7.4.2.2. Arithmetic Rules in Modular Arithmetic

So there are some interesting arithmetic rules which your modular arithmetic follows. So imagine that a modulo N is a’ and b modulo N is b’. Then a + b modulo N will be the same as a’ + b’ modulo N, a - b modulo N will be same as a’ - b’ modulo N.

In modular arithmetic, when you add or subtract two numbers (let's call them a and b), you can first reduce these numbers modulo N (i.e., find their remainders when divided by N) before performing the operation. For instance, if you find a mod N and b mod N (resulting in a' and b'), you can then add or subtract these results. The interesting property here is that the remainder of the sum (or difference) will be the same as the sum (or difference) of the remainders, which simplifies calculations significantly.

Imagine you have two bags of candies where each bag can hold a maximum of 10 candies (like our modulus N = 10). If you have 15 candies in one bag (a = 15) and 8 in another (b = 8), combining them gives you 15 + 8 = 23, which overflows a single bag. However, if you take the modulo 10 of 15 first, you find you only have 5 in the bag (15 mod 10 = 5). Adding the second bag then gives 5 + 8 = 13, which again overflows. When you take modulo 10 again, you'll see you effectively have 3 candies left after overflow.

So let us prove these rules, we will just prove it for the case of addition, for the other operations you can prove in the similar way. ... and you will get r.

To prove that (a + b) modulo N is the same as (a' + b') modulo N, first understand that if a modulo N is a', then 'a' can be expressed as a = q1 * N + a', where q1 is the quotient when 'a' is divided by N. Similarly, for b, we have b = q2 * N + b'. Thus, when we add a and b together, we get (q1 * N + a') + (q2 * N + b'). Factoring out the modulus, we find that the multiple of N will vanish when taking mod N, leaving us with the remainder as a' + b' mod N.

Think of adding two pieces of rope where one is 300 cm long and another is 140 cm long, but you can only fit pieces of 100 cm onto a shelf (as if the shelf is our modulus N). Instead of just adding the lengths (which gives 440 cm, overflowing the 100 cm limit repeatedly), you could find how much you can fit from each piece first: 0 from 300 cm (300 mod 100 = 0) and 40 from 140 cm (140 mod 100 = 40). When you stick them together, the mod calculation shows you'd fit 40 cm on the shelf.

So, what will be the interpretation of these arithmetic rules the interpretation here will be the following that you can first reduce the operands namely a and b modulo N and then you can perform the plus operation, subtraction operation, multiplication operation.

In practical applications of modular arithmetic, before performing operations such as addition or multiplication, you can simplify your calculations by reducing each operand modulo N. This means you can work with smaller numbers rather than larger, potentially cumbersome calculations. It streamlines operations and avoids overflow, similar to how basic math might keep things manageable by simplifying fractions.

Consider a cooking recipe needing 440 grams of flour, but your measuring cup holds only 100 grams. Instead of measuring out 440 grams and possibly making a mess, you could measure out 40 grams (440 mod 100 = 40) directly, making your task easier and cleaner! The same principle applies in modular arithmetic: working with reduced values makes calculations simpler and more efficient.

Now, we have seen the rules for addition, subtraction, multiplication, what about division? ... it does not necessarily mean that a and b are also congruent modulo N.

In modular arithmetic, division is more complicated than addition, subtraction, or multiplication. Unlike these operations, where congruences hold relatively straightforwardly, division by an integer b can lead to fractions, meaning we cannot always find an integer result. This means that a divided by b modulo N does not always equal (a' divided by b') modulo N, particularly if b is not a divisor of N. Hence, one must be careful when trying to 'cancel' terms in modulo operations.

If you have 12 cookies to share among 5 friends, each person gets 2 cookies with 2 left over. In regular division, it’s straightforward; however, in our setup of modulo, we might work with fractions of cookies which don't make sense in real life when you want whole cookies. This is similar to how division in modular arithmetic might yield non-integers and thus becomes complex.

So, now let us discuss some algorithms for modular arithmetic. ... all of them will need polynomial in n number of operations: $O(poly(n))$.

When performing modular arithmetic, especially in applications like cryptography, efficiency is key. Algorithms for addition, subtraction, multiplication, and even modular exponentiation must be assessed for their complexity based on the size of the numbers involved. Efficient algorithms will perform operations in polynomial time relative to the number of bits in the representations of those numbers. This ensures that computations remain manageable even for large inputs.

Think about finding the fastest route between cities on a map. Just as a driver would prefer using a route that minimizes travel time, what we seek in modular arithmetic is a way to perform calculations that minimizes the number of steps or operations required. Efficient algorithms lead to faster computations which is crucial when dealing with large datasets in real time.

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