Rosen
Section 1.3
Ans # 34
a) counter example : x = 0
b) counter example : x = √2
c) counter example : x = 0
Ans # 36
a) There exists a system which is in open state.
b) All the systems are either malfunctioning or in diagnostic state.
c) There exists at least one system which is in open state or there exists at
least one system which is in diagnostic state.
d) There exists a system which in not available state.
e) All systems are not in working state.
Partial credit is given who missed to mention system and states in the
answer.
Ans # 38
Marks have been deducted for wrong usage of quantifiers, usage of wrong
predicates and logical connectives.
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Ans # 44
Since A doesn’t involve any quantifiers, under any interpretation, it is either true
or false. If A is true, both sides are true. However, if A is false, then both sides are
equivalent to for all x P(x) for part (a), and there exists x P(x) for part (b). Hence
both sides are equivalent for both parts.
P(x) is not a proposition so avoid using truth tables. Partial credit is given to
those who used truth table to verify this.
Ans # 48
a.
b.
c.
d.
False: there is more than one x such that x>1.
False: there are two such values of x, 1 and -1.
True: the unique value is x=3.
False: there are no values of x satisfying this predicate.
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Section 3.3
Ans # 6
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Ans # 28
First, we note that the inequality holds for 0 and 1, but it does not hold for 2 and
3. (Equality holds at 1; at all other integers where the inequality holds, it is a strict
inequality.) Since it holds for 4 and 5, and the factorial increases faster than the
square, we assume that it holds for all n ≥ 4. The basis step in this case is at n = 4.
We note that 16 = 42 < 24 = 4!. For our inductive step, we must show that if n2 <
n!, then (n+1)2 < (n+1)!, when n ≥ 4. Writing (n+1)2 = n2 +2n+1 and noting that
by inductive assumption, we have n2 < n!, it follows that (n + 1)2 < n! + 2n + 1 <
n + 3n < n! + n2 < n! + n · n! = (n + 1)n! = (n + 1)!, which proves our result.
Ans # 34
Theorem: It takes n -1 breaks to separate the chocolate bar into single squares,
no matter what choices are made.
Proof: We use strong induction on n. If n = 1, the chocolate bar is already a single
square, and the number of breaks required is n - 1 = 0.
If n > 1, then after the first break we will have two bars, say with r and s squares
respectively. Since r and s are less than n, we can assume by induction that it
takes r – 1 breaks to separate the bar with r squares and s – 1 breaks to separate
the bar with s squares.
Counting the first break, we need (r – 1)+(s – 1 )+1 breaks in all, and
(r – 1)+(s – 1)+1 = r + s – 1 = n – 1.
Ans # 36
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Ans # 50
For n = 1, the formula is (1 + ½)2 = (3/2)2 = (9/8) ≠ 1.
The base case is not true.
Ans # 52
The proof assumes that for the max function, x and y must be the same number,
although it is possible that x ≠ y. Also, the inductive step does not follow because
(x-1) and (y-1) are not necessarily positive.
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