SOLUTIONS MANUAL
Algebra & Geometry: An Introduction to
University Mathematics 2nd Edition by Mark
Lawson
TABLE OF CONTENTS
Chapter 1 The Nature of Mathematics
Chapter 2 Proofs
Chapter 3 Foundations
Chapter 4 Algebra Redux
Chapter 5 Number Theory
Chapter 6 Complex Numbers
Chapter 7 Poly...
, SOLUTIONS MANUAL
Algebra & Geometry: An Introduction to
University Mathematics 2nd Edition by Mark
Lawson
TABLE OF CONTENTS
Chapter 1 The Nature of Mathematics
Chapter 2 Proofs
Chapter 3 Foundations
Chapter 4 Algebra Redux
Chapter 5 Number Theory
Chapter 6 Complex Numbers
Chapter 7 Polynomials
Chapter 8 Matrices
Chapter 9 Vectors
Chapter 10 The Principal Axes Theorem
Chapter 11 What are the Real Numbers?
, CHAPTER 2
Proofs
Exercises 2.3
1. (a) C was a knight. The argument runs as follows. If you ask a knight what
he is he will say he is a knight. If you ask a knave what he is he is obliged
to lie and so also say that he is a knight. Thus no one on this island can
say they are a knave. This means that B is a knave. Hence C was correct
in saying that B lies and so C was a knight.
(b) A is a knave, B a knight and C a knave. The argument runs as follows. If
all three were knaves then C would be telling the truth which contradicts
the fact that he is a knave. Thus at least one of the three is a knight and
at most two are knights. It follows that C is a knave. Suppose that there
were exactly two knights. Then both A and B would be knights but they
contradict each other. It follows that exactly one of them is a knight.
Hence A is knave and B is a knight.
2. Sam drinks water and Mary owns the aardvark. The following table shows all
the information you should have deduced.
1 2 3 4 5
House Yellow Blue Red White Green
Pet Fox Horse Snails Dog Aardvark
Name Sam Tina Sarah Charles Mary
Drink Water Tea Milk Orange juice Coffee
Car Bentley Chevy Oldsmobile Lotus Porsche
The starting point is the following table.
1 2 3 4 5
House
Pet
Name
Drink
Car
1
SOURCE: Browsegrades.net
, 2 □ Solutions to Algebra & Geometry: Second Edition
Using clues (h), (i) and (n), we can make the following entries in the table.
1 2 3 4 5
House Blue
Pet
Name Sam
Drink Milk
Car
There are a number of different routes from here. I shall just give some exam-
ples of how you can reason. Clue (a) tells us that Sarah lives in the red house.
Now she cannot live in the first house, because Sam lives there, and she cannot
live in the second house because that is blue. We are therefore left with the
following which summarizes all the possibilities so far.
1 2 3 4 5
House Yellow? White? Green? Blue Red? Red? Red?
Pet
Name Sam Sarah? Sarah? Sarah?
Drink Milk
Car
Clue (b) tells us that Charles owns the dog. It follows that Sam cannot own the
dog. We are therefore left with the following possibilities.
1 2 3 4 5
House Yellow? White? Green? Blue Red? Red? Red?
Pet Fox? Horse? Snails? Aardvark?
Name Sam Sarah? Sarah? Sarah?
Drink Milk
Car
Both Questions 1 and 2 demonstrate some of the logic needed in mathematics.
3. I shall prove that the sum of an odd number and an even number is odd. The
other cases are proved similarly. Let m be even and n be odd. Then m = 2r and
n = 2s + 1 for some integers r and s. Thus m + n = 2r + 2s + 1 = 2(s + r) + 1.
Thus m + n is odd.
4. Draw a diagonal across the quadrilateral dividing the figure into two triangles.
The sum of the interior angles of the figure is equal to the sum of the angles in
the two triangles. This is 360◦.
√ √
5. (a) Two applications of Pythagoras’ theorem give 22 + 32 + 72 = 62.
SOURCE: Browsegrades.net
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