then p and q are even

and in proof A and B they already supposed that p and q are even this is not correct .

in proof C contraposition (if p the q is equivalent to if not q than not p)

so if p or q are not even then 2(p+q) is not multi[ple of 4

in this one case if p and q both are odd then its not true p=2k+1,q=2k\'+1 p+q= 2(k+k\')+2

2(p+q) =4(k+k\'+1) is multiple of 4

D ) second line 2(p+q)=2(2m+2n) implies only p+q= 2m+2n not p=2m and q=2n

this also not correct

and the given statement is not correct you can see by counter example take p=3,q=5 any integers

2(p+q)=16=4(4) multiple of 4 bur p and q are not even