Answer to Question #191561 in Real Analysis for Amara

Question

For each of the following statements give the converse, the contrapositive and the negation of the statement.

(i) I take plastic bags when I go shopping.

(ii) x∈Bimpliesx∈/Xorx∈/Y.

(iii) Ifx∈A∩Bthenx∈Aand x∈B.

Let S = {a1, a2, a3, ...an}be a set of test scores. Prove using the the indirect method of proof

that if the average of this set of test scores is greater than 90, then at least one of the scores is greater than 90.

Answer 1

i)

converse: I go shopping when I take plastic bags.

contrapositive: I do not go shopping when I do not take plastic bags.

negation: I take plastic bags and I do not go shopping.

ii)

converse: x∉X or x∉Y implies x∈B

contrapositive: x∈X and x∈Y implies x∉B

negation: x∈B implies x∈X and x∈Y

iii)

converse: If x∈A and x∈B then x∈ A∩B

contrapositive: If x∉A or x∉B then x∉A∩B

negation: x∈ A∩B and x∉A or x∉B

We can use contrapositive rule to prove it,

for example P$\rightarrow Q$ ,the contrapositive of this will be $\sim Q\rightarrow \sim P$

we can covert " if the average of this set of test scores is greater than 90, then at least one of the scores is greater than 90 "into proposition logic $P\rightarrow Q$ contrapositive of this will be $( \sim Q\rightarrow \sim P )$ both are equal.

so we can prove it through contrapositive rule "none of them scores is greater than 90 then the average of this set of test scores is not greater than 90"

means $a_{1}< 90,a_{2}< 90,a_{3}< 90,.....a_{n}< 90\: (\text{ none of them is greater than 90)}$

then sum of these will be $a_{1}+a_{2}+a_{3}+.....a_{n}< 90n \text{ and average will be } \dfrac{a_{1}+a_{2}+a_{3}+.....a_{n}}{n}< \frac{90n}{n}$

Hence the average will be les than 90