Download number theory and methods of proof

Bridge of Don Academy – Department of Mathematics
Advanced Higher: Unit 3 – Number Theory and Proof
NUMBER THEORY AND METHODS OF PROOF
Learning Outcomes:
1.
Know the terms “necessary condition”, “sufficient condition”, “if and only if”, “converse”,
“negation” and “contrapositive”.
2.
Use further methods of mathematical proof: some simple examples involving natural
numbers.
3.
Direct methods of proof: sums of certain series and other straightforward results.
4.
Further proof by contradiction.
5.
Further proof by contrapositive.
6.
Further proof by mathematical induction e.g. prove the following result
n
r
2
r 1
 1 n(n  1)(2n  1), n  N .
6
n
7.
Know the result
r
r 1
3
 1 n 2 (n  1)2 .
4
n
8.
Apply the above results and the one for
 r to prove by direct methods results concerning
r 1
other sums.
9.
Know the division algorithm and proof.
10.
Use Euclidean algorithm to find the greatest common divisor (g.c.d.) of two positive
integers.
11.
Know how to express the g.c.d. as a linear combination of the two integers [A/B].
12.
Use the division algorithm to write integers in terms of bases other than 10 [A/B].
Page 1 of 21
Bridge of Don Academy – Department of Mathematics
Advanced Higher: Unit 3 – Number Theory and Proof
DEFINITIONS:
Statements and Negations
p is a statement (e.g. a rectangle has four sides) and can be true or false.
~p (pronounced “not p”) is the negation of a statement (e.g. a rectangle does not have four sides)
and can be false or true.
If a statement is true then the negation is false.
If a statement is false then the negation is true.
Contrapositive
If p  q then the contrapositive is ~p  ~q.
If p  q is true then the contrapositive ~p  ~q must be true also.
Similarly, If p  q is false then the contrapositive ~p  ~q must be false also.
Converse
We are already familiar with this concept – think of the Converse of the Theorem of Pythagoras:
Theorem of Pythagoras states
Converse of Pythagoras states
[right-angled triangle]  [c2 = a2 + b2]
[c2 = a2 + b2]  [right-angled triangle]
i.e. if a statement states that p  q then the converse will state that q  p.
Note that if p  q is true it does not always hold that q  p is true also.
For example:
x = 3  x2 = 9
however
x2 = 9

x = 3.
Necessary and Sufficient (and if and only if)
If a statement and its converse are true then the implication () works both ways ()
i.e. if p  q and q  p then p  q and we say that p is true if and only if q is true.
If p  q then we say that p is a sufficient condition for q.
If q  p then we say that p is a necessary condition for q.
If p  q and q  p then we say that p is a necessary and sufficient condition for q.
Page 2 of 21
Bridge of Don Academy – Department of Mathematics
Advanced Higher: Unit 3 – Number Theory and Proof
Need some more practice?
Exercise
1.
2.
3.
Write down the negation of each statement:
(a)
For all real x, x 2 is positive.
(b)
Some pupils find maths easy.
(c)
There exists a positive integer x such that x + 3 > 0.
(d)
All numbers of the form 2n  1 , where n is an integer, are prime.
State the converse of each statement and show by counter example that the converse is false.
(a)
If a number ends in zero, it is divisible by 5.
(b)
All primes greater than 2 are odd numbers.
(c)
If two numbers are odd then their sum is even.
Which of the following statements are necessary, sufficient or neither for the statement
q: “natural number n is divisible by 6” to be correct?
(a)
p: “n is divisible by 3”
(b)
p: “n is divisible by 12”
(c)
p: “ n 2 is divisible by 12”
(d)
p: “n = 384”
(e)
p: “n is even and divisible by 3”
(f)
p: “n = m(m + 1)(m + 2) where m is
some natural number”
Page 3 of 21
Bridge of Don Academy – Department of Mathematics
4.
Advanced Higher: Unit 3 – Number Theory and Proof
Prove by contradiction:
(a)
that if x and y are integers such that x + y is odd, then one of them must be odd and
the other must be even.
(b)
that if m and n are integers such that mn 2 is even, then at least one of m or n is even.
Page 4 of 21
Bridge of Don Academy – Department of Mathematics
Advanced Higher: Unit 3 – Number Theory and Proof
Induction
This process was examined in Unit 2. We are now going to look at some more examples:
n
1.
 r  12 n(n  1)
Remember:
see Unit 2 notes
LEARN
r 1
n
2.
Prove, by induction, that
r
 1 n(n  1)(2n  1) .
6
2
r 1
Step 1:
For n  1,
1
r
2
 12  1 and
r 1
LEARN
1 n(n  1)(2n  1)  1  1  2  3  1
6
6
 true for n  1
Step 2:
Assume for n  k 
k
r
r 1
Step 3:
2
 1 k (k  1)(2k  1).
6
For n = k + 1,
k 1
k
r  r
2
r 1
2
 (k  1) 2
r 1
C is whatever
number is at
the front
 1 k (k  1)(2k  1)  (k  1) 2
6
[now take out a common factor of C( k  1), as always]
 1 (k  1)  k (2k  1)  6(k  1) 
6
 1 (k  1)  2k 2  7k  6 
6
 1 (k  1)(k  2)(2k  3)
6
 1 (k  1)  (k  1)  1  2(k  1)  1
6
Step 4 (conclusion): Hence, if true for n = k then also true for n = k + 1
and since it is true for n = 1 it is true n.
n
r
r 1
2
 1 n(n  1)(2n  1) is true n.
6
Page 5 of 21
Bridge of Don Academy – Department of Mathematics
Advanced Higher: Unit 3 – Number Theory and Proof
n
3.
Prove, by induction, that

