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Way back in elementary school, we used remainders in division problems. For example, the equation $5\div2=2$ r $1$ means "$5$ divided by $2$ is $2$ with a remainder of $1$." It's commonly understood that, in this context, $2$ r $1$ is a pair of numbers, with $2$ being the quotient and $1$ being the remainder. However, the way the equation is written makes it look like $2$ r $1$ is one number. This annoyed me for the longest time, but now, I've been able to come up with a new number system that actually makes sense of quantities like $2$ r $1$ as numbers in their own right. I call them, "remainder numbers."

Remainder Numbers 1.0: A prototype

A remainder number is of the form $a_1$ r $a_2$, where $a_1$ and $a_2$ are both integers. Addition and subtraction of remainder numbers are done component-wise, and multiplication and division are defined as follows:

For any three integers $a$, $b_1$, and $b_2$: $a\times(b_1$ r $b_2)=ab_1+b_2$.

For any two integers $a$ and $b$: $a\div b=\lfloor\frac{a}{b}\rfloor$ r $(a$ mod $b)$.

(In this convention, fractions represent standard division, and $\div$ represents remainder number division. Also, $a$ mod $b$ is defined to be between $0$ and $b$, including $0$ but not $b$, allowing $a$ mod $b$ to be negative when $b$ is negative.)

Let's look at some examples of multiplication and division:

$5\div2=\lfloor\frac52\rfloor$ r $(5$ mod $2)=2$ r $1$

$2\times(2$ r $1)=2\cdot2+1=5$

$5\div-3=\lfloor-\frac53\rfloor$ r $(5$ mod $-3)=-2$ r $-1$

$-3\times(-2$ r $-1)=(-3\cdot-2)-1=5$

As you can see, multiplication "undoes" division, which is a very nice property to have. Unfortunately, this number system has a glaring problem: multiplication and division aren't closed over the remainder numbers. Let's fix that.

Remainder Numbers 2.0: Improving Multiplication and Division

Fixing multiplication is pretty simple:

$(a_1$ r $a_2)\times(b_1$ r $b_2)=(a_1$ r $a_2)\times b_1+b_2=a_1b_1+a_2+b_2$

This derivation isn't exactly rigorous, but it's perfectly consistent with what multiplication represents in the remainder numbers, and it's a pretty reasonable definition. To multiply two remainder numbers, simply multiply the integer parts and add the remainders.

Fixing division is more difficult, so I'll build up a comprehensive definition case by case.

The first case is $a\div(b_1$ r $b_2)$. Since I want multiplication to undo division, this is essentially asking, "What number, when multiplied by $b_1$ r $b_2$, results in $a$?" Because of the way multiplication is defined, this is also equivalent to asking, "What number, after multiplying by $b_1$ and adding $b_2$, results in $a$?" This effectively reduces the problem to $(a-b_2)\div b_1$, which is already defined.

Since the divisor's remainder can easily be removed, the only real case left to consider is when the dividend has a non-zero remainder: $(a_1$ r $a_2)\div b$. Unfortunately, this reveals another glaring problem, which is that multiplication only returns integers.

Remainder Numbers 3.0: Final Version

To make division closed, I'll have to drastically update the definition of remainder numbers. From now on, a remainder number can be defined recursively as $a_1$ r $a_2$, where $a_1$ is an integer and $a_2$ is a remainder number. Alternatively, remainder numbers can be embedded as sequences of integers. Basically, think of a remainder number as a decimal expansion, but without a specified number base. Addition and subtraction are still done component-wise, and the definitions of multiplication and division are extended as follows:

Recursive notation: $(a_1$ r $a_2$ r $a_3)\times(b_1$ r $b_2$ r $b_3)=(a_1b_1+a_2+b_2)$ r $(a_3+b_3)$, where $a_3$ and $b_3$ are remainder numbers.

