(Quantifiers.) (a) Without using any words of negation (i.e. without using the w

Question

(Quantifiers.) (a) Without using any words of negation (i.e. without using the words not, no, nor, neither,. . . ), write down a sentence that describes the negation of the following: “If a book on my bookshelf has a page with more than fifty words on it, then the first letter of every word on that page is a vowel.”

(b) A famous quote of Abraham Lincoln goes as follows: “You can fool all the people some of the time, and some of the people all the time, but you cannot fool all the people all the time.” Let P be the set of all people, T be the set of all times, and f(p, t) a function that takes in a person p and time t, and outputs either “fooled by you” or “not fooled by you.” Write this statement using quantifiers and logical/set theory notation.

(c) Let S be a subset of R. Consider the following two propositions: • A: There is a real number M such that for any x S, we have |x| M. • B: For any x S, there is a real number M such that |x| M. Write both of these statements using quantifiers. Is is true that A B? Is is true that B A?

Explanation / Answer

(a)-If a book on my bookshelf has a page with more than fifty words on it, then the first letter of every word on that page isn't a vowel.

(b) : (p)(t)(f(p, t)) (p)(t)(f(p, t)) ¬(p)(t)(f(p, t))

(c)- . true that A B (by Archimedian Property).

Then there exists x R such that |x| M for all x R. This means that the set R is bounded from above. Since the set R is non-empty (1 R), the completeness axioms implies that a := sup S R. Applying Proposition ("Let A R be bounded from below. Then a = inf A if and only if a x for any x A, and for any > 0 there exists x A such that x < a" ) = 1, we find m A satisfying m a m + 1. Since m + 1 R we obtain contradiction with the fact that a is an upper bound of R

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