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3.2: Modulo Arithmetic

( \newcommand{\kernel}{\mathrm{null}\,}\)

In this section, we will explore arithmetic operations in a modulo world.

Let n \in \mathbb{Z_+}.

Theorem 5 :

Let a, b, c,d, \in \mathbb{Z} such that a \equiv b (mod\,n) and c \equiv d (mod\, n). Then (a+c) \equiv (b+d)(mod\, n).

Proof:

Let a, b, c, d \in\mathbb{Z}, such that a \equiv b (mod\, n) and c \equiv d (mod \,n).

We shall show that (a+c) \equiv (b+d) (mod\, n).

Since a \equiv b (mod\, n) and c \equiv d (mod\, n), n \mid (a-b) and n \mid (c-d)

Thus a= b+nk, and c= d+nl, for k and l \in \mathbb{Z}.

Consider (a+c) -( b+d)= a-b+c-d=n(k+l), k+l \in \mathbb{Z}.

Hence (a+c)\equiv (b+d) (mod \,n).\Box

Theorem 6:

Edit section

Let a, b, c,d, \in \mathbb{Z} such that a \equiv b (mod \, n) and c \equiv d (mod \,n). Then (ac) \equiv (bd) (mod \,n).

Proof:

Let a, b, c, d \in \mathbb{Z}, such that a \equiv b (mod\, n) and c \equiv d (mod \, n).

We shall show that (ac) \equiv (bd) (mod\, n).

Since a \equiv b (mod \,n) and c \equiv d (mod\, n), n \mid (a-b) and n \mid (c-d)

Thus a= b+nk, and c= d+nl, for k and l \in \mathbb{Z}.

Consider (ac) -( bd)= ( b+nk) ( d+nl)-bd= bnl+dnk+n^2lk=n (bl+dk+nlk), where (bl+dk+nlk) \in \mathbb{Z}.

Hence (ac) \equiv (bd) (mod \,n).\Box

Theorem 7:

Let a, b \in \mathbb{Z} such that a \equiv b (mod \, n) . Then a^2 \equiv b^2 (mod \, n).

Proof:

Let a, b \in \mathbb{Z}, and n \in \mathbb{Z_+}, such that a \equiv b (mod \, n).

We shall show that a^2 \equiv b^2 (mod \, n).

Since a \equiv b (mod \, n), n\mid (a-b).

Thus (a-b)= nx, where x \in \mathbb{Z}.

Consider (a^2 - b^2) = (a+b)(a-b)=(a+b)(nx), = n(ax+bx), ax+bx \in \mathbb{Z}.

Hence n \mid a^2 - b^2, therefore a^2 \equiv b^2 (mod \, n). \Box

Theorem 8:

Let a, b \in \mathbb{Z} such that a \equiv b (mod \, n). Then a^m \equiv b^m (mod \, n), \forall \in \mathbb{Z}.

Proof:

Exercise.

By using the above result we can solve many problems. Some of them are discussed below: 

Example \PageIndex{1}: mod\, 3 Arithmetic

Let n = 3.

Addition

+ [0] [1] [2]
[0] [0] [1] [2]
[1] [1] [2] [0]
[2] [2] [0] [1]

Multiplication

x [0] [1] [2]
[0] [0] [0] [0]
[1] [0] [1] [2]
[2] [0] [2] [1]

Example \PageIndex{2}: mod\, 4 Arithmetic

Let n=4.

x [0] [1] [2] [3]
[0] [0] [0] [0] [0]
[1] [0] [1] [2] [3]
[2] [0] [2] [0] [2]
[3] [0] [3] [2] [1]

Example \PageIndex{3}:

Find the remainder when (101)(103)(107)(109) is divided by 11.

Answer

101 \equiv 2 (mod \, 11)

103 \equiv 4 (mod \, 11)

107 \equiv 8 (mod \, 11)

109 \equiv 10 (mod \,11).

Therefore,

(101)(103)(107)(109) \equiv (2)(4)(8)(10) (mod \,11) \equiv 2 (mod \,11).

Example \PageIndex{4}:

Find the remainder when 7^{1453} is divided by 8.

Answer

7^0 \equiv 1 (mod \, 8)

7^1 \equiv 7 (mod \, 8)

7^2 \equiv 1 (mod \, 8)

7^3 \equiv 7 (mod \, 8),

As there is a consistent pattern emerging and we know that 1453 is odd, then 7^{1453} \equiv 7 (mod \, 8). Thus the remainder is 7.

Example \PageIndex{5}:

Find the remainder when 7^{2020} is divided by 18.

Answer

7^0 \equiv 1 (mod \, 18)

7^1 \equiv 7 (mod \, 18)

7^2 \equiv 13 (mod \, 18)

7^3 \equiv 1 (mod \, 18),

As there is a consistent pattern emerging and we know that 2020=(673)3+17^{2020}= 7^{(673)3+1}=\left( 7^3\right)^{673}7^1 \equiv 7 (mod \, 18). Thus the remainder is 7.

