Is there any way to define arithmetical multiplication as other thing than repeated addition? For example, how could you define $a\cdot b$ as other thing than $\underbrace{a+a+\cdots+a}_{b \text{-times}}$ or $\underbrace{b+b+\cdots+b}_{a \text{-times}}$?
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asked May 9, 2015 at 18:55
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4$\begingroup$ Would you allow real numbers and logarithms in a definition? $\endgroup$– GEdgarMay 9, 2015 at 18:59
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7$\begingroup$ If you can define $a^2$ then you can define $a\cdot b=[(a+b)^2-a^2-b^2]/2$, $\endgroup$– WillOMay 9, 2015 at 18:59
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1$\begingroup$ ....And you can in fact define $a^2$ as a sum of an initial segment of the odd numbers. $\endgroup$– WillOMay 9, 2015 at 19:00
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3$\begingroup$ You can write down the multiplication algorithm for binary numbers. Presumably computers do this, and not $a+a+\dots+a$ taken $b$ times or something. $\endgroup$– GEdgarMay 9, 2015 at 19:00
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$\begingroup$ Related but not identical question: math.stackexchange.com/questions/64488/… Some of the answers address tangentially this question. $\endgroup$– jgonMay 9, 2015 at 19:06
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11 Answers
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Given two sets $A$ and $B$ of cardinality $a$ and $b$, respectively, the cardinality of the cartesian product $A\times B$ is called the product of $a$ and $b$, and is denoted by $a\cdot b$.
Update
When I wrote this answer I didn't have infinite sets in mind. I just wanted to convey a mental picture of multiplication that does not involve repeated addition.
answered May 9, 2015 at 19:09
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4$\begingroup$ This potentially defines multiplication "by infinity" and also does not define multiplication of finite numbers that aren't non-negative integers. Neither of those observations is meant as a criticism of this definition, which I actually quite like a lot. $\endgroup$ May 10, 2015 at 1:17
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4$\begingroup$ @Todd If addition is defined by the successor function, then presumably it's already only dealing with non-negative integers. $\endgroup$– cpastMay 10, 2015 at 4:43
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1$\begingroup$ Suppose $A=\{1\}$ and $B=\mathbb{N}$. Then $a=1$ and $b=\aleph_0$ and the product of $a$ and $b$ is $\aleph_0$. It's like someone asking "What's one times infinity?' and answering "infinity" with some rigor behind it. $\endgroup$ May 10, 2015 at 7:00
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1$\begingroup$ @ToddWilcox, I'm not following. Recall that we're trying to define multiplication of natural numbers $n$ and $m$. Christian's approach is to take two sets $N$ and $M$ of cardinality $n$ and $m$ respectively, and to define that $n \cdot m = |N \times M|$. (And then we have to prove that this is well-defined.) Now since $n$ and $m$ are natural numbers, hence $N$ and $M$ will necessarily be finite. Ergo, we're just taking products of finite sets. There's no infinity lurking here, except perhaps for the observation that the category $\mathbf{FinSet}$ is infinite. $\endgroup$ May 10, 2015 at 11:21
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2$\begingroup$ Oh I missed the part of Christian's answer where it says $A$ and $B$ are finite sets. I figured since the Cartesian cross product is well defined regardless of the cardinality of the sets involved, then this product was also well defined for transfinite cardinal numbers. I'll have to make an appointment with my eye doctor because I still can't find the mention of natural numbers in either the question of this answer. $\endgroup$ May 10, 2015 at 14:43
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$a\cdot b$ is the the value of $f_a(b)$, where $f_a$ is the unique endomorphism of $\mathbb N$ (under addition) satisfying $f_a(1)=a$.