Sequence notation: $(a_1,a_2,a_3,a_4,\ldots)\times(b_1,b_2,b_3,b_4,\ldots)=(a_1b_1+a_2+b_2,a_3+b_3,a_4+b_4,\ldots)$

I'll skip a few steps in deriving the division formula, but here it is:

Recursive notation: $(a_1$ r $a_2)\div(b_1$ r $b_2$ r $b_3)=\lfloor\frac{a_1-b_2}{b_1}\rfloor$ r $[(a_1-b_2)$ mod $b_1]$ r $(a_2-b_3)$

Sequence notation:

$(a_1,a_2,a_3,\ldots)\div(b_1,b_2,b_3,b_4,\ldots)=(\lfloor\frac{a_1-b_2}{b_1}\rfloor,(a_1-b_2)$ mod $b_1,a_2-b_3,a_3-b_4,\ldots)$

Analysis and Final Thoughts

Admittedly, this number system looks incredibly ugly. Multiplication isn't associative, it doesn't distribute over addition, it's not one-to-one, and it doesn't even have an identity element. Also, the number system has zero divisors. (But hey, at least multiplication undoes division and is commutative, so I guess it has that going for it.) So, yeah, it's pretty clear why we use rational numbers instead of remainder numbers. Having said that, though, do you see any interesting properties or potential applications of this number system, or is it just the useless mess that it appears to be? Also, if you can find a way to modify this number system to make it more mathematically pleasing, that would be nice.

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    $\begingroup$ Note representing an integer $n$ by $\{ n \bmod p^k,p \text{ prime }, k \ge 1\}$ is the idea behind the adele ring $\mathbb{A}_\mathbb{Q}$ which is useful to observe the unique factorization of integers and the local-global principle. $\endgroup$ – reuns Sep 4 '17 at 0:44
  • $\begingroup$ Multiplication undoes division but not conversely: $((1\text{ r }1)\times1)\div1=2\div1=2$. $\endgroup$ – stewbasic Sep 4 '17 at 1:06
  • $\begingroup$ This is going to take a lot of effort to rewrite as something understandable. I've read it several times and don't understand what you mean. Also there is some weird stuff like "[multiplication] isn't one-to-one". Can you just express your v3 in terms of how to multiply and divide two elements of some appropriate set? $\endgroup$ – rschwieb Sep 4 '17 at 2:28
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    $\begingroup$ It would also help if there was some motivating construction. Right now it's just a mess of stuff that looks possibly made up, it is hard to see how (or even if) it works as advertised. $\endgroup$ – rschwieb Sep 4 '17 at 2:32
  • $\begingroup$ When I said that multiplication isn't one-to-one, I meant that there are instances where $a\times b=a\times c$, but $b\neq c$. This is an instance of multiplication by $a$ not being one-to-one. For example, $1\times(1$ r $1)=1\times2=2$. It's an important thing to point out, as it implies that multiplication can't be undone by division, as stewbasic pointed out. $\endgroup$ – Tyler Borgard Sep 4 '17 at 2:45
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According to me we can't say this is a new number system. We are seeing $\mathbb Z$ just from a different point, you are representing the integers (not a new number) on basis of three parameters divisor, quotient and remainder and making the known operations from that point of view. Formally, we can say a number new if it's totally different from the previous number systems(if you consider boolean algebra from abstract point of view) or a number system that contains a operation which was not in previous one(subtraction can be done in $\mathbb Z$ but not in $\mathbb N$) and contains the previous number system as a isomorphic copy. For this reason we can say $\mathbb C$ is a new number system than $\mathbb R$, but if we consider the elements of $\mathbb R$ in decimal representation rather than normal format($1.414....$ in place of $\sqrt 2$) and perform operations, it's not a new system, but new point of view of reals.......However if some system can allow division by zero then that's will be new.

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  • $\begingroup$ I'm not sure you understand how remainder numbers work. Every sequence of integers represents a unique remainder number, whereas every digit in the decimal expansion of a real number is limited to one of finitely many values (depending on the base). For example, $(1,2,3,4,\ldots)$ is a remainder number, but it can't be the decimal expansion of any real number regardless of base since the integer sequence is unbounded. Also, $\mathbb{Z}$ is a proper subset of the remainder numbers, so it's not just "seeing $\mathbb{Z}$ from a different point." $\endgroup$ – Tyler Borgard Sep 5 '17 at 16:56

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