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C H A P T E R
25
E
Proof and number
PL
Objectives
To understand implication
To understand converse and equivalence
To understand the purpose of counter examples
To construct proofs
To understand the principle of mathematical induction
To solve linear Diophantine equations
To apply the Euclidean algorithm to find the highest common factor of two
numbers
25.1
M
To apply the Euclidean algorithm in the solution of linear Diophantine equations
An introduction to proof
Implication, converse and equivalence
SA
For two statements p and q, ‘ p ⇒ q’ is read ‘p implies q’.
For example,
x −3=4⇒ x =7
x is divisible by 3 ⇒ 2x is divisible by 6
x = 3 ⇒ x 2 = 9.
In the first two examples the converse statements also hold.
q ⇒ p is the converse of p ⇒ q
The converse of the two examples are
x =7⇒ x −3=4
2x is divisible by 6 ⇒ x is divisible by 3.
Combining p ⇒ q and q ⇒ p we can write p ⇔ qwhich may be read ‘p is equivalent to q’
Thus we can write
x −3=4⇔ x =7
x is divisible by 3 ⇔ 2x is divisible by 6.
607
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608
Essential Advanced General Mathematics
E
An equivalent symbol for ⇔ is iff which is read ‘if and only if’.
However, x 2 = 9 does not imply that x = 3, i.e., the converse does not hold.
In fact, x 2 = 9 ⇔ x = 3 or x = −3
There are many other examples where the implication only holds one way.
For example,
x > 3 ⇒ x 2 > 9, but x 2 > 9 does not imply x > 3
x rational ⇒ x 2 rational, but x 2 rational does not imply x rational
x and y even ⇒ x + y even, but x + y even does not imply x and y even
a = b ⇒ a 2 = b2 , but a 2 = b2 does not imply a = b
Quadrilateral ABCD is a square ⇒ the sides of quadrilateral ABCD are of equal length,
but the sides of the quadrilateral ABCD are of equal length does not imply quadrilateral
ABCD is a square.
Counter example
PL
P1: FXS/ABE
Consider numbers of the form n 2 + n + 11 where n is a natural number.
n
n 2 + n + 11
1
13
2
17
3
23
4
31
5
41
6
53
7
67
8
83
M
This table of values might lead to the conjecture:
‘Numbers of the form n 2 + n + 11 are prime numbers’.
However, if n = 11 it is evident that n 2 + n + 11 is divisible by 11 and hence n 2 + n + 11 is
not prime. The value of n 2 + n + 11 when n = 11 provides a counter example to our
conjecture. The conjecture has been shown to be false.
Consider the conjecture:
SA
Consider the conjecture:
Consider the conjecture:
(a + b)2 = a 2 + b2 for all a and b ∈ R.
The values 2 and 3 for a and b respectively provide a counter
example.
x 2 > x for all x ∈ R.
1
The value for x provides a counter example.
3
The cube of a natural number is greater than the natural number.
The natural number 1 provides a counter example.
This final conjecture can be ‘fixed up’ by making the conjecture ‘the cube of a natural number
greater than 1 is greater than the natural number.’
‘Fix up’ the first two conjectures.
Proof
In Year 10 you may have come across proofs in geometry or used them in your problem
solving. Proofs are a very important part of Mathematics. If we have a conjecture which we
suspect to be true and for which a counter example cannot be found, then we try to construct a
‘chain of reasoning’ which will enable us to deduce the result from assumptions which are as
simple as possible.
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Chapter 25 — Proof and number
609
Example 1
Consider this pattern.
42 − 32 + 22 − 12 = 10
52 − 42 + 32 − 22 = 14
62 − 52 + 42 − 32 = 18
d 2 − c2 + b2 − a 2 = d + c + b + a’.
Prove this conjecture is true.
Solution
E
From this it could be conjectured that
‘If a, b, c, d are consecutive natural numbers with a < b < c < d then
PL
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The result is not immediately obvious.
Let the numbers a, b, c, d be n − 1, n, n + 1, n + 2 respectively.
Then
and
d +c+b+a =n+2+n+1+n+n−1
= 4n + 2
2
2
2
2
d − c + b − a = (n + 2)2 − (n + 1)2 + n 2 − (n − 1)2
= n 2 + 4n + 4 − n 2 − 2n − 1 + n 2 − n 2 + 2n − 1
= 4n + 2
M
The result has been proved. In fact, it has been proved for any four consecutive
integers.
