The document discusses mathematical induction as a method for proving statements about natural numbers, emphasizing two key steps: verifying the base case and proving the inductive step. It also outlines divisibility rules for integers, providing methods for checking divisibility by numbers like 2, 3, 4, 5, 6, 7, and 13. The document includes examples to illustrate the application of these principles in both mathematical induction and divisibility tests.

1. Mathematical Induction
And
Divisibility Rules

2. UNIVERSITY OF AZAD JAMMU & KASHMIR
MUZAFFARABAD
Department of CS & IT
 Dawood Faheem Abbasi 05

3. Mathematical induction
INTRODUCTION:
The principal of mathematical induction is useful tool for proving that a
certain predicate is true for all natural numbers.
It can not be used to discover theorems.
In mathematical induction, we try to prove that L.H.S is equal to R.H.S
It has only 2 steps
Step 1. Show it is true for first one.
Step 2. Show that if any one is true then the next one is true..
Then all are true.

4. Have you heard of “dominos effect”?
 Step1. The first domino falls.
 Step2. When any domino falls, the next domino falls.
So.. All dominos falls.

5. In the world of numbers we say;
Step 1. Show that it is true for n=1.
Step 2. Show that if n=k is true then n =k+1 is also true.
How to do it?
Step 1 is usually easy, we just have to prove it is true for n=1.
Step 2 is done by;
Assume it is true for n=k.
Prove it is true for n=k+1

6. Types of questions solve by induction
Statements giving expressions about summation or multiplication of
special series
Statements to show divisibility of an expression by a certain natural
number.
Statements containing signs of inequality.

8. Adding up all odd numbers.
1+3+5+……….+(2n-1)=n2
 Step 1. show it is true for n=1
1=12 is true
 Step 2. assume it is true for n=k
1+3+5+…..+(2k-1)=k2 is true
 Now, prove it is true for “k+1”
1+3+5+…..+(2k-1)+(2(k+1)-1)=(k+1) 2
 We know that 1+3+5+…(2k-1)=k2 so,
K2+(2(k+1)-1)=(k+1) 2
 Expanding;
k2+2k+2-1=k2+2k+1
K2+2k+1=k2+2k+1 they are same! So it is true.

11. Prove that 2n>n for all positive integers n
Let p(n)be the given statement
P(n):2n>n
 step1. when n=1
21>1
Hence , p(1)is true.
 Step 2. assume that p(k) is true for any positive integer k i.e
2k>k
 We shall now prove that pk+1is true.
Multiplying both sides of 1 by 2.
2.2k>2k i.e
2k+1>2k
K+k>k+1
Therefore, pk+1 is true when p(k+1) is true.
So, P(n) is true for every positive integer n.

12.  n < 2n
 for all positive integers n.
 Solution: Let P(n) be the proposition that n < 2n.
 BASIS STEP: P(1) is true, because 1 < 21 = 2. This completes the basis step.
 INDUCTIVE STEP: We first assume the inductive hypothesis that P(k) is true for anarbitrary
 positive integer k. That is, the inductive hypothesisP(k) is the statement thatk < 2k.To complete
 the inductive step, we need to show that if P(k) is true, then P(k + 1), which is the statement
 that k + 1 < 2k+1, is true. That is, we need to show that ifk < 2k, then k + 1 < 2k+1. To show
 320 5 / Induction and Recursion
 that this conditional statement is true for the positive integer k, we first add 1 to both sides of
 k < 2k, and then note that 1 ≤ 2k. This tells us that
 k + 1
 IH
 <2k + 1 ≤ 2k + 2k = 2 · 2k = 2k+1.
 This shows that P(k + 1) is true, namely, that k + 1 < 2k+1, based on the assumption that P(k)
 is true. The induction step is complete.
 Therefore, because we have completed both the basis step and the inductive step, by
 the principle of mathematical induction we have shown that n < 2n is true for all positive
 integers n.

