Solutions Manual For Algebra & Geometry An Introduction to
University Mathematics, 2e by Mark Lawson (All Chapters ,100%
Original Verified, A+ Grade)
This is The Only Solutions Manual For Algebra & Geometry An
Introduction to University Mathematics, 2e by Mark Lawson All
Other Files in The Market are Fake/Old/Wrong Edition.
The Textbook This Document Fully Covers
Algebra & Geometry An Introduction to University
Mathematics, 2e by Mark Lawson
• Publisher : Chapman and Hall/CRC
• Publication date : June 22, 2021
• Edition : 2nd
• Language : English
• Print length : 424 pages
• Format : Print Replica
• ISBN-10 : 9781000402506
• ISBN-13 : 978-1000402506
,Solutions Manual for Algebra & Geometry An Introduction
Solutions Manual For Algebra & Geometry An Introduction to University Mathematics, 2e by Mark Lawson
to University Mathematics, 2e by Mark Lawson (All
Chapter, Ch 1 Missing)
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
Solutions Manual For Algebra & Geometry An Introduction to University Mathematics, 2e by Mark Lawson
,Solutions Manual For Algebra & Geometry An Introduction to University Mathematics, 2e by Mark Lawson
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.
Solutions Manual For Algebra & Geometry An Introduction to University Mathematics, 2e by Mark Lawson
, Solutions Manual For Algebra & Geometry An Introduction to University Mathematics, 2e by Mark Lawson
Proofs 3
(b) Draw a diagonal of the square and then construct the square with side
that diagonal. This has twice the area by Pythagoras’ theorem.
(c) The area of such a triangle is 21 xy but we are told that it equals 14 z2 . Hence
1 1 2 2 2 2
2 xy = 4 z . By Pythagoras’ theorem z = x + y . Substituting this value
1 1 2
for z we get that 2 xy = 4 (x + y ). Rearrange this to get (x − y)2 = 0.
2 2
Hence x − y = 0 and so x = y. It follows that the triangle is isosceles.
6. (a) Every natural number n can be written n = 10q + r where 0 ≤ r ≤ 9. It
follows that n2 = 10(10q2 + 2qr) + r2 . Thus the last digit of n2 is equal to
the last digit of r2 . Direct calculation now shows that this can only be one
of 0, 1, 4, 5, 6, 9. The converse is not true because 14 is a natural number
ending in 4 but is not a perfect square.
(b) There are two statements to prove (1) if n is even then its last digit is even
and (2) if the last digit of n is even then n is even. We prove (1). We may
write n = 10q + r. Thus r = n − 10q. But n is even and 10q is even so it
follows that r is even. We prove (2). We may write n = 10q + r where r
is even. Thus r = 2s. It follows that n = 10q + 2s. It is now clear that n is
even.
(c) Again there are two statements to prove (1) if n is divisible by 9 then the
sum of the digits in n is divisible by 9 and (2) if the sum of the digits
in n are divisible by 9 then n is divisible by 9. The proof is based on
the following observation. Each power of 10 leaves remainder 1 when
divided by 9. Thus 10 = 9 + 1 and 100 = 99 + 1 etc. It follows that n can
be written as a multiple of 9 plus the sum of its digits. The proofs of (1)
and (2) now follow easily.
√
7. The proof follows exactly the same script as that for showing 2 is irrational,
except that you need to prove the following result: if 3 divides a2 then 3 divides
a. To prove this we need the following. We may write a = 3q + r where r =
0, 1, 2. It follows that a2 = 3(3q2 + 2r) + r. Thus a2 is divisible by 3 if and only
if a is divisible by 3.
8. Show that the angles AD̂B = AĈB using Proposition III.21. Deduce that trian-
gle AXD is similar to triangle ABC. Show that the angles DĈA = DB̂A using
Proposition III.21. The angles DÂC = BÂX from what we already know and
how the angles are formed. Thus the triangles AXB and ACD are similar. The
two equations follow immediately from the fact that we have two sets of simi-
lar triangles. The remainder of the proof is just routine algebra using these two
equations where we eliminate e.
9. (a) Lines in the plane have equations of the form y = tx + s for some t and
s. Our lines passes through the point (−1, 0) and so t = s. Thus it has the
form y = t(x + 1).