It can be seen from the proof that a more general result is true.
If a, b, c, d are consecutive integers then
d 2 − c2 + b2 − a 2 = a + b + c + d
SA
The proof has actually added to the understanding of the problem by leading to a
generalisation of the original conjecture.
Example 2
Consider the following.
321
941
− 123 − 149
198
792
+ 297
+ 891
1089
1089
987
− 789
198
+ 891
1089
980
− 089
891
+ 198
1089
Conjecture
Take any three digit number whose digits decrease as you read them from left to right. Make
another number by reversing the order of the digits and subtract the smaller from the larger.
Reverse the order of the digits of the difference and add the number so formed to the
difference. The result will always be 1089.
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610
Essential Advanced General Mathematics
Solution
Let the number be a × 102 + b × 10 + c.
Reverse the digits to get the number c × 102 + b × 10 + a.
It will be assumed a > c without loss of generality.
a × 102 + b × 10 + c − c × 102 − b × 10 − a = (a − c) × 102 + (c − a)
E
Now c − a is negative, but (10 + c − a) is positive.
Write (a − c) × 102 + (c − a) = (a − c − 1) × 102 + 90 + (10 + c − a)
The digits are now correctly displayed.
Reverse the order to obtain (10 + c − a) × 102 + 90 + (a − c − 1) and add.
PL
(10 + c − a + a − c − 1) × 102 + (90 + 90) + (10 + c − a + a − c − 1)
= 9 × 102 + 180 + 9
= 900 + 180 + 9
= 1089
We can attempt to generalise in this case. The question arises ‘What happens when a
four digit number is considered?’ The proof also indicates that the base chosen is
important. What is the result for different bases?
M
Exercise 25A
1 Insert ⇒ or ⇐ to make the following into true statements about integers.
a p is even . . . pq is even
c x = 0 . . . xy = 0
b p + q is odd . . . pq is even
d ab = ac . . . b = c
2 State with reasons whether the following statements are true or false.
SA
a n (A) = 5 and n (B) = 3 ⇒ n (A ∪ B) = 8
c A ∩ B = ∅ ⇒ A = ∅ or B = ∅
b A⊂ B ⇒ A∩B = A
d A = ∅ ⇐ A = 3 Write down the converses of the statements given in 2 and state whether each new
statement is true or false.
4 State the converse of each of the following statements and also state whether the converse
is true or false.
a If n is odd then n 2 is odd.
c x 2 > 4 ⇐ x < −2
b N is divisible by 3 ⇒ N 2 is divisible by 9
5 The sum of two consecutive odd numbers is divisible by 4. Can you make similar
statements about:
a the sum of three consecutive odd numbers
b the sum of four consecutive odd numbers?
Prove your assertions.
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Chapter 25 — Proof and number
611
6 Prove that the sum of the squares of five consecutive integers is divisible by 5.
(Take n − 2, n − 1, n, n + 1, n + 2 as the integers.)
7 For each of the following statements give a counter example which proves that the given
statement is false.
E
a The sum of the squares of two numbers is equal to the square of the sum of these two
numbers.
b If a number is even then it is not divisible by 7.
√
√
√
a + b = a + b for all a, b ∈ R
c
b+c
1 a+b
1
a+
=
+ c for all a, b, c ∈ R
d
2
2
2
2
e The sum of two prime numbers is a prime number.
1 1
1
= + for all s, t ∈ R\ {0}
f
s+t
s
t
PL
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8 Write the converse of each of the following and state whether it is true or false.
a If a − b is positive then a > b
b If x = 0 and y = 0 then x = y
c If x + y = 0 then x = −y
d If x is even and y is odd then xy is even.
e The square of an even number is even.
9 For each of the following conjectures for the set of natural numbers either prove or
provide a counter example.
A number N has an odd number of divisors if and only if it is a perfect square.
For any odd number, N, there is a number with exactly N divisors.
For any number, N, there is a number with exactly N divisors.
There are infinitely many numbers with exactly N divisors (N = 1).
M
a
b
c
d
SA
10 a Show that a 2 + b2 ≥ 2ab for all real a and b.