13.  Solution: To construct the proof, let P(n) denote the proposition: “7n+2 + 82n+1 is divisible by
 57.”
 BASIS STEP: To complete the basis step, we must show that P(0) is true, because we want
 to prove that P(n) is true for every nonnegative integer. We see that P(0) is true because
 70+2 + 82·0+1 = 72 + 81 = 57 is divisible by 57. This completes the basis step.
 INDUCTIVE STEP: For the inductive hypothesis we assume that P(k) is true for an arbitrary
 nonnegative integer k; that is, we assume that 7k+2 + 82k+1 is divisible by 57. To complete the
 inductive step, we must show that when we assume that the inductive hypothesis P(k) is true,
 then P(k + 1), the statement that 7(k+1)+2 + 82(k+1)+1 is divisible by 57, is also true.
 The difficult part of the proof is to see howto use the inductive hypothesis.To take advantage
 of the inductive hypothesis, we use these steps:
 7(k+1)+2 + 82(k+1)+1 = 7k+3 + 82k+3
 = 7 · 7k+2 + 82 · 82k+1
 = 7 · 7k+2 + 64 · 82k+1
 = 7(7k+2 + 82k+1) + 57 · 82k+1.
 We can now use the inductive hypothesis, which states that 7k+2 + 82k+1 is divisible by
 57. We will use parts (i) and (ii) of Theorem 1 in Section 4.1. By part (ii) of this theorem, and
 the inductive hypothesis, we conclude that the first term in this last sum, 7(7k+2 + 82k+1), is
 divisible by 57. By part (ii) of this theorem, the second term in this sum, 57 · 82k+1, is divisible
 by 57. Hence, by part (i) of this theorem, we conclude that 7(7k+2 + 82k+1) + 57 · 82k+1 =
 7k+3 + 82k+3 is divisible by 57. This completes the inductive step.
 Because we have completed both the basis step and the inductive step, by the principle of
 mathematical induction we know that 7n+2 + 82n+1 is divisible by 57 for every nonnegative
 integer n.

14. Strong Induction
 Before we illustrate how to use strong induction, we state this principle again.
 STRONG INDUCTION To prove that P(n) is true for all positive integers n, where P(n)
is a propositional function, we complete two steps:
 BASIS STEP: We verify that the proposition P(1) is true.
 INDUCTIVE STEP: We show that the conditional statement [P(1) ∧ P(2) ∧ · · · ∧
P(k)] → P(k + 1) is true for all positive integers k.

15.  Note that when we use strong induction to prove that P(n) is true for all positive integers n,
our inductive hypothesis is the assumption that P(j) is true for j = 1, 2, . . . , k. That is, the
inductive hypothesis includes all k statements P(1), P(2), . . . , P (k). Because we can use all k
statements P(1), P(2), . . . , P (k) to prove P(k + 1), rather than just the statement P(k) as in a
proof by mathematical induction,
 strong induction is a more flexible proof technique. Because
of this, some mathematicians prefer to always use strong induction instead of mathematical
induction, even when a proof by mathematical induction is easy to find.
 You may be surprised that mathematical induction and strong induction are equivalent.
 That is, each can be shown to be a valid proof technique assuming that the other is valid.

17. Divisibility rules
for numbers

18. Introduction:
The rules given below transform a given number into a generally
smaller number, while preserving divisibility by the divisor of
interest. Therefore, unless otherwise noted, the resulting number
should be evaluated for divisibility by the same divisor.

19. Divisibility by 2
 First, take any number (for this example it will be 376) and note the last digit
in the number, discarding the other digits. Then take that digit (6) while
ignoring the rest of the number and determine if it is divisible by 2. If it is
divisible by 2, then the original number is divisible by 2.

