Instructor Jared Weinstein
1075 Evans Hall
www.math.berkeley.edu/
∼
jared
[email protected]
Math 55
∼
Discrete Mathematics
U.C. Berkeley Dept. of Mathematics
Summer 2007
Homework #1 Solutions
1.1.20
Write each of the following sentences in “if . . . then” form, in English.
Answer
: (a) “If you do not send me an email message, then I will not remember to send you the address.” Or: “If
I remember to send you the address, then you will have sent me an email message.” Or “If I’m going to remember
to send you the address, then you must send me an email message.”
(b) “If you were born in the United States then you are a citizen of this country.”
(f) “If there is a storm then the beach erodes.” (But notice how the original form “The beach erodes whenever there
is a storm” conveys the meaning with greater clarity. My answer seems to leave open the possibility that there are
two storms and the beach erodes once, while the origianl formulation makes it clear that each storm results in erosion
and more directly suggests cause and effect. That’s why we would say it that way.)
(g) “If someone logs onto the server then they must have a valid password.”
1.1.32(a)
Construct a truth table for the compound proposition (
p
∨
q
)
∨
r
.
Solution
:
p
q
r
p
∨
q
(
p
∨
q
)
∨
r
T
T
T
T
T
T
T
F
T
T
T
F
T
T
T
T
F
F
T
T
F
T
T
T
T
F
T
F
T
T
F
F
T
F
T
F
F
F
F
F
1.1.44
A barber shaves those who do not shave themselves, but no one else. Is this possible?
Answer
: No. Let
b
be the barber. Let
S
be the set of all people who do not shave themselves. Then
b
shaves
p
if
and only if
p
∈
S
.
Now either
b
shaves
b
, or not. Case (1): If
b
shaves
b
then by assumption
b
∈
S
. But this means that
b
does not
shave himself. Thus
b
doesn’t shave
b
, a contradiction. Case (2): If
b
doesn’t shave
b
then
b
doesn’t shave himself,
so
b
∈
S
, so
b
shaves
b
by hypothesis. This is again a contradiction. Since one of the two cases must occur, and both
cases lead to contradictions, this situation is impossible. (Note that this is essentially the same paradox as the set of
all sets that are not elements of themselves.)
1.1.46a
A group of people come in two types: those who always lie, and those who always tell the truth. By asking
one of them exactly one question, we must determine which of the two types he/she is. (a) If we ask “Are you a
liar?” then the truthtellers will answer “No”, and liars will answer “No” as well, so we won’t gain any information
from their responses. (b)
First solution.
If we ask “Is 1 + 1 = 2?” then truthtellers must answer “Yes”, while liars
must answer “No.” So we can tell them apart.
Alternative solution.
We ask “If you are a truthteller, do you always
tell the truth?”. A truthteller would answer “Yes”. If a liar is answering, then the correct answer is “Yes”, because
this ifthen statement is automatically true, since its premise is false. The liar must therefore answer lie and answer
“No”, thereby distinguishing herself/himself from truthtellers.
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 Spring '11
 MichealAlain
 Math, Logic, Negative and nonnegative numbers, Logical connective

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