MT1_1997S sln

MT1_1997S sln - Answer to Midterm Question 1 Determine...

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Answer to Midterm Question 1 Determine whether or not [not(q) --> not(p)] --> (p --> q) is a tautology. If it is a tautology, prove it without using a truth table. If not, give a counter-example. [not(q) --> not(p)] --> (p --> q) <=> not{not[not(q)] or not(p)} or [not(p) or q] def. of --> <=> not[q or not(p)] or [not(p) or q] double negation <=> not[not(p) or q] or [not(p) or q] commutativity <=> T not(x) or x = T You got one third of the credit for correct use of each of: 1) the definition of --> 2) techniques to simplify the expression and 3) not(x) or x = T. If you used the fact that a statement is logically equivalent to its contrapositive in your argument, you got five points for remembering that fact, but that's all since such arguments employ circular reasoning. ------------------------------- Answer to Midterm Question 2 Classify each of the following functions f as one-to-one, onto , both, or neither. R denotes the set of all real numbers, and R+denotes the set of nonnegative real numbers 2.1: There is no problem 2.1. Everybody got 0/0 on it. :-). 2.2: f:R->R. f = g o h, where g(x) = sqrt(x) maps R+ to R and h(x) = x^2 maps R to R+. Hence f(x) = sqrt(x^2). Thus f(x) = x for all x >= 0, and f(x) = -x for all x<0. So f(x) = |x| for all x. Thus f is neither one-to-one (since f(-1) = f(1) = 1, for instance) nor onto (since there is no x for which f(x) = -1, for example). Grading policy: 0 for saying that g(x) was not a function. 0 for saying that f was both one-to-one and onto. 2 for saying that f was one-to-one, but not onto. 2 for saying that f was onto, but not one-to-one. 5 for saying that f was neither one-to-one nor onto. 2.3: f:R->R. f(x) = 3(x^1/3) + 13. f is one-to-one, since it's a strictly increasing function. f is onto, since, for any y in the real numbers,
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MT1_1997S sln - Answer to Midterm Question 1 Determine...

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