r 3  1 n 2 (n  1)2
4
r 1

or



2
1 n(n  1)  .

2

LEARN
Page 6 of 21
Bridge of Don Academy – Department of Mathematics
Advanced Higher: Unit 3 – Number Theory and Proof
n
4.
Prove, by direct methods, that
 (r  2)(r  3)  13 n(n
2
 9n  26) .
r 1
n

(r  2)(r  3) 
r 1
n
 (r
2
 5r  6)
r 1

n

r 1
n

r2  5
n
1
r6
r 1
r 1
[each of these has been proven already by induction so ....]


 1 n(n  1)(2n  1)  5 1 n(n  1)  6n
6
2
 1 n (n  1)(2n  1)  15(n  1)  36
6
 1 n  2n 2  3n  1  15n  15  36 
6
 1 n  2n 2  18n  52 
6
 1 n  n 2  9n  26 
3
Other types of problems:
5.
Prove that 5n + 3 is always divisible by 4, where n  N.
51 + 3 = 8 and 8 is divisible by 4 i.e. 8  4 = 2  true for n = 1.
(i)
For n = 1,
(ii)
Assume true for n = k
(iii)
For n = k + 1,
 5k + 3 = 4p where p  N.
 5k = 4p – 3
5k 1  3  5k  51  3
 (4 p  3)  5  3
 20 p  15  3
 20 p  12
 4(5 p  3)
(iv)

