Homework Solutions Week 1

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```							                      Homework Solutions Week 1
Anne Thomas
January 18, 2009

1. Glossary: Ex. 1
Solution.

1. A statement.

2. A statement.

3. A statement. Statements do not have to be true.

4. A statement. We do not have to be able to verify the truth of a statement.

5. Not a statement. A question cannot be a statement.

6. Not a statement. This is not a declarative sentence.

7. Not a statement. A statement is either true or false. This sentence is a paradox: if it
is true, then by what it says, it is false, but then its negation is that “This sentence is
true”, etc.

8. Not a statement. This is not a meaningful sentence.

2. The Propositional Calculus: Ex. 2
Solution.

• If f is a diﬀerentiable function, then f is continuous.

• If α and β are two right angles, then α = β.

• If M is a matrix with a zero eigenvalue, then M is not invertible.

1
P   ¬P   P ∨ ¬P
0    1      1
1    0      1

Table 1: Truth table for The Propositional Calculus: Ex. 7(a)

P   Q    P ⇒Q     P ∧ (P ⇒ Q) (P ∧ (P    ⇒ Q)) ⇒ Q
0   0      1            0                1
0   1      1            0                1
1   0      0            0                1
1   1      1            1                1

Table 2: Truth table for The Propositional Calculus: Ex. 7(e)

P   Q    P ⇒Q     ¬Q ⇒ ¬P    (P ⇒ Q) ⇒ (¬Q ⇒ ¬P )
0   0      1         1                1
0   1      1         1                1
1   0      0         0                1
1   1      1         1                1

Table 3: Truth table for The Propositional Calculus: Ex. 7(f)

P   Q    R   P ⇒Q    Q⇒R      (P ⇒ Q) ∧ (Q ⇒ R)      P ⇒R     ((P ⇒ Q) ∧ (Q ⇒ R)) ⇒ (P ⇒ R)
0   0    0     1      1               1                1                    1
0   0    1     1      1               1                1                    1
0   1    0     1      0               0                1                    1
0   1    1     1      1               1                1                    1
1   0    0     0      1               0                0                    1
1   0    1     0      1               0                1                    1
1   1    0     1      0               0                0                    1
1   1    1     1      1               1                1                    1

Table 4: Truth table for The Propositional Calculus: Ex. 7(h)

2
3. Solow: Ex. 2.9

(a) How can I show that one real number is less than or equal to another real number?
How can I show that the sum of the squares of two non-negative real numbers is less
than or equal to than the square of their sum?

(b) How can I show that two lines are parallel? How can I show that two lines do not
intersect? How can I show that if two lines have equations in ‘ax + b’ form with the
same x–coeﬃcient, then their slopes are the same?

4. Solow: Ex. 2.21
Solution. Let A be the statement: RST is a triangle such that SU is a perpendicular
bisector of RT . Let B be the statement: triangle SU R is congruent to triangle SU T . We
are trying to show that A ⇒ B.
A key question is: how can I show that two triangles are congruent? Moving forward from
A, because SU is a perpendicular bisector of RT , we obtain statement A1: RU = U T .
Also moving forward from A, because SU is a perpendicular bisector we obtain statement
A2: ∠RU S = ∠T U S = 90◦ . An answer to the key question is: show that two sides of
the triangles and the included angle are equal. A speciﬁc answer is: show that RU = U T ,
SU = SU and ∠RU S = ∠T U S. But SU = SU is clear, so we are done.
5. Solow: Ex. 3.14
Analysis of Proof. The forward-backward method gives rise to the key question “How
can I show that a real number is rational?”. One answer is to use the deﬁnition of a rational
number as a number which can be expressed as a ratio of integers with nonzero denominator.
So we must show that

B1 : a + b can be expressed as a ratio of integers with nonzero denominator.

Turning to the forward process, we can use the deﬁnition of a rational number to obtain
statements
p
A1 : a = where p and q are integers and q = 0
q
and
r
A2 : b =     where r and s are integers and s = 0.
s
Hence
p r
A3 : a + b =    + .
q s

3
Continuing forward from this, by carrying out the addition of fractions we obtain
ps + rq
A4 : a + b =           .
qs
Returning to the backward process, it is now enough to establish statement
ps + rq
B2 :           is a ratio of integers with nonzero denominator.
qs
Going forward again, combining the facts about p, q, r, s in statements A1 and A2, we obtain

A5 : ps + rq is an integer.

and
A6 : qs is a nonzero integer.
Statements A5 and A6 together establish statement B2, and the proof is complete.
Proof. By the deﬁnition of rational numbers, since a is a rational number, there are integers
p
p and q, with q = 0, such that a = . Similarly, there are integers r and s, with s = 0, such
q
r
that b = . We now compute:
s
p r
a+b =      +
q s
ps + rq
=         .
qs
Now ps + rq and qs are both integers, since p, q, r and s are all integers. Moreover, qs = 0
since q and s are both nonzero. Hence a + b can be expressed as the ratio of two integers
such that the denominator is not zero. We conclude that a + b is rational.

4

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