20. Examples
 376 (The original number)
 37 6 (Take the last digit)
 6 ÷ 2 = 3 (Check to see if the last digit is divisible by 2)
 376 ÷ 2 = 188 (If the last digit is divisible by 2, then the whole number is
divisible by 2)

21. Divisibility by 3 or 9
First, take any number (for this example it will be 492) and add
together each digit in the number (4 + 9 + 2 = 15). Then take
that sum (15) and determine if it is divisible by 3. The original
number is divisible by 3 (or 9) if and only if the sum of its digits
is divisible by 3 (or 9).
If a number is a multiplication of 3 consecutive numbers then
that number is always divisible by 3. This is useful for when the
number takes the form of (n × (n − 1) × (n + 1))

22. Example
 492 (The original number)
 4 + 9 + 2 = 15 (Add each individual digit together)
 15 is divisible by 3 at which point we can stop. Alternatively we can continue
using the same method if the number is still too large:
 1 + 5 = 6 (Add each individual digit together)
 6 ÷ 3 = 2 (Check to see if the number received is divisible by 3)
 492 ÷ 3 = 164 (If the number obtained by using the rule is divisible by 3, then
the whole number is divisible by 3)

23. Divisibility by 4
The basic rule for divisibility by 4 is that if the number formed by the last two
digits in a number is divisible by 4, the original number is divisible by 4; this is
because 100 is divisible by 4 and so adding hundreds, thousands, etc. is
simply adding another number that is divisible by 4. If any number ends in a
two digit number that you know is divisible by 4 (e.g. 24, 04, 08, etc.), then
the whole number will be divisible by 4 regardless of what is before the last
two digits.
Alternatively, one can simply divide the number by 2, and then check the
result to find if it is divisible by 2. If it is, the original number is divisible by 4.
In addition, the result of this test is the same as the original number divided
by 4

24. Example
 1720 (The original number)
 1720 ÷ 2 = 860 (Divide the original number by 2)
 860 ÷ 2 = 430 (Check to see if the result is divisible by 2)
 1720 ÷ 4 = 430 (If the result is divisible by 2, then the original number is
divisible by 4)

25. Divisibility by 5
 Divisibility by 5 is easily determined by checking the last digit in the number
(475), and seeing if it is either 0 or 5. If the last number is either 0 or 5, the
entire number is divisible by 5.

26. Conti….
 If the last digit in the number is 5, then the result will be the remaining digits
multiplied by two (2), plus one (1). For example, the number 125 ends in a 5,
so take the remaining digits (12), multiply them by two (12 × 2 = 24), then add
one (24 + 1 = 25). The result is the same as the result of 125 divided by 5
(125/5=25

27. Example
 If the last digit in the number is 5, then the result will be the remaining digits
multiplied by two (2), plus one (1). For example, the number 125 ends in a 5,
so take the remaining digits (12), multiply them by two (12 × 2 = 24), then
add one (24 + 1 = 25). The result is the same as the result of 125 divided by
5 (125/5=25).

28. If the last digit is 5
 85 (The original number)
 8 5 (Take the last digit of the number, and check if it is 0 or 5)
 8 5 (If it is 5, take the remaining digits, discarding the last)
 8 × 2 = 16 (Multiply the result by 2)
 16 + 1 = 17 (Add 1 to the result)
 85 ÷ 5 = 17 (The result is the same as the original number divided by 5)

29. Divisibility by 6
 Divisibility by 6 is determined by checking the original number to see if it is both
an even number (divisible by 2) and divisible by 3. This is the best test to use.
 If the number is divisible by six, take the original number (246) and divide it by
two (246 ÷ 2 = 123). Then, take that result and divide it by three (123 ÷ 3 = 41).
This result is the same as the original number divided by six (246 ÷ 6 = 41)

30. Example
 324 (The original number)
 324 ÷ 3 = 108 (Check to see if the original number is divisible by 3)
 324 ÷ 2 = 162 OR 108 ÷ 2 = 54 (Check to see if either the original number or the
result of the previous equation is divisible by 2)
 324 ÷ 6 = 54 (If either of the tests in the last step are true, then the original
number is divisible by 6. Also, the result of the second test returns the same
result as the original number divided by 6)

31. Divisibility by 7
 Divisibility by 7 can be tested by a recursive method. A number of the form 10x + y
is divisible by 7 if and only if x − 2y is divisible by 7. In other words, subtract twice
the last digit from the number formed by the remaining digits. Continue to do this
until a number known to be divisible by 7 is obtained. The original number is
divisible by 7 if and only if the number obtained using this procedure is divisible by
7. For example, the number 371: 37 − (2×1) = 37 − 2 = 35; 3 − (2 × 5) = 3 − 10
= −7; thus, since −7 is divisible by 7, 371 is divisible by 7