Example \PageIndex{6}:

Find the remainder when 26^{1453} is divided by 3.

Odd and Even integers:

An integer n is even iff n\equiv 0(mod \, 2).

An integer n is odd iff n\equiv 1(mod \, 2).

Two integers a and b are said to have some parity if they are both even or both odd otherwise a and b are said to have different parity.

Example \PageIndex{6}:

Show that the sum of an odd integer and an even integer is odd.

Answer

Proof: Let a be an odd integer and let b be an even integer. We shall show that a+b is odd by using the language of congruency.

Since a is odd, a \equiv 1 (mod \, 2).

Since b is even, b \equiv 0 (mod \, 2).

Then (a+b) \equiv (1+0)(mod \, 2),

(a+b) \equiv 1 (mod \, 2).

Hence a+b is odd.\Box

Example \PageIndex{7}:

Show that the product of an odd integer and an even integer is even.

Answer

Proof: Let a be an odd integer and let b be an even integer. We shall show that ab is even by using the language of congruency.

Since a is odd, a \equiv 1 (mod \, 2).

Since b is even, b \equiv 0 (mod \, 2).

Then (ab) \equiv (1)(0)(mod \, 2),

(ab) \equiv 0 (mod \, 2).

Hence ab is even.\Box

Example \PageIndex{8}:

Show that n^2+1 is not divisible by 3 for any integer n.

Answer

Proof: Let n \in \mathbb{Z} . We shall show that (n^2+1) is not divisible by 3 using the language of congruency.

We shall show that (n^2+1)(mod \,3) \not \equiv 0 by examining the possible cases.

Case 1: n \equiv 0 (mod \, 3).

n^2 \equiv 0^2 (mod \,3).

(n^2+1) \equiv 1 (mod \,3).

Hence n^2+1 is not divisible by 3.

Case 2: n \equiv 1 (mod \, 3).

n^2 \equiv 1^2 (mod \,3).

(n^2+1) \equiv 1 (mod \,3).

Hence n^2+1 is not divisible by 3.

Case 3: n \equiv 2 (mod \, 3).

n^2 \equiv 2^2 (mod \,3) \equiv 1 (mod \,3).

(n^2+1) \equiv 2 (mod \,3).

Hence n^2+1 is not divisible by 3.

Since none of the possible cases is congruent to 0 (mod \, 3), n^2+1 is not divisible by 3. \Box

Example \PageIndex{9}:

Show that 5 \mid a^5+4a for any integer a.

Answer

Notice that 4 \equiv -1 (mod \, 5). Therefore, 5 \mid a^5+4a iff 5 \mid a^5-a for all integer a.

Let .

We shall show that .

Then and .

We will proceed by examining the 5 possible cases of \( . \) Specifically, .

Case : If then

.

Thus .

Case : If then

.

Thus .

Case : If then

.

Thus .

Case : If then

.

Thus .

Case : If then

.

Thus .

Having examined all possible cases, .◻

 

Answer

Let where and .

Then .

Thus . Rearranging, we obtain .

Consider

.

Let .

Since , then .

Thus .

Clearly, . We shall examine the possible cases of .

Case b=0: If then . Thus .

Case b=1: If then and where q .

Thus .

Case b=2: If then and .

Thus .

Case b=3: If then and .

Thus .

Case b=4: If then and .

Thus .

Having examined all possible cases, .◻

Example \PageIndex{10}

Answer

Proof:

Let be the statement .

says that .

says that . Since and , is true.

Assume is true for some .

We will show that is true.

Specifically, we will show that .

Consider that

.

Since , .

Let .

Thus .

Having shown the inductive step, for every positive integer , is divisible by .

 

ISBN Check Digit

ISBN is the abbreviation for the International Standard Book Number. ISBN numbers are 10 digits in length. In an ISBN of the form X-XX-XXXXXX-X:

  • The first block of digits on the left represents the language of the book (0 is used to represent English).
  • The second block of digits represents the publisher.
  • The third block of digits represents is the number assigned to the book by the publishing company.
  • The fourth block consists of the check digit.

Consider the ISBN 0-13-190190-?

To determine the check digit for this Book Number, follow the below steps.

Step 1: Multiply all nine assigned digits by weighted values. The weighted values are 1 for the first digit from the left, 2 for the second digit, three for the third digit, etc.

1*0 + 2*1 + 3*3 + 4*1 + 5*9 + 6*0 + 7*1 + 8*9 + 9*0 = 139.

Step 2: ISBN uses (mod \, 11) to determine the check digit. So 139 = 11(11) + 7 so 139 \equiv 7 (mod \,11). Hence the check digit is 7. Note that X is used for 10.

If someone orders a book by the ISBN number then the check digit is a way of ensuring that the number was reported and recorded correctly so that the correct book is obtained.


This page titled 3.2: Modulo Arithmetic is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pamini Thangarajah.

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