√
u+v
≥ uv
b Hence, or otherwise, prove that if u ≥ 0, v ≥ 0, then
2
1 1
+
≥4
11 Prove that for all positive integers a and b, (a + b)
a
b
12 Prove each of the following for a, b ∈ Z .
a
c
e
f
If a and b are even, a + b is even.
b If a and b are odd, a + b is even.
If a is even and b is odd, ab is even.
d If a and b are odd, ab is odd.
a + b is even if and only if a − b is even.
a + b even and a − b even implies ab is a difference of perfect squares.
13 The floor of a rectangular room is covered with square tiles. The room is m tiles wide and
n tiles long with m ≤ n.If exactly half of the tiles are on the perimeter, find all possible
values of m and n.
14 Prove that no positive integer (except 1) all of whose digits are 1s is a perfect square.
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612
Essential Advanced General Mathematics
15 Find all right-angled triangles which have integer length sides and for which the area has
the same numerical value as the perimeter.
16 If a, b and c are positive integers such that no integer greater than 1 divides them all and
1
1 1
+ = , prove that a + b is a perfect square.
a
b
c
Note: This is a difficult question.
25.2
The principle of mathematical induction
E
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Suppose that P(k) is a statement for each positive integer k.
For example, the statement could be ‘k 2 − 2k + 1 ≥ 0 for each positive integer k’ or ‘k 2 − k
is an even integer for each positive integer k’.
The method of mathematical induction is used to prove P(k) for all k as follows.
PL
i Show P(1) is true.
ii Show P(k) ⇒ P(k + 1) for every positive integer k.
From this it can be seen that if both of the statements hold, then
P(1) ⇒ P(2), P(2) ⇒ P(3), P(3) ⇒ P(4)
M
and continuing in this fashion it can be seen that P(k) is true for every positive integer k.
The method of proof by induction can be illustrated by a line of dominoes (starting with
domino 1) stretching away without an end. To be sure that all dominoes will be knocked over it
is enough to know that
i the first domino is knocked over and
ii if one domino falls it will certainly knock over the next.
Another integer, besides 1, may be chosen for the starting value.
SA
Example 3
Prove that the sum of the first n integers is
n (n + 1)
.
2
Solution
Let P(n) be the statement that the sum of the first n integers is
1×2
2
k (k + 1)
i.e. that P(k) is true
1 + 2 + ··· + k =
2
k (k + 1)
1 + 2 + ··· + k + k + 1 =
+ (k + 1)
2
(k + 1)(k + 2)
=
2
P(1) is certainly true as
Assume
then
n (n + 1)
2
1=
∴ P(k + 1) is true
and the principle of induction gives that P(n) is true for all n.
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Chapter 25 — Proof and number
613
It is sometimes convenient to begin an induction at a point other than k = 1. In the following
example, the starting point is at k = 5.
Example 4
Prove that 2n > 1 + n 2 for all n > 4, n ∈ N .
Solution
Let P(n) be the statement that 2n > 1 + n 2 .
If n = 5, 25 = 32 and 1 + 52 = 26
As 25 > 26, P(5) is true.
Assume true for k.
E
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PL
i.e.
2k > 1 + k 2
For
2k+1 = 2.2k > 2(1 + k 2 ) = 2 + k 2 + k 2
1
2
But k > 2k for k > 2
Therefore
2 + k 2 + k 2 > 2 + 2k + k 2 = 1 + 1 + 2k − k 2
= 1 + (1 + k)2
2
From inequalities 1 and 2
M
2k+1 > 1 + (1 + k)2
∴ P(k + 1) is true
and the principle of induction gives that P(n) is true for all n ≥ 5.
SA
Example 5
Prove that 32n − 1 is divisible by 8 for all n ∈ N .
Solution
Let P(n) be the statement that 32n − 1 is divisible by 8.
If n = 1, 32 − 1 = 8 is divisible by 8
i.e. P(l) is true.
Assume true for k.
i.e. P(k) is true and 32k − 1 is divisible by 8
Consider
32(k+1) − 1 = 32k+2 − 1 = 32 · 32k − 32 + 32 − 1
= 32 (32k − 1) + 8
Since 32k − 1 is divisible by 8 and 8 is divisible by 8,
32(k+1) − 1 is divisible by 8.
∴ P (k + 1) is true, ∴ P(n) is true for all n ∈ N .
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Essential Advanced General Mathematics
Example 6
Prove that 2n > n 3 for n ≥ 10.