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$\begingroup$ Damn I was going to post something similar, but this is much better worded than anything I would have written. (+1) (It might be helpful to point out that the way this endomorphism is extended to the rest of $\Bbb{N}$ is of course repeated addition) $\endgroup$– jgonMay 9, 2015 at 19:09
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1$\begingroup$ @YoTengoUnLCD, because $(\mathbb{N},+,0)$ can be defined as the additively denoted commutative monoid freely generated by $\{1\}$. This just pushes the question back though; how do we know that [insert your favourite definition of $\mathbb{N}$ here] is the same as the commutative monoid freely generated by $\{1\}$? $\endgroup$ May 9, 2015 at 19:55
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$\begingroup$ @YoTengoUnLCD Because $f(n)=f(1+1+...+1)=f(1)+...+f(1)$ is completely determined by the value of $f(1)$. $\endgroup$– Jack MMay 9, 2015 at 20:02
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3$\begingroup$ @JackM If you say that $f(n)=f(1+1+\dots+1)$, isn't that the same thing as the normal definition for multiplication? $\endgroup$– KitegiMay 9, 2015 at 20:12
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Here is an answer that requires an $x$ and $y$ axis. Let us say that we would like to multiply $A$ and $B$. Then we locate the point, $(1,A)$ and make the line determined by $(0,0)$ and $(1,A)$. Then locate the point $(B,0)$, and draw the vertical line that goes through this point. Then find the intersection of the vertical line just formed and the line formed by connecting the origin and $(1,A)$. You get a point, $(B,C)$, and the point $C$ is equal to $AB$.
What I like about this definition is that it works with the real numbers, the fact that it is Euclidean in spirit, and that it makes clear that multiplication is (de)magnification.
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We can define $\mathbb{N}$ as the initial semiring. In this approach, not only do we not have to define multiplication as repeated addition, but, in fact, we do not have to define multiplication at all.
:)
answered May 9, 2015 at 19:51
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1$\begingroup$ But then you'll have to show that this thing exists. $\endgroup$ May 10, 2015 at 15:36
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$\begingroup$ @PaŭloEbermann, true. However in my opinion, a good foundation should have innate support for inductive types. Furthermore, saying that $\mathbb{N}$ is the initial semiring is the same as defining it as a particular inductive type. So, existence is immediate. $\endgroup$ May 10, 2015 at 16:10
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Combining my comments into an answer:
First define $a^2=\sum_{i=1}^a (2i-1)$.
Then define $a\cdot b={(a+b)^2-a^2-b^2\over 2}$.
(You could of course argue that this is equivalent to repeated addition, but the same would be true of any valid definition.)
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9$\begingroup$ I think the symbol $\sum$ means repeated addition, so this is not only equivalent to repeated addition, but also using repeated addition in a different way. $\endgroup$– JiKMay 9, 2015 at 23:13
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$\begingroup$ If you're going to use squaring to obtain a product, and both $a$ and $b$ are positive (or both negative), it's more efficient to use $a\cdot b = \frac{a^2+b^2-(a-b)^2}2$, as $a-b$ will be significantly smaller than $a+b$. (just thought I'd note that, not saying there's anything wrong with your answer) $\endgroup$– Glen OMay 11, 2015 at 14:07
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If you want a definition of the usual operation of multiplication that does not reduce it to repeated addition, you can always define it axiomatically the same way that we were taught it:
- If $a, b$ are between 0 and 9, $a\cdot b$ is given by the multiplication table.
- Otherwise, the operation is defined recursively by specifying the rules of digit-by-digit multiplication, carry, and positive/negative sign. [I don't see the point of actually formulating the definitions here; I trust you get the idea.]
This wouldn't be particularly insightful, since the rules appear completely arbitrary; but it is well-defined and constructive. It is, after all, the system we learned and internalized as children.
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$\begingroup$ There is still lots of addition going on. (I was considering writing something similar, though.) $\endgroup$ May 11, 2015 at 9:04
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$\begingroup$ Indeed: handling the carry requires addition. I'd have preferred to write "a definition that makes no reference to addition", but that's not the case. But the repeated-addition aspect is well and truly gone. $\endgroup$– alexisMay 11, 2015 at 10:38
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Define set $S_n$ of multiples of $n$ as intersection of all sets $S\subseteq\Bbb N$ such that $0\in S,\forall m\in S:m+n\in S$. Define least common multiple of $a,b$ as the least positive element of $S_n\cap S_m$. Define $n^2$ as a number $n$ less than lcm of $n,n+1$. Define $a\cdot b$ as half of the number $(a+b)^2-a^2-b^2$.
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The binary operation $\cdot$ is multiplicative if
- There exists an $x$ such that, for every $y$, $x\cdot y=y$. That is, there is a (left?) identity element.
- There exists an additive binary operation $+$ such that for every $x$, $y$, and $z$, $x\cdot(y+z)=x\cdot y+x\cdot z$. That is, $\cdot$ is (left?) distributive over $+$. This, of course, requires a definition of an additive operation.
