Modular arithmetic

1. An introduction

When we divide two integers we will have an equation that looks like the

following:

A/B = Q remainder R

$A$  is the dividend

$B$ is the divisor

$Q$ is the quotient

$R$ is the remainder

Sometimes, we are only interested in what the remainder is when we divide $A$ by $B$.
For these cases there is an operator called the modulo operator (abbreviated as mod).

Using the same $A$, $B$, $Q$, and $R$ as above, we would have: $A\text{ mod }B=R$.We would say this as $A$ modulo $B$ is congruent to $R$. Where $B$ is referred to as the modulus.

For example:

13/5 = 2 remainder 3

13 mod 5 = 3

If we have  $A\text{ mod }B$  and we increase  $A$  by a  multiple of  B , we will end up in
the same result:

$A\text{ mod }B=(A+K\cdot B)\text{ mod }B$ for any integer K.

2. Congruence Modulo

You may see an expression like:

A≡B(mod C)equivalent

This says that $A$ is congruent to $B$ modulo $C$.


Let's imagine we were calculating mod 5 for all of the integers:

Suppose we labelled 5  slices  0, 1, 2, 3, 4. Then, for each of the integers, we

put it into a slice that matched the value of the integer mod 5.Think of these slices as buckets, which hold a set of numbers. For example, 26would go in the slicelabeled 1,  because 26 mod 5=1.It would be useful to have a way of expressing that numbers belonged in the same slice. (Notice 26 is in the same slice as 1, 6, 11, 16, 21 in above example).

A common way of expressing that two values are in the same slice, is to say they are in the same equivalence class.The way we express this mathematically for mod C is: $A\equiv B(\text{mod }C)$.

Examining the expression closer:

1.$\equiv$ is the symbol for congruence, which means the values $A$ and $B$ are in
the same equivalence class.

2.$(\text{mod }C)$ tells us what operation we applied to $A$ and $B$.

3.when we have both of these, we call “$\equiv$” congruence modulo $C$.

e.g. $26\equiv 11(\text{mod }5)$

$we\; can\; make\; a\; key\; observation:$

The values in each of the slices are equal to the label on the slice plus or minus some multiple of  C .

This means the difference between any two values in a slice is some multipleof  C .

Before proceeding it’s important to remember the following statements are equivalent:

$A\equiv B(\text{mod }C)$

$A\text{ mod }C\equiv B\text{ mod }C$

C ∣ (A−B) The | symbol means divides, or is a factor of)

A=B+K⋅C (where KK is some integer)

For example the following are equivalent:

$13\equiv 23(\text{mod }5)$

$13\text{ mod }5\equiv 23\text{ mod }5$

5 ∣ (13−23), (5 \ |\ -10(5 ∣ −10, which is true since  5×−2=−10)
13=23+K⋅5. We can satisfy this with  K=−2 :
13=23+(−2)×5


 Congruence Modulo is an Equivalence Relation

The Quotient Remainder Theorem
When we want to prove some properties about modular arithmetic we oftenmake use of the Quotient Remainder Theorem.It is a simple idea that comes directly from long division.

The Quotient Remainder theorem says:

Given any integer A, and a positive integer B, there exist unique integers Q
and R such that

A= B * Q + R where 0 ≤ R < B

We can see that this comes directly from long division. When we  divide A by B in long division, Q is the quotient and  R is the remainder . If we can write a number in this form then  A mod B = R


A = 7, B = 2 

7 = 2 * 3 + 1 
7 mod 2 = 1 

A = 8, B = 4 

8 = 4 * 2 + 0 
8 mod 4 = 0 

A = 13, B = 5 

13 = 5 * 2 + 3 
13 mod 5 = 3 

A = -16, B = 26

-16 = 26 * -1 + 10
-16 mod 26 = 10

Modular addition andsubtraction

(A + B) mod C = (A mod C + B mod C) mod C

Proof:

From the quotient remainder theorem we can write A and B as:

A = C * Q1 + R1 where 0 ≤ R1 < C and Q1 is some integer. A mod C = R1

B = C * Q2 + R2 where 0 ≤ R2 < C and Q2 is some integer. B mod C = R2

(A + B) = C * (Q1 + Q2) + R1+R2

(A + B) mod C = ( C *（Q1 + Q2) + R1 + R2) mod C = (R1 + R2) mod C

(A mod C + B mod C) mod C = (R1 + R2) mod c

A very similar proof holds for modular subtraction

(A - B) mod C = (A mod C - B mod C) mod C

Let's explore the multiplication property of modular arithmetic:


(A * B) mod C = (A mod C * B mod C) mod C




Finally, let's explore the exponentiation property:

A^B mod C = ( (A mod C)^B ) mod C

Often we want to calculate  A^B mod C  for  large values of B .

Unfortunately, A^B becomes very large for even modest sized values for B.

These huge values cause our calculators and computers to


return overflowerrors.Even if they didn't, it would take a long time 


to find the mod of these hugenumbers directly.

Suppose we want to calculate 2^90 mod 13, but we have a calculator that
can't hold any numbers larger than 2^50.

Here is a simple divide and conquer strategy:

Step 1: Split A^B into smaller parts using exponent rules

2^90 = 2^50 * 2^40

Step 2: Calculate the mod C for each part

2^50 mod 13 = 1125899906842624 mod 13 = 4

2^40 mod 13 = 1099511627776 mod 13 = 3

Step 3: Use modular multiplication properties to combine the parts

2^90 mod 13 = (2^50 * 2^40) mod 13

2^90 mod 13 = (2^50 mod 13 * 2^40 mod 13) mod 13

2^90 mod 13 = ( 4 * 3 ) mod 13

2^90 mod 13 = 12 mod 13 = 12




https://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/what-is-modular-arithmetic