Solution
P(10) is true as 210 > 103 (1024 > 1000)
Assume P(k) is true i.e. 2k > k 3 for k > 9
2k+1 = 2 × 2k > 2 × k 3
2k+1 > (k + 1)3 + k 3 − 3k 2 − 3k − 1
2k+1 > (k + 1)3 + k 3 − 3k 2 − 3k − 459
2k+1 > (k + 1)3 + (k 2 + 6k + 51)(k − 9)
2k+1 > (k + 1)3 for k > 9 (Note: k 2 + 6k + 51 > 0 for all k)
P (k + 1) is true and P(n) is true for all n ≥ 10
E
then
∴
∴
∴
∴
∴
PL
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Exercise 25B
Sums
SA
M
1 Prove each of the following by induction, for all natural numbers n.
1
a 1 + 2 + · · · + n = n (n + 1)
2
1
b 12 + 22 + · · · + n 2 = n (n + 1) (2n + 1)
6
1
1
n
1
+
+ ··· +
=
c
(2n − 1) (2n + 1)
1×3 3×5
2n + 1
1
d 1 × 2 + 2 × 3 + 3 × 4 + · · · + n(n + 1) = n(n + 1)(n + 2)
3
e 1 × 4 + 2 × 7 + 3 × 10 + · · · + n(3n + 1) = n(n + 1)2
f 1 + 2 × 21 + 3 × 22 + · · · + n × 2n−1 = 1 + (n − 1) 2n
1
1
1
1
1
1 1
+
+
··· +
−
g
=
(2n + 1) (2n + 3)
3×5 5×7 7×9
2 3 2n + 3
h 13 + 33 + 53 + · · · + (2n − 1)3 = n 2 (2n 2 − 1)
1
i 2 × 1 + 3 × 2 + · · · + n(n − 1) = n(n 2 − 1)
3
Divisibility
2 Prove by induction that, for all natural numbers,
a
c
e
g
n(n + 1)(n + 2) is divisible by 3
8n + 2 × 7n − 1 is divisible by 7
23n − 1 is divisible by 22
5n 3 − 3n 2 − 2n is divisible by 6
b 4n 3 − 4n is divisible by 3
d n(n 2 + 2) is divisible by 3
f 8n − 5n is divisible by 3
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Chapter 25 — Proof and number
615
3 a Prove that 4n + 5n is divisible by 9 for all positive odd integers, n.
b Prove that 3n − 1 is divisible by 8 for all positive even integers, n.
c Prove that 6n+1 − 5 (n + 1) − 1 is divisible by 25 for all positive integers, n.
Inequalities
E
4 a Prove that 3n > n 3 for all positive integers greater than 3.
n4
for all positive odd integers.
b Prove that 13 + 23 + · · · + n 3 >
4
n
2
c Prove that 2 > n for all positive integers greater than 4.
d Prove that 2n > 3n for all positive integers greater than 3.
e Prove that 2n ≤ n! for all n ≥ 4.
PL
5 Prove that
1
1
1
1
n
a
+
+
+ ··· +
=
,n ∈ N
1×2 2×3 3×4
n (n + 1)
n+1
1
b 13 + 23 + · · · + n 3 = n 2 (n + 1)2 , n ∈ N
4
c 33n+1 + 9 × 2n+3 is divisible by 25, n ∈ N
6 Prove by induction that if S is a set of n elements, then S has 2n subsets.
25.3
Linear Diophantine equations
M
Consider the equation 3x + 4y = 1. This equation defines a straight line. If the values of x and
y are integers, a family of solutions may be described. This equation is called a linear
Diophantine equation.
On the graph, a family of solutions is illustrated.
y
(–5, 4)
SA
4
3
2
(–1, 1) 1
–5 – 4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
–1
(3, –2)
–2
–3
–4
–5
(7, –5)
–6
–7
–8
(11, –8)
–9
–10
–11
(15, –11)
x
Notice that as the integer solutions for x increase by 4, the y integer solutions decrease by 3.
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The solutions may be built up in the following way using (−1, 1) as the starting point.
x
−1 + 4
−1 + 2 × 4
−1 + 3 × 4
y
1−3
1−2×3
1−3×3
i.e.
x
3
7
11
y
−2
−5
−8
The family of solutions may be described as
x = −1 + 4t, y = 1 − 3t where t ∈ Z
Using set notation the solution is
E
P1: FXS/ABE
{(x, y) : x = −1 + 4t, y = 1 − 3t, t ∈ Z }
PL
If a linear Diophantine equation has one solution then it has infinitely many.
