hw2sol - CSE 20: Discrete Mathematics Fall 2010 Problem Set...

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

View Full Document Right Arrow Icon
CSE 20: Discrete Mathematics Fall 2010 Problem Set 2 Answers Instructor: Daniele Micciancio Due on: Wed. Oct. 13, 2010 Problem 1 (6 points) . 1. x Z . y Z .x < y (a) For every integer, there is an integer which is larger. (b) ¬ ( x Z . y Z .x < y ) ⇔ ∃ x Z . ¬ ( y Z .x < y ) ⇔ ∃ x Z . y Z . ¬ ( x < y ) ⇔ ∃ x Z . y Z .x y (c) The original statement is true, since for any integer x , there exists integer y = x + 1 such that x < x + 1 = y . 2. y Z . x Z .x < y (a) There is an integer which is larger than all integers. (b) ¬ ( y Z . x Z .x < y ) ⇔ ∀ y Z . ¬ ( x Z .x < y ) ⇔ ∀ y Z . x Z . ¬ ( x < y ) ⇔ ∀ y Z . x Z .x y (c) The negation of the original statement is true, since for any integer y , if we set x = y then x = y y , so such an integer always exists. Problem 2 (8 points)
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/08/2011 for the course CSE cse105 taught by Professor Cs during the Fall '10 term at UCSD.

Page1 / 2

hw2sol - CSE 20: Discrete Mathematics Fall 2010 Problem Set...

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

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