(b) To find the coordinates of the point P, we have to solve the equation
2
x2 +t 2 (x + 1)2 = 1 for x. This yields x = 1−t
1+t 2
since we are excluding x =
Solutions Manual For Algebra & Geometry An Introduction to University Mathematics, 2e by Mark Lawson
University Mathematics, 2e by Mark Lawson (All Chapters ,100%
Original Verified, A+ Grade)
This is The Only Solutions Manual For Algebra & Geometry An
Introduction to University Mathematics, 2e by Mark Lawson All
Other Files in The Market are Fake/Old/Wrong Edition.
The Textbook This Document Fully Covers
Algebra & Geometry An Introduction to University
Mathematics, 2e by Mark Lawson
• Publisher : Chapman and Hall/CRC
• Publication date : June 22, 2021
• Edition : 2nd
• Language : English
• Print length : 424 pages
• Format : Print Replica
• ISBN-10 : 9781000402506
• ISBN-13 : 978-1000402506
,Solutions Manual for Algebra & Geometry An Introduction
Solutions Manual For Algebra & Geometry An Introduction to University Mathematics, 2e by Mark Lawson
to University Mathematics, 2e by Mark Lawson (All
Chapter, Ch 1 Missing)
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
Solutions Manual For Algebra & Geometry An Introduction to University Mathematics, 2e by Mark Lawson
,Solutions Manual For Algebra & Geometry An Introduction to University Mathematics, 2e by Mark Lawson
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.
Solutions Manual For Algebra & Geometry An Introduction to University Mathematics, 2e by Mark Lawson
, Solutions Manual For Algebra & Geometry An Introduction to University Mathematics, 2e by Mark Lawson
Proofs 3
(b) Draw a diagonal of the square and then construct the square with side
that diagonal. This has twice the area by Pythagoras’ theorem.
(c) The area of such a triangle is 21 xy but we are told that it equals 14 z2 . Hence
1 1 2 2 2 2
2 xy = 4 z . By Pythagoras’ theorem z = x + y . Substituting this value
1 1 2
for z we get that 2 xy = 4 (x + y ). Rearrange this to get (x − y)2 = 0.
2 2
Hence x − y = 0 and so x = y. It follows that the triangle is isosceles.
6. (a) Every natural number n can be written n = 10q + r where 0 ≤ r ≤ 9. It
follows that n2 = 10(10q2 + 2qr) + r2 . Thus the last digit of n2 is equal to
the last digit of r2 . Direct calculation now shows that this can only be one
of 0, 1, 4, 5, 6, 9. The converse is not true because 14 is a natural number
ending in 4 but is not a perfect square.
(b) There are two statements to prove (1) if n is even then its last digit is even
and (2) if the last digit of n is even then n is even. We prove (1). We may
write n = 10q + r. Thus r = n − 10q. But n is even and 10q is even so it
follows that r is even. We prove (2). We may write n = 10q + r where r
is even. Thus r = 2s. It follows that n = 10q + 2s. It is now clear that n is
even.
(c) Again there are two statements to prove (1) if n is divisible by 9 then the
sum of the digits in n is divisible by 9 and (2) if the sum of the digits
in n are divisible by 9 then n is divisible by 9. The proof is based on
the following observation. Each power of 10 leaves remainder 1 when
divided by 9. Thus 10 = 9 + 1 and 100 = 99 + 1 etc. It follows that n can
be written as a multiple of 9 plus the sum of its digits. The proofs of (1)
and (2) now follow easily.
√
7. The proof follows exactly the same script as that for showing 2 is irrational,
except that you need to prove the following result: if 3 divides a2 then 3 divides
a. To prove this we need the following. We may write a = 3q + r where r =
0, 1, 2. It follows that a2 = 3(3q2 + 2r) + r. Thus a2 is divisible by 3 if and only
if a is divisible by 3.
8. Show that the angles AD̂B = AĈB using Proposition III.21. Deduce that trian-
gle AXD is similar to triangle ABC. Show that the angles DĈA = DB̂A using
Proposition III.21. The angles DÂC = BÂX from what we already know and
how the angles are formed. Thus the triangles AXB and ACD are similar. The
two equations follow immediately from the fact that we have two sets of simi-
lar triangles. The remainder of the proof is just routine algebra using these two
equations where we eliminate e.
9. (a) Lines in the plane have equations of the form y = tx + s for some t and
s. Our lines passes through the point (−1, 0) and so t = s. Thus it has the
form y = t(x + 1).
(b) To find the coordinates of the point P, we have to solve the equation
2
x2 +t 2 (x + 1)2 = 1 for x. This yields x = 1−t
1+t 2
since we are excluding x =
Solutions Manual For Algebra & Geometry An Introduction to University Mathematics, 2e by Mark Lawson
