Solutions Manual For Mathematical Proofs A Transition To Advanced Mathematics 3rd Edition by Chartrand
Solutions Manual For Mathematical Proofs A Transition To Advanced Mathematics 3rd Edition by Chartrand
Solutions Manual For Mathematical Proofs A Transition To Advanced Mathematics 3rd Edition by Chartrand
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2.6 (a) A can be any of the sets ∅, {1}, {2}, {1, 2}, that is, A is any subset of {1, 2, 4} that does not
contain 4.
(b) A can be any of the sets {1, 4}, {2, 4}, {1, 2, 4}, {4}, that is, A is any subset of {1, 2, 4} that
contains 4.
(c) A = ∅ and A = {4}.
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√
2.11 (a) 2 is not a rational number.
(b) 0 is a negative integer.
(c) 111 is not a prime number.
P Q ∼P ∼Q
T T F F
T F F T
F T T F
F F T T
√
2.13 (a) The real number r is greater than 2.
(b) The absolute value of the real number a is at least 3.
(c) At most one angle of the triangle is 45o .
(d) The area of the circle is less than 9π.
(e) The sides of the triangle have different lengths.
(f) The point P lies on or within the circle C.
2.16 (a) True. (b) False. (c) False. (d) True. (e) True.
P Q ∼Q P ∧ (∼ Q)
T T F F
T F T T
F T F F
F F T F
P Q ∼ P P ⇒ Q (P ⇒ Q) ⇒ (∼ P )
T T F T F
T F F F T
F T T T T
F F T T T
√
2.21 (a) P ⇒ Q: If 2 is rational, then 22/7 is rational. (True)
√
(b) Q ⇒ P : If 22/7 is rational, then 2 is rational. (False)
√
(c) (∼ P ) ⇒ (∼ Q): If 2 is not rational, then 22/7 is not rational. (False)
√
(d) (∼ Q) ⇒ (∼ P ): If 22/7 is not rational, then 2 is not rational. (True)
√ √
2.22 (a) (P ∧ Q) ⇒ R: If 2 is rational and 23 is rational, then 3 is rational. (True)
√ √
(b) (P ∧ Q) ⇒ (∼ R): If 2 is rational and 32 is rational, then 3 is not rational. (True)
√ √
(c) ((∼ P ) ∧ Q) ⇒ R: If 2 is not rational and 23 is rational, then 3 is rational. (False)
√ √
(d) (P ∨ Q) ⇒ (∼ R): If 2 is rational or 32 is rational, then 3 is not rational. (True)
2.25 (a) true. (b) false. (c) true. (d) true. (e) true.
2.33 (a) True for (x, y) = (3, 4) and (x, y) = (5, 5) and false for (x, y) = (1, −1).
(b) True for (x, y) = (1, 2) and (x, y) = (6, 6) and false for (x, y) = (2, −2).
(c) True for (x, y) ∈ {(1, −1), (−3, 4), (1, 0)} and false for (x, y) = (0, −1).
2.34 (a) If the x-coordinate of a point on the straight line with equation 2y + x − 3 = 0 is an integer,
then its y-coordinate is also an integer. Or: If −2n + 3 ∈ Z, then n ∈ Z.
(b) If n is an odd integer, then n2 is an odd integer.
(c) Let n ∈ Z. If 3n + 7 is even, the n is odd.
2.41 True if n = 3.
2.42 True if n = 3.
2.43 P (1) ⇒ Q(1) is false (since P (1) is true and Q(1) is false).
Q(3) ⇒ P (3) is false (since Q(3) is true and P (3) is false).
P (2) ⇔ Q(2) is true (since P (2) and Q(2) are both true).
2.46 The compound statement P ⇒ (P ∨ Q) is a tautology since it is true for all combinations of truth
values for the component statements P and Q. See the truth table below.
P Q P ∨Q P ⇒ (P ∨ Q)
T T T T
T F T T
F T T T
F F F T
2.47 The compound statement (P ∧(∼ Q))∧(P ∧Q) is a contradiction since it is false for all combinations
of truth values for the component statements P and Q. See the truth table below.
P Q ∼Q P ∧Q P ∧ (∼ Q) (P ∧ (∼ Q)) ∧ (P ∧ Q)
T T F T F F
T F T F T F
F T F F F F
F F T F F F
2.48 The compound statement (P ∧ (P ⇒ Q)) ⇒ Q is a tautology since it is true for all combinations of
truth values for the component statements P and Q. See the truth table below.
P Q P ⇒Q P ∧ (P ⇒ Q) (P ∧ (P ⇒ Q)) ⇒ Q
T T T T T
T F F F T
F T T F T
F F T F T
2.49 The compound statement ((P ⇒ Q) ∧ (Q ⇒ R)) ⇒ (P ⇒ R) is a tautology since it is true for all
combinations of truth values for the component statements P , Q and R. See the truth table below.