If ax + by = c is a linear Diophantine equation in two unknowns and (x0 , y0 ) is found to be
one solution, the general solution is given by
b
t
d
a
y = y0 − t, where t ∈ Z
d
and d is the highest common factor of a and b.
x = x0 +
M
Proof
Suppose x1 , y1 are solutions to the equation.
Then
ax1 + by1 = c
1
and
ax0 + by0 = c
2
SA
Subtracting 2 from 1
a(x1 − x0 ) = b(y0 − y1 )
Divide both sides by d
a
b
(x1 − x0 ) = (y0 − y1 )
d
d
a
b
and have no common factors.
d
d
b
Hence x1 − x0 must be divisible by
d
and
Similarly
b
t
d
a
y1 = y0 − t
d
x1 = x0 +
It can be proved by substitution that
x = x0 +
b
a
t, y = y0 − t, is a solution of the equation for any t ∈ Z .
d
d
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Chapter 25 — Proof and number
617
Example 7
A man has $200 in his wallet. This is made up of $50 and $20 notes. What are the possible
numbers of each of these types of notes?
Solution
Let x, y be the number of $50 and $20 notes respectively.
The linear Diophantine equation is
50x + 20y = 200
5x + 2y = 20
By inspection a solution is
PL
x = 4, y = 0
E
P1: FXS/ABE
The general solution will be
x = 4 + 2t, y = 0 − 5t, t ∈ Z
M
We are only interested in the case where x, y ≥ 0
Thus 4 + 2t ≥ 0 and 0 − 5t ≥ 0
Hence
−2 ≤ t ≤ 0
For t = −2, x = 0, y = 10
t = −1, x = 2, y = 5
t = 0, x = 4, y = 0
Hence the man can have ten $20 notes
or two $50 notes and five $20 notes
or four $50 notes.
SA
Exercise 25C
1 Find all solutions of the following Diophantine equations
a 11x + 3y = 1
d 22x + 6y = 2
b 2x + 7y = 2
e 2x + 7y = 22
c 24x + 63y = 99
f 10x + 35y = 110
2 For a, c, e in 1 find the solutions for which x, y are both positive.
3 Prove that if ax + by = c and the highest common factor of a and b does not divide c,
then there is no solution to the Diophantine equation.
4 A student puts a number of spiders (with eight legs) and a number of beetles (with six
legs) in a box. She counted 54 legs in all.
a Form a Diophantine equation.
b Find the number of spiders and the number of beetles in the box.
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618
Essential Advanced General Mathematics
5 Helena has a number of coins in her wallet. They are all either 20c or 50c coins. The total
value of the coins is $5.00. What are the possible numbers of each type of coin?
6 One of the solutions of the equation 19x + 83y = 1983 in positive integers x and y is
obviously x = 100, y = 1. Show that there is only one other pair of positive integers
which satisfy this equation and find it. Consider the equation 19x + 98y = 1998.
E
7 A man has $500 in his wallet made up of $50 and $10 notes. Find the possible
combinations of notes that he could have.
PL
8 There are seven coconuts and 63 heaps of pineapples. Each heap has exactly the same
number of pineapples. The fruit is to be divided equally between 23 people.
Let x be the number of pineapples in each heap and y the number of pieces of fruit that
each person receives.
Form a Diophantine equation and find the possible values for x and y.
9 A dealer spent $10 000 buying cattle, some at $410 each and the rest at $530 each. How
many of each sort did she buy?
10 Find the smallest positive number which, when divided by 7, leaves a remainder of 6, and
when divided by 11 leaves a remainder of 9. Also find the general form of such numbers.
11 Given a 3 litre jug and a 5 litre jug can I measure exactly 7 litres of water? If it is possible,
explain how this may be done as efficiently as possible.
12 The Guadeloupe Post Office has only 3c and 5c stamps. What amounts of postage can the
post office sell?
M
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13 A man spent $29.60 buying party hats. There were two types of party hat. Type A cost
$1.70 while type B cost $1.00. How many of each type did he buy?
