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ArsDigita University
Month 2: Discrete Mathematics  Professor Shai Simonson
Problem Set 1 – ANSWERS
By: Dimitri Kountourogiannis
(1) Logic Proofs
(a) Prove that
a
→
b
is equivalent to
¬
b
→ ¬
a
using a truth table.
ANSWER:
We compute the truth table values and observe they are the same.
Note that
¬
b
→ ¬
a
is called the contrapositive of
a
→
b
. (Check out
http://logik.phl.univie.ac.at/
∼
chris/formularukzentral.html
for an automated truthtable generator)
a
b
¬
a
¬
b
a
→
b
(
¬
b
)
→
(
¬
a
)
T
T
F
F
T
T
T
F
F
T
F
F
F
T
T
F
T
T
F
F
T
T
T
T
(b) Prove it using
algebraic
identities.
ANSWER:
Using the equivalence rules on pages 1718 of Rosen,
a
→
b
⇔ ¬
a
∧
b
⇔
b
∧ ¬
a
⇔
(
¬¬
b
)
∧ ¬
a
⇔ ¬
b
→ ¬
a
(c) Prove that
a
→
b
is not equivalent to
b
→
a
.
ANSWER:
This is the sort of thing that is easy to show with a truth table
(semantically) and is extremely diﬃcult to show using the algebraic identities
(syntactically). If
a
→
b
were equivalent to
b
→
a
, then the two would have
to agree for all possible
true/false
assignments of
a
and
b
. But this is not the
case: if we let
a
be
true
and
b
be
false
, then
a
→
b
evaluates to
false
, and
b
→
a
evaluates to
true
. So the two formulas are not equivalent.
(2) Aristotle’s Proof that the Square Root of Two is Irrational.
(a) Prove the
lemma,
used by Aristotle in his proof, which says that if
n
2
is even,
so is
n
. (Hint: Remember that
a
→
b
is equivalent to
¬
b
→ ¬
a
.
ANSWER:
It is easiest to prove the contrapositive of the lemma, that if
n
is
odd, then
n
2
is odd. An odd number is one that can be written in the form of
2
k
+ 1 for some other whole number
k
. So suppose that
n
= 2
k
+ 1. Then
n
2
= (2
k
+ 1)
2
= 4
k
2
+ 4
k
+ 1 = 2(2
k
2
+ 2
k
) + 1
And so
n
2
is odd. This proves the contrapositive and hence the lemma.
(b) Prove that the square root of 3 is irrational using Aristotle’s techniques. Make
sure to prove the appropriate lemma.
ANSWER:
We need the following lemma:
1
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Lemma:
For any whole number
n
,
n
2
is divisible by
3
implies that
n
is divisible by
3
Proof of Lemma: Again it is easier to prove the contrapositive:
n
is not divisible by
3
implies that
n
2
is not divisible by 3.
So assume that
n
is not divisible by 3 then
n
is of the form
n
= 3
k
+ 1, or
n
= 3
k
+ 2 for some whole number
k
. Then in the ﬁrst case
n
2
= 9
k
2
+ 6
k
+ 1 = 3(3
k
2
+ 2
k
) + 1
which is not divisible by 3. The second case is similar. Q.E.D. Lemma.
Proof that the square root of 3 is irrational:
Suppose not, and that we can write
√
3 =
p
q
with
p
q
in lowest terms. Then
3
q
2
=
p
2
And so
p
2
is divisible by 3. By the lemma it follows that
p
is divisible by 3, say
p
= 3
p
1
for some whole number
p
1
. Plugging back in to the formula, we get
q
2
= 3
p
2
1
Now the tables are turned!
q
2
is divisible by 3 and so by the lemma
q
is divisible
by 3, say
q
= 3
q
1
for some whole number
q
1
. This implies that
p
q
=
3
p
1
3
q
1
=
p
1
q
1
This is the contradiction we were looking for: We assumed that
p
q
was in lowest
terms, which we can always do with a fraction, but the assumption that
p
q
=
√
3
implies that
p
q
is
not
in lowest terms. Contradiction! Therefore
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This document was uploaded on 10/01/2011.
 Spring '09

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