# Modular Arithmetic

An introduction to the fundamentals

Last updated

An introduction to the fundamentals

Last updated

The Basics

Modular Arithmetic is an incredibly important aspect of practically all asymmetric cryptography. A common way to refer to it is as **clock arithmetic** - imagine it's eleven o'clock right now. Three hours later it'll be two o'clock, right? Well we do the same thing in modular arithmetic:

$11 + 3 = 14 \equiv 2 \mod 12$

Essentially we take the sum of $11$ and $3$ and then divide by $12$ and keep the remainder. Here, the $12$ is called the **modulus**. The triple equals $\equiv$ denotes a *congruence*, and we say $14$ is congruent to $2$ $\mod 12$.

We can think of this another way and say that two numbers $a$ and $b$ are congruent modulo $m$ if their difference $a-b$ is divisible by $m$. For example, $26$ and $2$ are congruent modulo $12$ because $26-2=24$ which is divisible by $12$.

Modular (Multiplicative) Inverses

In normal maths, every number has an **inverse** - another number that, when multiplied with them, equals $1$. In maths terms, we would say that for every number $a$ there is another $b$ where $ab=1$. Generally, we can say that $b=\frac{1}{a}$ as that fits the equation - the inverse of $7$, for example, is $\frac{1}{7}$.

In modular arithmetic, there are no fractions. Instead of finding $b$ such that $ab=1$, we instead find $b$ such that $ab \equiv 1 \mod m$. This means that for different values of $m$ there are different inverses for $a$. For example, the inverse of $3 \mod 5$ is $2$, because $3 \cdot 2 = 6 \equiv 1 \mod 5$. This inverse is called the *modular multiplicative inverse*. We can therefore say that $3^{-1} \equiv 2 \mod 5$.

Yet not every number $a$ has an inverse modulo $m$ - for example, $2$ has no inverse modulo $6$. Why? This is because $gcd(a,m)=1$. Let's prove this in the general case and see why that is.

Proof

Assume there is $b$ such that $a \cdot b \equiv 1 \mod m$. This would mean that there also exists $c$ such that $ab = cm + 1$ and therefore $ab - cm = 1$. It follows therefore that both sides must be divisible by $gcd(a,m)$, meaning $gcd(a,m)=1$.

As a result, if there is an inverse of $a$ modulo $m$, such an inverse can only exist if $gcd(a,m)=1$ $\square$

An Intuitive Approach

I suggest you spend some time reflecting on this fact, as it is a very important result.

Uses

The uses of modular multiplicative inverses are similar to those of regular inverses - as division. Normally, if we divide two numbers, we can treat the number being divided by as *multiplying by its inverse.*

If you're struggling to visualise this proof, think back to the previous example of $2$ having no inverse modulo $6$. Why is this? Well, regardless of how much you multiply $2$ with, the sequence will go $0, 2, 4, 0, 2, 4, ...$ and loop around *because* it is a factor of 6.

Actually, the fact that a number it is not **coprime **with the modulus shows that it can never have an inverse. Again, why? Think about it this way: if they have a common factor, the remainder when one is divided by the other will also have that factor - and therefore not be $1$. If we choose $4$ rather than $2$, we can see this - if there is a common factor (as there is of $2$ in this case) there is no value $x$ for which $4x$ will return $1$ modulo $6$. Any number you can multiply $4$ by will always give a remainder divisible by $2$ when divided by 6.

$8 \div 2 = 8 \cdot 2^{-1} = 8 \cdot \frac{1}{2} = 4$

We can use the modular multiplicative inverse the same way - say we want to divide $3$ by $2$, modulo $7$:

$3 \div 2 \equiv 3 \cdot 2^{-1} \equiv 3 \cdot 4 \equiv 12 \equiv 5 \mod 7 \\$

Note that $4$ is the inverse of $2$ as $2 \cdot 4 \equiv 1 \mod 7$.

And just like that we realise that dividing $3$ by $2$ modulo $7$ gives us $5$.