The Euclidean algorithm
SA
25.4
The Euclidean algorithm provides a method for finding the highest common factor of two
numbers and also a method for solving linear Diophantine equations.
Theorem 1
If a, b are integers with a > 0, then there are unique integers q, r such that b = aq + r with
0 ≤ r < a.
Proof
Suppose there exists another pair of integers, q1 and r1 with
b = aq1 + r1 and 0 ≤ r1 < a.
Suppose that r > r1 . Then by subtraction
∴
∴
0 = aq1 + r1 − (aq + r )
0 = a(q1 − q) + (r1 − r )
r − r1 = a(q1 − q)
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Chapter 25 — Proof and number
619
Now since the right hand side is an integer and is a multiple of a, then a divides the left hand
side. But the left hand side is an integer which is greater than zero and less than a.
Therefore the assumption, that r > r1 , must be false.
But if it is assumed r < r1 , then consider
0 = (aq + r ) − (aq1 + r1 )
Thus
so
r = r1 . In that case
r − r1 = 0 = a(q1 − q)
q1 = q and r1 = r
and the uniqueness of integers q, r has been proved.
Example 8
E
and a similar contradiction will arise
PL
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Express −45 in the form 6q + r where 0 ≤ r < 6.
Solution
M
Here −45 = 6(−8) + 3
Note that −45 = 6(−7) − 3 is not a correct answer since the remainder −3 is less
than zero.
In the following, (a, b) denotes the highest common factor of the integers a and b.
Theorem 2
SA
If a and b are two integers, a = 0 and b = aq + r where q, r are integers, then (a, b) = (a, r ).
(This theorem may be used to determine the highest common factor of any two given integers.)
Proof
If d is a common divisor of a and r, then d divides the right-hand side of equation b = aq + r
and so d divides b.
This proves that all common divisors of a and r will be common divisors of a and b. But
(a, r ) is a common divisor of a and r, and so (a, r ) must divide a and b. It follows that (a, r )
must divide (a, b).
That is
(a, b) = (a, r )x where x is an integer
Now rewrite equation
b = aq + r as r = b − aq
1
All common divisors of a and b divide the right hand side of this relation and so divide r.
Thus (a, b) must divide r as well as a. Hence (a, b) must divide (a, r ).
It has now been proved
(a, r ) = (a, b)y, where y is an integer
2
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620
Essential Advanced General Mathematics
From equations 1 and 2 obtain
(a, r ) = [(a, r )x]y
1 = xy
Hence
This equation in integers x, y is possible only if both x, y are +1 or −1.
Hence (a, b) = +(a, r ), since −1 is inappropriate to this problem.
Since the integer r is less than the integer b, the calculation of (a, r ) is easier than the
calculation of (a, b).
E
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Example 9
Find (1271, 3875).
PL
3875 = 1271 × 3 + 62
Solution
and
(1271, 3875) = (1271, 62) by theorem 2.
Now
1271 = 62 × 20 + 31
∴
(62, 1271) = (62, 31) again by theorem 2.
As
62 = 31 × 2 + 0
we have (1271, 3875) = 31 by using the theorem a final time.
This procedure (algorithm) is called the Euclidean algorithm.
M
Method for finding a solution of a linear
Diophantine equation
The method presented here uses the Euclidean algorithm.
Example 10
SA
Find a, b ∈ Z such that 22a + 6b = 2
Solution
Apply the division algorithm to 22 and 6
i.e.
22 = 3 × 6 + 4
6=1×4+2
4=2×2
(22, 6) = 2
1
2
3
Using these results
∴
∴
∴
∴
or
2=6−1×4
from 2
2 = 6 − 1(22 − 3 × 6) from 1
2 = 6 − 22 + 3 × 6
2 = 4 × 6 − 1 × 22
−1 × 22 + 4 × 6 = 2
A solution is (−1, 4)
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Chapter 25 — Proof and number
621
Example 11
Find a, b ∈ Z such that 125a + 90b = 5
Solution
Divide by 5.
25a + 18b = 1
Apply the division algorithm.