P Q ∼P ∼Q P ⇒Q (∼ P ) ⇒ (∼ Q)
T T F F T T
T F F T F T
F T T F T F
F F T T T T
Since P ⇒ Q and (∼ P ) ⇒ (∼ Q) do not have the same truth values for all combinations of
truth values for the component statements P and Q, the compound statements P ⇒ Q and
(∼ P ) ⇒ (∼ Q) are not logically equivalent. Note that the last two columns in the truth table
are not the same.
(b) The implication Q ⇒ P is logically equivalent to (∼ P ) ⇒ (∼ Q).
P Q ∼P ∼Q P ∨Q ∼ (P ∨ Q) (∼ P ) ∨ (∼ Q)
T T F F T F F
T F F T T F T
F T T F T F T
F F T T F T T
Since ∼ (P ∨ Q) and (∼ P ) ∨ (∼ Q) do not have the same truth values for all combinations
of truth values for the component statements P and Q, the compound statements ∼ (P ∨ Q)
and (∼ P ) ∨ (∼ Q) are not logically equivalent.
(b) The biconditional ∼ (P ∨ Q) ⇔ ((∼ P ) ∨ (∼ Q)) is not a tautology as there are instances when
this biconditional is false.
2.53 (a) The statements P ⇒ Q and (P ∧ Q) ⇔ P are logically equivalent since they have the same
truth values for all combinations of truth values for the component statements P and Q. See
the truth table.
P Q P ⇒Q P ∧Q (P ∧ Q) ⇔ P
T T T T T
T F F F F
F T T F T
F F T F T
(b) The statements P ⇒ (Q ∨ R) and (∼ Q) ⇒ ((∼ P ) ∨ R) are logically equivalent since they
have the same truth values for all combinations of truth values for the component statements
P , Q and R. See the truth table.
P Q R ∼P ∼Q Q∨R P ⇒ (Q ∨ R) (∼ P ) ∨ R (∼ Q) ⇒ ((∼ P ) ∨ R)
T T T F F T T T T
T F T F T T T T T
F T T T F T T T T
F F T T T T T T T
T T F F F T T F T
T F F F T F F F F
F T F T F T T T T
F F F T T F T T T
2.54 The statements Q and (∼ Q) ⇒ (P ∧ (∼ P )) are logically equivalent since they have the same truth
values for all combinations of truth values for the component statements P and Q. See the truth
table below.
P Q ∼P ∼Q P ∧ (∼ P ) (∼ Q) ⇒ (P ∧ (∼ P ))
T T F F F T
T F F T F F
F T T F F T
F F T T F F
2.55 The statements (P ∨ Q) ⇒ R and (P ⇒ R) ∧ (Q ⇒ R) are logically equivalent since they have the
same truth values for all combinations of truth values for the component statements P , Q and R.
See the truth table.
P Q R P ∨Q (P ∨ Q) ⇒ R P ⇒R Q⇒R (P ⇒ R) ∧ (Q ⇒ R)
T T T T T T T T
T F T T T T T T
F T T T T T T T
F F T F T T T T
T T F T F F F F
T F F T F F T F
F T F T F T F F
F F F F T T T T
2.56 If S and T are not logically equivalent, there is some combination of truth values of the component
statements P, Q and R for which S and T have different truth values.
2.57 Since there are only four different combinations of truth values of P and Q for the second and third
rows of the statements S1 , S2 , S3 , S4 and S5 , at least two of these must have identical truth tables
and so are logically equivalent.
2.58 (a) The statement P ∨ (Q ∧ R) is logically equivalent to (P ∨ Q) ∧ (P ∨ R) since the last two
columns in the truth table in Figure 4 are the same.
P Q R P ∨ Q . P ∨ R . Q ∧ R . P ∨ (Q ∧ R). (P ∨ Q) ∧ (P ∨ R)
... ... ... ...
... ... ... ...
T T T T .. T .. T .. T . T
...........................................................................................................................................................................................................................................................................................................................................
... ... ... ... .
.. .. .. .. ...
... ... ...
T F T T . T . F . T .. T
.................................................................................................................................................................................................................................................................................................................................................
... ... ... ...
... ... ... ...
F T T T ... T ... T ... T ... T
..............................................................................................................................................................................................................................................................................................................................................
... ... ... ...
... ... ...
F F T F . T .. F . F ...
. F
..............................................................................................................................................................................................................................................................................................................................................
... ... ... ...
T T F T ... T .
. .
.
.. F
.
.
.. T
.
.
... T
...............................................................................................................................................................................................................................................................................................................................................
... ... ... ...
T F F T .... T ..
. F ..
. T .... T
..............................................................................................................................................................................................................................................................................................................................................
..... .... .... .....
F T F T .
.. F .
.
. F .
.
. F .
.
. F
..................................................................................................................................................................................................................................................................................................................................................
.. ... ... ...
F F F F ..... F ..
.. F ..
.. F ...
.. F
(b) The statement ∼ (P ∨ Q) is logically equivalent to (∼ P ) ∧ (∼ Q) since the last two columns
in the truth table in Figure 5 are the same.
