# Josh Kayani

Hey, I'm Josh Kayani, a student and fledgling software developer. Here, I rant and rave about tech and other topics.

# Explaining modulus to a friend

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Last semester, a friend and I were taking Discrete Math (counting, sets, logic, etc.), and he asked me a question about how modulus worked with negative numbers. It took me a bit to come up with a good answer, and I think it provides a nice framework for understanding how that operation works as a whole, so here goes.

When someone’s asked to calculate $a \bmod b$ , they’re being asked what the remainder would be when dividing $a$ by $b$. For example, calculating $7 \bmod 3$ would be $1$, since $3$ can only “go into” $7$ twice, and there will be $1$ left over $(7 - (3 * 2) = 1$.

We can make this a little more mathematical (it comes in handy when extending this to negative numbers): when calculating $a \bmod b$, you’re trying to find the value $c$ such that $c = a - (b * k)$, where $k$ is the largest possible integer such that $a$ is greater than $(b * k)$.  Again, consider calculating $7 \bmod 3$: here, we’re trying to find the largest $k$ that when multiplied by $3$, is still less than $7$. If $k = 3$, then $(b * k)$ equals $9$, and $7$ is not greater than $9$. However, if $k = 2$, then $(b * k)$ equals $6$, and $7$ is greater than $6$.  Plugging this into our equation, we find that $c = 7 - (3 * 2)$, which means $c$ equals $1$.

To understand where that model came from, think about what divisibility means. If 2 numbers, $a$ and $b$ are divisible, it means that $a / b$ yields no remainder; in other words, there is some integer $k$ such that $a = (b * k)$. Using that same model, we can calculate the remainder of any division by taking $a - (b * k)$, since this tells us how far $a$ was from $b * k$.

Now to answer the question: how do we calculate $-a \bmod b$? Our model remains the same; we’re still trying to find $c$ using the equation above, and $k$ should still be the largest possible integer so that $-a > (b * k)$.  The key is in this last expression; for a negative integer $x$ to be greater than another integer $y$, $y$ should be negative but have a larger absolute value. For example, $-1$ is larger than all the integers in the range $-2$ to $-\infty$, but all those numbers have a larger absolute value than $1$ ($2 > 1, 3 > 1, … \infty > 1$). With negative numbers, larger absolute values mean the number gets smaller – exactly the opposite of what happens with positive numbers.

Taking this into account, consider $7 \bmod 3$ versus $-7 \bmod 3$. The largest $k$ that can be multiplied by $3$ to be less than 7 is $k = 2$, giving us $7 - (3 * 2) = 1$. But the largest $k$ that can be multiplied by $2$ and be less than $-7$ is $k = -3$, since $-7$ is greater than $-9$ ($-3 * 3$), so our answer becomes $-7 - (-3 * 3) = -7 - (-9) = -7 + 9 = 2$.  From this, we can see that there is no shortcut like: calculate $a \bmod b$ and then multiply it by $-1$.

This is a lot of math to do a basic operation, so here’s a shortcut. When calculating $a \bmod b$, and $a$ is positive, you can simply find the closest multiple of $b$ that is still less than $\lvert a \rvert$(absolute value of $a$), call it $d$, and do $a - d$ to get your answer. When $a$ is negative however, you should still find the closest multiple of $b$ that is less than $\lvert a \rvert$, call it $d$, but then do $a + (d + b)$ to get your answer. This just comes from applying the above model.

You can also think of this in terms of going “under” and “over”; with $a \bmod b$, you’re going “under” $\lvert a \rvert$ to find $k$, and taking the difference $a - (b * k)$. With $-a \bmod b$, you’re going “over” $\lvert a \rvert$ to find $k$, and again taking the difference $a - (b * k)$.

Hopefully I managed to articulate all of that well!