This is related to the definition of a linear operator. This definition holds even for somewhat exotic definitions of multiplication, such as that for octonions (which is non-commutative and non-associative). It holds for ordinal numbers, cardinal numbers, vector spaces, projective spaces, the circle group, etc. These have interesting properties. Note that ordinal addition is itself non-commutative, for example, while the circle group possesses no zero element.
Come to think of it, where may one find examples of non-associative addition? Perhaps special relativity?
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1$\begingroup$ How about concatenation of curves: If $\beta, \gamma: [0,1]\to X$ and $\beta(1)=\gamma(0)$, then $$(\beta+\gamma)(t)=\begin{cases} \beta(2t) & t\le \frac12 \\ \gamma(2t-1) & t>\frac12\end{cases} $$ Of course this is normally only used where we quotient out reparameterization equivalence such that it does become associative, so it might not really count as an example. $\endgroup$ May 10, 2015 at 2:45
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I like the definition of $ab$ as the area of a rectangle with side lengths $a$ and $b$. Then it's clear that multiplication is commutative, for example.
Incidentally, this can sometimes be useful in communicating with non-mathematicians. For example, you can say that a pile of stones has a prime number of stones in it if there's no way to arrange them into a rectangle apart from making a long line of stones. And Goldbach's conjecture becomes: "You have a pile containing an even number of stones. Is it possible to split it into two smaller piles such that there's no way of making a rectangle out of either of the smaller piles?"
answered May 11, 2015 at 3:39
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$\begingroup$ I thought of this, but then realized that rigorously defining area is decidedly nontrivial. I upvoted you anyway. $\endgroup$ May 11, 2015 at 3:52
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$\begingroup$ Yes... I see what you mean. Maybe it's not so bad in the discrete case, but then this is the same as the top answer. $\endgroup$ May 11, 2015 at 4:00
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$\begingroup$ I recall reading about this many years ago. iirc, you start with a square of area one and build from this. $\endgroup$ May 11, 2015 at 4:19
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$$a\times b = \begin{cases} \frac{a}{2} \times \left(2\times b\right) &\text{ if } a \text{ is even}\\ b +\frac{a-1}{2}\times\left(2\times b\right)& \text{ if } a \text{ is odd} \end{cases}$$
answered May 10, 2015 at 2:52
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$\begingroup$ This doesn't work. You don't define division by 2 (which is ordinarily defined with multiplication), whenever we follow the odd branch, we're doing repeated addition, and the recursion never terminates. For example, say we try 2x2. 2x2 = (2/2)x(2x2) = (2/2)x((2/2)x(2x2)) = (2/2)x((2/2)x((2/2)x(2x2))) = ..... $\endgroup$ May 10, 2015 at 10:01
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$\begingroup$ @MichaelShaw $\frac a2$ is the unique $x\in\mathbb N_0$ with $x+x=a$ (in fact the existence of such $x$ is the definiotn of "$a$ is even"). And $2\times b$ should be replaced with $b+b$. $\endgroup$ May 10, 2015 at 20:06
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$\begingroup$ @HagenvonEitzen: It still doesn't quite work: 2x2 = 1x4 = 4 + 0x8 = 4 + 0x16 .... We can define multiplication by zero, but then we're left with what is a fancy way of saying "add b to itself a times", which is exactly what the OP is not looking for. $\endgroup$ May 10, 2015 at 20:37
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Here's another way to define multiplication that I found quite interesting, from one of Norman Wildberger's videos:
Suppose we have four points $\{0,1,a,b\}$ on a line $L$, where $0$ is distinct from $1$, and wish to construct a point $ab$ on $L$ such that
$$1:a=b:ab$$
That is, the ratio of the length of the line segment $\overline{(0,1)}$ to the length of $\overline{(0,a)}$ is equal to the ratio of the length of $\overline{(0,b)}$ to the length of $\overline{(0,ab)}$.
First, draw an auxiliary line $L'$ through $0$ distinct from $L$. Draw a line from $1$ to some point $1'$ on $L'$ distinct from $0$. Draw a parallel line between $a$ and its corresponding point $a'$ on $L'$. At this point, note that
$$1 : a = 1' : a'$$
Draw a line from $1'$ to $b$. Draw a parallel line from $a'$ to the corresponding point $ab$ on $L$. Then
$$1' : a' = b : ab$$
By transitivity,
$$1 : a = b : ab$$
GeoGebra demo: https://ggbm.at/cdgzzkvz