P Q ∼ P ∼ Q P ∨ Q ∼ (P ∨ Q) (∼ P ) ∧ (∼ Q)
T T F F T F F
T F F T T F F
F T T F T F F
F F T T F T T
2.62 The statement [(P ∨ Q)∧ ∼ (P ∧ Q)] is logically equivalent to ∼ (P ⇔ Q) since the last two columns
in the truth table below are the same.
P Q P ∨Q P ∧Q ∼ (P ∧ Q) P ⇔Q (P ∨ Q)∧ ∼ (P ∧ Q) ∼ (P ⇔ Q)
T T T T F T F F
T F T F T F T T
F T T F T F T T
F F F F T T F F
2.68 (a) There exists a rational number r such that 1/r is not rational.
(b) For every rational number r, r2 6= 2.
2.72 (d) implies that (∼ P (x)) ⇒ Q(x) is false for some x ∈ S (in fact, for all x ∈ S).
2.73 (b) and (c) imply that P (x) ⇒ Q(x) is true for all x ∈ T .
2.74 (a) For all real numbers x, y and z, (x − 1)2 + (y − 2)2 + (z − 2)2 > 0.
(b) False, since P (1, 2, 2) is false.
(c) ∃x, y, z ∈ R, (x − 1)2 + (y − 2)2 + (z − 2)2 ≤ 0. (∃x, y, z ∈ R, ∼ P (x, y, z).)
(d) There exist real numbers x, y and z such that (x − 1)2 + (y − 2)2 + (z − 2)2 ≤ 0.
(e) True, since (1 − 1)2 + (2 − 2)2 + (2 − 2)2 = 0.
2.76 (a) For every circle C1 with center (0, 0), there exists a circle C2 with center (1, 1) such that C1
and C2 have exactly two points in common.
(b) ∃C1 ∈ A, ∀C2 ∈ B, ∼ P (C1 , C2 ).
(c) There exists a circle C1 with center (0, 0) such that for every circle C2 with center (1, 1), C1
and C2 do not have exactly two points in common.
2.77 (a) There exists a triangle T1 such that for every triangle T2 , r(T2 ) ≥ r(T1 ).
(b) ∀T1 ∈ A, ∃T2 ∈ B, ∼ P (T1 , T2 ).
(c) For every triangle T1 , there exists a triangle T2 such that r(T2 ) < r(T1 ).
2.78 (a) For every a ∈ A, there exists b ∈ B such that a/b < 1.
(b) For a = 2, let b = 4. Then a/b = 1/2 < 1.
For a = 3, let b = 4. Then a/b = 3/4 < 1.
For a = 5, let b = 6. Then a/b = 5/6 < 1.
2.80 (a) Two lines in the plane are defined to be perpendicular if they intersect at right angles.
Two lines in the plane are perpendicular if and only if (1) one line is vertical and the other is
horizontal or (2) the product of the slopes of the two lines is −1.
(b) A rational number is a real number that can be expressed as a/b, where a, b ∈ Z and b 6= 0.
A real number is rational if and only if it has a repeating decimal expansion.
2.82 Only (f) is a characterization; (a), (c) and (e) are implications only; (b) is a definition; and (d) is
false.
2.83 (a)-(d) are characterizations. (d) is the Pythagorean theorem. (e) is not a characterization. (Every
positive number is the area of some rectangle.)
P Q ∼P Q ⇒ (∼ P ) P ∧ (Q ⇒ (∼ P ))
T T F F F
T F F T T
F T T T F
F F T T F
2.87 P ∨ (∼ Q)
2.91 (a) See the truth table below. (b) can be similarly verified.
P Q R ∼Q ∼R P ∧Q (P ∧ Q) ⇒ R P ∧ (∼ R) (P ∧ (∼ R)) ⇒ (∼ Q)
T T T F F T T F T
T F T T F F T F T
F T T F F F T F T
F F T T F F T F T
T T F F T T F T F
T F F T T F T T T
F T F F T F T F T
F F F T T F T F T
n2 +3n
P (n) : 2 is odd. Q(n) : (n − 2)2 > 0. R(n) : (n + 1)n−1 is odd.
The statement P (n) is true for n = 2, 3; Q(n) is true for n = 1, 3; and R(n) is true for n = 1, 2.
Thus the implications P (1) ⇒ Q(1), Q(2) ⇒ R(2) and R(3) ⇒ P (3) are true and their respective
converses are false.
2.96 No. Since Q(a) ⇒ P (a), R(b) ⇒ Q(b) and P (c) ⇒ R(c) are false, it follows that
P (a), Q(b) and R(c) are false and Q(a), R(b) and P (c) are true.
At least two of the three elements a, b and c are the same. If a = b, then Q(a) and Q(b) are both
true and false. This is impossible for a statement. If a = c, then P (c) and P (a) are both true and
false, again impossible. If b = c, then R(b) and R(c) are both true and false, which is impossible.
2.98 (a) For every x ∈ A and y ∈ B, there exists z ∈ C such that P (x, y, z).
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