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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 counterexample.
[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 onetoone,
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 onetoone (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 onetoone and onto.
2 for saying that f was onetoone, but not onto.
2 for saying that f was onto, but not onetoone.
5 for saying that f was neither onetoone nor onto.
2.3: f:R>R.
f(x) = 3(x^1/3) + 13.
f is onetoone, since it's a strictly increasing
function.
f is onto, since, for any y in the real numbers,
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 Spring '08
 STRAIN

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