MT1_2000

# MT1_2000 - Math 55 Fall 2000 Midterm 1 Answers Complete...

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Math 55 - Fall 2000 - Midterm 1 - Answers Complete answers to all versions of each question are shown. Q1) (10 pts) Determine whether or not the following expression is a tautology: xor means exclusive or. If it is, prove it using rules for simplifying logical expressions, without using a truth table. If not, give a counterexample. 1) (p or not q) -> ( (p xor not q) or (p and not q) ) [<=> not(p or not q) or (p xor not q) or (p and not q) definition of -> <=> (not p and q ) or (p xor not q) or (p and not q) DeMorgan <=> (not p and q ) or (p and q) or (not p and not q) or (p and not q) definition of xor <=> [ q and (not p or p) ] or [ (not p or p ) and not q] distributive law (twice) <=> q or not q not p or p = T <=> T q or not q = T ] 2) (not p or q) -> ( (not p xor q) or (not p and q) ) [<=> not(not p or q) or (not p xor q) or (not p and q) definition of -> <=> (p and not q ) or (not p xor q) or (not p and q) DeMorgan <=> (p and not q ) or (not p and not q) or (p and q) or (not p and q) definition of xor, not not cancels <=> [ not q and (p or not p) ] or [ (p or not p ) and q] distributive law (twice) <=> not q or q p or not p = T <=> T not q or q = T ] 3) (not p or not q) -> ( (not p xor not q) or (not p and not q) ) [<=> not(not p or not q) or (not p xor not q) or (not p and not q) definition of -> <=> (p and q) or (not p xor not q) or (not p and not q) DeMorgan <=> (p and q ) or (p and not q) or (not p and q) or (not p and not q) definition of xor <=> [ p and (q or not q) ] or [ not p and (q or not q)] distributive law (twice) <=> q or not q not p or p = T <=> T q or not q = T ] Q2) (10 pts) Classify f as one-to-one, onto, both, or neither. R+ is the set of nonnegative real numbers. Justify your answers. 1a) h:R->R+, h(x) = x^4, g:R+->R, g(x) = x^(1/8), f(x) = g o h [neither: note that f:R->R f(x) = (x^4)^(1/8) = sqrt(abs(x)), f(4) = f(-4) = 2, so f is not one-to-one

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since f(x) >= 0 for all x, and the codomain includes negative numbers, f is not onto] 1b) h:R->R+, h(x) = x^4, g:R+->R, g(x) = x^(1/8), f(x) = h o g [both: note that f:R+->R+, f(x) = (x^(1/8))^4 = sqrt(x) since the equation y=f(x) = sqrt(x) has a unique solution x = y^2 for every y in the codomain R+, f(x) is a bijection] 1c) h:R+->R, h(x) = x^(1/8), g:R->R, g(x)=x^4, f(x) = g o h [one-to-one: note that h:R+->R, f(x) = (x^(1/8))^4 = sqrt(x) since the equation y=f(x) = sqrt(x) has a unique solution
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## This note was uploaded on 03/08/2010 for the course MATH 55 taught by Professor Strain during the Spring '08 term at Berkeley.

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MT1_2000 - Math 55 Fall 2000 Midterm 1 Answers Complete...

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