25 = 1 × 18 + 7
18 = 2 × 7 + 4
7=1×4+3
4=1×3+1
3=3×1
PL
1
2
3
4
5
E
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from 4
from 3
from 2
from 1
M
i.e., (25, 18) = 1
1=4−1×3
∴
1 = 4 − 1(7 − 1 × 4)
∴
1=2×4−1×7
∴
1 = 2 × (18 − 2 × 7) − 1 × 7
∴
1 = 2 × 18 − 5 × 7
∴
1 = 2 × 18 − 5(25 − 18)
∴
1 = 7 × 18 − 5 × 25
∴
a = −5 and b = 7 is one solution.
∴ Solution is given by a = −5 + 18t and b = 7 − 25t; t ∈ Z .
SA
Exercise 25D
1 For the following, express b in the form b = aq + r with 0 ≤ r < a and show in each case
(a, b) = (a, r ).
a a = 5, b = 43
b a = 13, b = 39
c a = 17, b = 37
d a = 16, b = 128
2 If d is a common factor of a and b, prove that d is a common divisor of a + b and a − b.
3 Use the Euclidean algorithm to find
a (4361, 9284)
b (999, 2160)
c (−372, 762)
d (5255, 716 485)
4 Solve in the integers the equations
a 804x + 2358y = 6
c 3x + 4y = 478
e 804x + 2688y = 12
b 18x + 24y = 6
d 3x − 5y = 38
f 1816x + 2688y = 8
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Essential Advanced General Mathematics
Chapter summary
E
For two statements p and q, ‘ p ⇒ q’ is read ‘p implies q’.
q ⇒ p is the converse of p ⇒ q
Combining p ⇒ q and q ⇒ p we can write p ⇔ q which may be read ‘p is equivalent to q’
An equivalent symbol for ⇔ is iff which is read ‘if and only if ’.
A counter example is an example which proves a conjecture to be false.
The method of mathematical induction is used to prove P(k) for all k as follows
i Show P(1) is true.
ii Show P(k) ⇒ P(k + 1) for every positive integer k.
M
PL
Consider the equation 3x + 4y = 1. This equation defines a straight line. The coefficients
of the left hand side of the equation and the right hand side are integers. If the values of x
and y are integers, a family of solutions may be described. This equation is called a linear
Diophantine equation.
If a linear Diophantine equation has one solution then it has infinitely many.
If ax + by = c is a linear Diophantine equation in two unknowns and (x0 , y0 ) is found to
b
a
be one solution, the general solution is given by x = x0 + t, y = y0 − t , where t ∈ Z
d
d
If a, b are integers with a > 0, then there are unique integers q, r such that b = aq + r with
0 ≤ r < a.
The Euclidean algorithm
In the following, (a, b) denotes the highest common factor of the integers a and b.
If a and b are two integers, a = 0 and b = aq + r where q, r are integers, then
(a, b) = (a, r ).
(This result may be used to determine the highest common factor of any two given integers,
and to solve any linear Diophantine equation.)
Multiple-choice questions
SA
Review
622
1 If m is a positive even integer and n is a positive odd integer the statement which is false is
A m + 2n is even
B m + n is odd
C 3m + 2n is even
2
D m × n is even
E m + n is even
2 The statement below which is true is
A x − 3 > 0 ⇒ (x − 3)(2 − x) > 0
C (x − 3)(2 − x) > 0 ⇔ x − 3 > 0
E x − 3 > 0 ⇔ (x − 3)(x − 2) > 0
B (x − 3)(2 − x) > 0 ⇒ x − 3 > 0
D x − 3 > 0 ⇒ (x − 3)(x − 2) > 0
3 If p and q are positive real numbers, and p > q with p + q = 1, the largest quantity from
the following is
1
1
1
C
D
E pq
B
A p
pq
p
q
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Chapter 25 — Proof and number
1
1
<
p
q
A always
B never
C only when p and q are positive
D for all p and q except when both are negative
E whenever pq > 0
E
5 The number of factors that the integer 2 p 3q 5r has is
( p + q + r )!
A
B pqr
C p+q +r
q!q!r !
