Solution to 1. (i) Prove that if m and n are integers and mn is even, then … - Sikademy
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Archangel Macsika

1. (i) Prove that if m and n are integers and mn is even, then m is even or n is even. (ii) Show that if n is an integer and n3 + 5 is odd, then n is even using (a) a proof by contraposition (b) a proof by contradiction (iii) Prove that if n is an integer and 3n + 2 is even, then n is even using (a) a proof by contraposition (b) a proof by contradiction (iv)ProvethepropositionP(0),whereP(n)istheproposition”ifnisapositiveintegergreater than 1, then n2 > n.” What kind of proof did you use? (v) Prove the proposition P(1), where P(n) is the proposition ” If n is a positive integer, then n2 ≥ n.” What kind of proof did you use?

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1.(i) Suppose mn is even. Then we can choose an integer k such that mn=2k. If m is even, then there is nothing more to proove, so suppose m is odd. Hence m=2j+1 for some integer j. We get


(2j+1)n=2k2jn+n=2kn=2(k-jn)

Since k-jn is an integer, it follows that n is even.


(ii)

a) a proof by contraposition

The contrapositive is “If n is odd, then n^3+5 is even.” Assume that n is odd. We can now write n=2k+1 for some integer k. Then


n^3+5=(2k+1)^3+5=8k^3+12k^2+6k+1+5==8k^3+12k^2+6k+6=2(4k^3+6k^2+3k+3).

Thus n^3+5 is two times some integer, so it is even by the definition of an even integer.

b) a proof by contradiction

Suppose that n^3+5 is odd and that n is odd. Since n is odd, the product of odd numbers is odd. So we can see that n^3 is odd. But if we subtract 5, then the difference between the two odd numbers n^3+5 and n^3 is even. Thus, our assumption was wrong and it is a contradiction. 


Therefore if n is an integer and n^3+5 is odd, then n is even.


(iii)

(a) a proof by contraposition 

(Contrapositive: If n is even, then 3n+2 is even) Suppose that the conclusion is false, i.e., that n is even. Then n=2k for some integer k.

Then 3n+2=3(2k)+2=6k+2=2(3k+1).

Thus 3n+2 is even, because it equals 2j for an integer j=3k+1. So 3n+2 is not odd. 

b) a proof by contradiction

Assume that the conclusion is false, i.e., that n is even, and that 3n+2 is odd. Then n=2k for some integer k and 3n+2=3(2k)+2=6k+2=2(3k+1).

Thus 3n+2 is even, because it equals 2j for an integer j=3k+1.This contradicts the assumption “3n+2 is odd”. This completes the proof by contradiction, proving that if 3n+2 is odd, then n is odd. 


(iv)

Note that P(0) is that "If is a positive integer greater than 1, then 0^2>0 ".

Since " is a positive integer greater than 1 " is false the the overall statement is true.

This is a vacuous proof.


(v)

Note that P(1) is that "If 1 is a positive integer, then 1^2\geq1".

Since 1^2=1, then P(1) is true.

This is a trivial proof.

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Question ID: mtid-5-stid-8-sqid-3890-qpid-2589