{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

homework1solnmat25

homework1solnmat25 - Homework 1 Solutions 1 Construct a...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Homework 1 Solutions 1. Construct a truth table for the statement: [ p ⇒ ( q ∧ ∼ q )] ⇔∼ p. Solution: First note that q ∧ ∼ q is a contradiction, so replace it with the variable c which only takes on the truth value F. Then we get the truth table: p c p ⇒ c ∼ p [ p ⇒ c ] ⇒∼ p [ p ⇒ c ] ⇐∼ p [ p ⇒ c ] ⇔∼ p T F F F T T T F F T T T T T Because the truth values of the statement are always T, this says the statement is a tautology. In other words, whenever you want to prove that p ⇒ c where c is always F (i.e. c is a contradiction), you can instead prove ∼ p , and vice-versa. 2. Consider the following statement: A function f is uniformly continuous on a set S if and only if for every > there is a δ > such that | f ( x )- f ( y ) | < whenever x and y are in S and | x- y | < δ . (a) Rewrite the statement using the logical symbols ⇒ , ⇔ , ∀ , ∃ , etc. Solution: [ f is uniformly continuous on a set S ] ⇔ [ ∀ > 0, ∃ δ > 0 so that ∀ x,y ∈ S and | x- y | < δ ⇒ | f ( x )- f ( y ) | ≤ .] (b) Write the negation of the statement in words and in logical symbols....
View Full Document

{[ snackBarMessage ]}

Page1 / 3

homework1solnmat25 - Homework 1 Solutions 1 Construct a...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online