D ( p + 1)(q + 1)(r + 1)
E p+q +r +1
6 If an integer of two digits is k times the sum of its digits, the number formed by
interchanging the digits is the sum of the digits multiplied by
A 9−k
B 10 − k
C 11 − k
D k−5
E k+8
PL
7 The number of pairs of integers (m, n) which satisfy the equation m + n = mn is
A 1
B 2
C 3
D 4
E more than 4
8 If a, b, c are any real numbers, and a > b, the statement which must be true is
1
1
1
1
E
<
>
B ac > bc
C a 2 > b2
D a+c >b+c
A
a
b
a
b
9 The number of solutions of the Diophantine equation 3x + 5y = 1008, where x and y are
positive integers, is
A 1
B 134
C 68
D 67
E infinite
M
10 If y = (n − 1)(n − 2)(n − 3), where n is a positive integer, then y is not always divisible by
A 6
B 5
C 3
D 2
E 1
SA
Short-answer questions (technology-free)
1 Find the highest common factor of 1885 and 365 using the Euclidean algorithm.
2 Consider 9x + 43y = 7. Solve for x and y where
a x ∈ Z, y ∈ Z
b x ∈ Z +, y ∈ Z +
3 Prove that the product of two consecutive odd integers is odd. (You may assume that the
sum and product of any two integers is an integer.)
4 Using the Euclidean algorithm, find the highest common factor of 10 659 and 12 121.
5 a Solve the Diophantine equation 5x + 7y = 1
b Hence solve the Diophantine equation 5x + 7y = 100
c Find {(x, y) : 5x + 7y = 1; y ≥ x, x, y ∈ Z }.
6 The sum of the ages of Tom and Fred is 63. Tom is twice as old as Fred was when Tom was
as old as Fred is now. What are the ages of Tom and Fred?
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Review
4 If p > q and pq = 0, it is true that
623
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Essential Advanced General Mathematics
Extended-response questions
1 The strips shown below are formed by joining an odd number of triangles together. The
triangles are formed on 1 cm isometric paper as shown.
E
A strip of length 3
A strip of length 5
A strip of length 7
PL
In the strip of length 3 there are two parallelograms as shown below.
M
In the strip of length 5 there are six parallelograms.
a
b
c
d
Find and draw all the possible parallelograms in a strip of length 7.
How many parallelograms are there in a strip of length 11?
Find the number of parallelograms in a strip of length n where n is an odd number.
Prove your result.
SA
Review
624
2 A common error for adding fractions is the following:
c
a+c
a
+ =
b d
b+d
Investigate this error for a, b, c, d ∈ N .
3 Twelve cubes are used to build a tower three
cubes high. This is shown with the aid of
isometric dot paper in the diagram.
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Chapter 25 — Proof and number
625
E
PL
4 You have an inexhaustible supply of 5¢ and 8¢ stamps.
a List all possible ways of obtaining a total value of 38¢ with these stamps.
b List all possible ways of obtaining a total of $1.20 with these stamps.
5 The digits of a three-digit number are interchanged so that none of the digits has retained its
original place. Then the new number is subtracted from the original. If the difference is a
two-digit number which is also a perfect square, find all such two-digit numbers.
M
6 Find the positive integer n such that n and n + 100 have an odd number of divisors.
1
7 Prove that if a > 0, then a + ≥ 2.
a
1 2
Hint: Use the result that a −
≥ 0 for all a ∈ R.
a
8 Show that the sum of three consecutive positive integers divides the sum of the cubes of
these three integers.
Hint: Consider the numbers n − 1, n, n + 1.
9 Add 1 to the product of four consecutive positive integers. Prove that the result is a perfect
square.
SA
10 Choose any two positive integers which are not divisible by 3. Prove the difference between
their squares is divisible by 3.
11 Prove that for every positive integer n, the expression n 2 (n 4 − 1) is divisible by 60.
12 Choose any six consecutive positive integers greater than 3. Prove that there are at most two
prime numbers among them.
13 Find all six-digit numbers with the property:
If the first and last digits are interchanged, then the new six-digit number is six times the
original number.
14 Prove by mathematical induction
1
a 12 + 42 + 72 + · · · + (3n − 2)2 = n(6n 2 − 3n − 1)
2
b 6n + 4 is divisible by 10 for any natural number n.
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Review
a How many cubes are needed to build a tower of this type which is eighty cubes high?
Work out an expression for the number of cubes needed to build a tower of this type
which is n cubes high.
b The number of cubes needed to build a tower n cubes high, Tn , is related to the number
of cubes needed to build a tower (n − 1) cubes high, Tn−1 , by the formula
Tn = Tn−1 + kn, where k is a constant.
What is the value of k?
c How many dots on the isometric dot paper are needed to draw a tower of this type n
cubes high?
d Investigate other types of towers.