32. Divisibility by 13
 Remainder Test 13 (1, −3, −4, −1, 3, 4, cycle goes on.) If you are not comfortable
with negative numbers, then use this sequence. (1, 10, 9, 12, 3, 4)
 Multiply the right most digit of the number with the left most number in the
sequence shown above and the second right most digit to the second left most
digit of the number in the sequence. The cycle goes on

33. Example
 Example: What is the remainder when 321 is divided by 13?
Using the first sequence,
Ans: 1 × 1 + 2 × −3 + 3 × −4 = −17
Remainder = −17 mod 13 = 9
 312 is divisible by 13?
Using first sequence
2 x 1+ 1 x -3+ 3x-4= -13 => 0

34. Divisor Divisibility Condition Examples
21 Subtract twice the last digit from the rest 168: 16 − (8×2) = 0, 168 is divisible.
1050: 105 − (0×2) = 105, 10 − (5×2) = 0, 1050 is
divisible.
23 Add 7 times the last digit to the rest. 3128: 312 + (8×7) = 368, 368 ÷ 23 = 16.
25 The number formed by the last two digits is divisible
by 25
134,250: 50 is divisible by 25.
27 Sum the digits in blocks of three from right to left. If
the result is divisible by 27, then the number is
divisible by 27.
2,644,272: 2 + 644 + 272 = 918 = 27×34.

35. Divisor Divisibility condition Examples
29 Add three times the last digit to the rest. 261: 1×3 = 3; 3 + 26 = 29
31 Subtract three times the last digit from the rest 837: 83 − 3×7 = 62
32
The number formed by the last five digits is divisible
by 32.[1][2] 25,135,520: 35,520=1110×32
32
If the ten thousands digit is even, examine the
number formed by the last four digits.
41,312: 1312.
33
Add 10 times the last digit to the rest; it has to be
divisible by 3 and 11
627: 62 + 7 × 10 = 132,
13 + 2 × 10 = 33.

36. Divisor Divisibility Condition Examples
35 Number must be divisible by 7 ending in 0 or 5. 700 is divisible by 7 ending in a 0
37 Take the digits in blocks of three from right to left and
add each block, just as for 27.
2,651,272: 2 + 651 + 272 = 925. 925 = 37×25.
39 Add 4 times the last digit to the rest 351: 1 × 4 = 4; 4 + 35 = 39
41 Subtract 4 times the last digit from the rest 738: 73 − 8 × 4 = 41

37. Divisor Divisible Condition Example
43 Add 13 times the last digit to the rest. 36,249: 3624 + 9 × 13 = 3741,
374 + 1 × 13 = 387,
38 + 7 × 13 = 129,
12 + 9 × 13 = 129 = 43 × 3
45 The number must be divisible by 9 ending in 0 or 5 495: 4 + 9 + 5 = 18, 1 + 8 = 9;
(495 is divisible by both 5 and 9.)
47 Subtract 14 times the last digit from the rest. 1,642,979: 164297 − 9 × 14 = 164171,
16417 − 14 = 16403,
1640 − 3 × 14 = 1598,
159 − 8 × 14 = 47.
49 Add 5 times the last digit to the rest. 1,127: 112+(7×5)=147.
147: 14 + (7×5) = 49

38. Divisor Divisible Condition Example
50 The last two digits are 00 or 50 134,250: 50.
51 Subtract 5 times the last digit to the rest 204: 20-(4×5)=0
53 Add 16 times the last digit to the rest. 3657: 365+(7×16)=477 = 9 × 53
55 Number must be divisible by 11 ending in 0 or 5. 935: 93 − 5 = 88 or 9 + 35 = 44

39. Divisor Divisible Condition Example
71 Subtract 7 times the last digit from the rest. 852: 85-(2×7)=71
81 Subtract 8 times the last digit from the rest 162: 16-(2×8)=0
100 Ends with at least two zeros 900 ends with 2 zeros
1000 Ends with at least three zeros. 2000 ends with 3 zeros