if true for n = k then also true for n = k + 1
and since true for n = 1, true for all n  N.
Now try:
Page 141, Exercise 3A – question 2, 12
Page 141, Exercise 3B – question 1, 7, 8
Page 7 of 21
Bridge of Don Academy – Department of Mathematics
6.
Advanced Higher: Unit 3 – Number Theory and Proof
Prove that, for sufficiently large n, 12 + 32 + 52 + ….. + (2n – 1)2 =
(2n  1) 2n (2n  1)
.
6
Page 8 of 21
Bridge of Don Academy – Department of Mathematics
7.
Advanced Higher: Unit 3 – Number Theory and Proof
Prove that, for all n, n  N, 6n + 4 is divisible by 10.
Page 9 of 21
Bridge of Don Academy – Department of Mathematics
Advanced Higher: Unit 3 – Number Theory and Proof
The Division Algorithm
203  8 = 25 remainder 3
 203 = 25  8 + 3
Given any two positive integers, a and b (b  0), there exists unique integers q and r where 0  r < b
such that
a = bq + r.
This is known as the Division Algorithm.
Examples:
8.
a = 193, b = 17
 193 = 17  11 + 6
9.
(i.e. q = 11 and r = 6)
75 = 12q + r where q and r are integers.
Use the division algorithm to find the values of q and r.
Page 10 of 21
Bridge of Don Academy – Department of Mathematics
Advanced Higher: Unit 3 – Number Theory and Proof
Euclidean Algorithm
The repeated application of the Division Algorithm is called the Euclidean Algorithm and is used to
find the greatest common divisor (g.c.d.) of two or more positive integers, where this cannot be
done simply.
Suppose we have two integers a and b (b  0) then the division algorithm states
a  bq1  r1
(0  r  b) and by the theorem (a, b)  (b, r1 )
 b  r1q2  r2
(0  r2  r1 )
 r1  r2 q3  r3
(0  r3  r2 )
 rn 2  rn 1qn  rn
(0  rn  rn 1 )
 rn 1  rn qn 1  0
(i.e. eventually r becomes 0)
Examples:
10.
Find the g.c.d. of 15 and 24.
[N.B. common sense tells us that the g.c.d. of 15 and 24 is 3]
However, let us look at the Euclidean algorithm in action
15  24  0  15
 24  15  1  9
 15  9  1  6
 9  6 1  3
 6  3 2  0
The last non-zero remainder was 3, therefore the g.c.d. of 15 and 24 is 3.
Page 11 of 21
Bridge of Don Academy – Department of Mathematics
11.
Advanced Higher: Unit 3 – Number Theory and Proof
Find the g.c.d. of 583 and 318.
We can express the g.c.d. as a linear combination of the two integers.
Example:
12.
Write the g.c.d of 1147 and 851 as a linear combination of 1147 and 851.
[alternative phrasing: Find the values of x and y when (1147, 851) = x1147 + y851.]
We can calculated that (1147, 851) = 37 as below
1147  1  851  296
(1)  296  1147  1  851
851  2  296  259
(2)  259  851  2  296
296  1  259  37
(3)  37  296  1  259
259  7  37  0
Start with (3), which gives us
37  296  1  259
then from (2) we get
 296  1  (851  2  296)
 296  1  851  2  296
 3  296  1  851
then from (1) we get
 3  (1147  1  851)  1  851
 3  1147  3  851  1  851
 3  1147  4  851
(1147, 851) = 3  1147 – 4  851
 x = 3 and y = – 4
Page 12 of 21
Bridge of Don Academy – Department of Mathematics
Advanced Higher: Unit 3 – Number Theory and Proof
Example:
13.
(a)
Find the g.c.d. of 7293 and 798.
(b)
Hence find the integers x and y to write this g.c.d. in the form x  7293 + y  798.
Now try:
Page 147, Exercise 5 – questions 4(a, b), 7
Page 13 of 21
Bridge of Don Academy – Department of Mathematics
Advanced Higher: Unit 3 – Number Theory and Proof
We can also use the division algorithm to express integers in terms of bases other than 10.
Remember that 235 base 10 means
23510 = 2  102 + 3  101 + 5  100
and 142 base 6 means
1426 = 1  62 + 4  61 + 2  60
=
To change 15810 to base6
158  6 = 26 remainder 2
26  6 = 4 remainder 2
4  6 = 0 remainder 4
start at the bottom
and read upwards
to get the answer
6210
 15810 = 4226
Examples:
14.
Express t81e12 in base 10, where t and e are digits representing 10 and 11 respectively.
15.
Express 33310 to base 4.
Now try:
Page 151, Exercise 7 – questions 1(a, c), 2(a, c, f)
Page 152, Review – questions 1 to 5
Page 14 of 21
Bridge of Don Academy – Department of Mathematics
Advanced Higher: Unit 3 – Number Theory and Proof
Further examples (from SQA exams):
16.
Use the Euclidean algorithm to find integers x and y such that
149 x  139 y  1 .
4
Page 15 of 21
Bridge of Don Academy – Department of Mathematics
17.
Advanced Higher: Unit 3 – Number Theory and Proof
Use the Euclidean algorithm to show that (231, 17) = 1 where (a, b) denotes the highest
common factor of a and b.
Hence find integers x and y such that 231x + 17y = 1.
4
Page 16 of 21
Bridge of Don Academy – Department of Mathematics
18.
Advanced Higher: Unit 3 – Number Theory and Proof
 2 1
A matrix A  
 . Prove by induction that
 1 0 
 n 1 n 
An  
,
 n 1  n 
where n is any positive integer.
6
Page 17 of 21
Bridge of Don Academy – Department of Mathematics
19.
(a)
Advanced Higher: Unit 3 – Number Theory and Proof
Prove by induction that for all natural numbers n  1
n
 3 r
2

 r  (n  1)n(n  1) .
r 1
4
40
(b)
Hence evaluate
 3 r
r 11
2

r .
2
Page 18 of 21
Bridge of Don Academy – Department of Mathematics
20.
Advanced Higher: Unit 3 – Number Theory and Proof
 
n
Prove by induction that d n xe x  ( x  n)e x for all integers n  1 .
dx
5
Page 19 of 21
Bridge of Don Academy – Department of Mathematics
21.
Advanced Higher: Unit 3 – Number Theory and Proof
Prove by contradicton that if x is an irrational number, then 2  x is irrational.
4
22.
For all natural numbers n, prove whether the following results are true or false.
(a)
n3  n is always divisible by 6.
(b)
n3  n  5 is always prime.
5
Page 20 of 21
Bridge of Don Academy – Department of Mathematics
23.
Advanced Higher: Unit 3 – Number Theory and Proof
The square matrices A and B are such that AB  BA .
Prove by induction that An B  BAn for all integers n  1 .
5
Page 21 of 21