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MACM 101 — Discrete Mathematics I
Exercises on Predicates and Quantiﬁers. Due:
Wednesday, October 13th (at the beginning of
the class)
1.
Use predicates and quantiﬁers to express this statement
“There is a customer in this coﬀee shop who has tried every
type of cookies produced by one of the suppliers.”
Solution:
∃
x
[
CustomerInThisCoﬀeeShop
(
x
)
∧
∀
o
[(
CookieType
(
o
)
∧∃
y
(
CookieSupplier
(
y
)
∧
Produces
(
y,o
)))
→
Tried
(
x,o
)]]
In practice we use
P
,
Q
,
R
, etc., but I kept long predicate names so
that you can understand better. Also, there are several other logically
equivalent solutions. Many people wrote much simpler statements whee
they simply used a predicate to denote say “x tried every type of cookies
produced by one of the suppliers”. That was OK too, but not as good
as the solution above. In general, you should try to extract as much
logical information as possible from a sentence.
2.
Determine the truth value of each of these statements if the
universe of each variable consists of (i) rational numbers, (ii)
all integers. Justify your answer.
(a)
∀
x
∀
y
∃
z
(
z
=
x
+
y
1000
)
;
(b)
∃
z
∀
x
∀
y
(
z
=
x
+
y
1000
)
;
(c)
∃
x
∀
y
(
y >
0
→ ∃
z
(
y
x
=
z
))
.
Solution:
(a)
∀
x
∀
y
∃
z
(
z
=
x
+
y
1000
);
1
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View Full Document(i) True for rationals. No matter which
x
and
y
we take, the
rational number
x
+
y
1000
) does exist. (ii) False for integers. Take
x
= 1,
y
= 2. Then
x
+
y
1000
) is not an integer.
(b)
∃
z
∀
x
∀
y
(
z
=
x
+
y
1000
);
(i, ii) False for rationals and for integers. There is no rational (or
integer) number
z
which simultaneously equals to all possible sums
of two arbitrary rational (or integer) numbers, divided by 1000! If
such a number existed, call it
a
, then
a
would be simultaneously
equal to say 2 (if
x
=
y
= 1000) and 1 (if
x
=
y
= 500), and thus
2 would be equal to 1, which is impossible.
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 Summer '08
 PEARCE

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