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1. Let the statement be “If n is not an odd integer then square of n is not odd.”, then if P(n) is “n is an not an odd integer” and Q(n) is “(square of n) is not odd.” For direct proof we should prove _________a) ∀nP ((n) → Q(n))b) ∃ nP ((n) → Q(n))c) ∀n~(P ((n)) → Q(n))d) ∀nP ((n) → ~(Q(n)))

Question

  1. Let the statement be “If n is not an odd integer then square of n is not odd.”, then if P(n) is “n is an not an odd integer” and Q(n) is “(square of n) is not odd.” For direct proof we should prove _________a) ∀nP ((n) → Q(n))b) ∃ nP ((n) → Q(n))c) ∀n~(P ((n)) → Q(n))d) ∀nP ((n) → ~(Q(n)))
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Solution 1

For a direct proof of the given statement, we should prove option a) ∀nP ((n) → Q(n)). This is because the statement is a conditional statement that applies to all integers n (denoted by the universal quantifier ∀n), stating that if P(n) is true (n is not an odd integer), then Q(n) is also true ((square of n) is not odd).

Solution 2

For a direct proof of the given statement, we should prove option a) ∀nP ((n) → Q(n)). This is because the statement is a conditional statement that applies to all integers n (denoted by the universal quantifier ∀n), stating that if P(n) is true (n is not an odd integer), then Q(n) is also true ((square of n) is not odd).

Similar Questions

3. Let the statement be “If n is not an odd integer then sum of n with some not odd number will not be odd.”, then if P(n) is “n is an not an odd integer” and Q(n) is “sum of n with some not odd number will not be odd.” A proof by contraposition will be ________a) ∀nP ((n) → Q(n))b) ∃ nP ((n) → Q(n))c) ∀n~(P ((n)) → Q(n))d) ∀n(~Q ((n)) → ~(P(n)))

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