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What is the algebraic difference between arithmetic operations, that prevents entities with different units from being summed or subtracted, but allows them to be multiplied or divided?

This looks more like a question for Physics, but lengths and areas, for example, are in the domain of pure mathematic.

Now, I cannot sum or subtract an area and a length, but I can multiply and divide an area with a length!

Reading Wikipedia, it looks like this is a property of the dimensions set. Does it just depend on the definition of the dimensions, or is it something intrinsic in the operations of add, subtract, multiply and divide?

Please explain with simple words, if possible.

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This was the basis for a very successful question on physics.SE. The accepted answer there is the same as @fgp's. –  Jack M May 12 at 22:34
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We can multiply and divide different quantities because we agreed on what would that mean (scale invariance had a large part in it, but intuitions from area, volume and other things had too). However, there is no such consensus for addition in general. Still, if we were to agree on something, there would be no problems with such expressions, for example, "fruits" is commonly understood as the lowest common ancestor of "apples" and "oranges", so quite a few may regard $$ 2\text{ apples} + 3\text{ oranges} = 5\text{ fruits}$$as true. –  dtldarek May 12 at 23:28
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See esp. KCd's nice answer, which surely deserves many more votes. –  Bill Dubuque May 12 at 23:53
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You should see if the AIR report on the comparability of apples and oranges is available anywhere online (May 1995). I still carry the chart from that article with me. (the author, Scott Sanford, then of NASA-Ames, prepared samples of both and generated spectral reflectance scans; they are quite similar) –  Carl Witthoft May 13 at 13:54
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Why is it that so many people answering this question seem to be so preoccupied by making sense out of adding apples and oranges? The core of this question is rather about the foundations and general nature of arithmetic of mathematical objects representing the physical world. Explaining the particular example about adding apples and oranges seem such a small question compared to the general aspects of this! I hope others will count the number of apples/oranges-comments before adding another one! –  String May 14 at 9:02

10 Answers 10

Apples and oranges are actually a rather bad example. The reason why it doesn't make sense to add quantities with different dimensions, but it does make sense to multiply (or divide) them is scale invariance.

Let U be the unit of some quantity $u$, and $V$ be the unit of another quantity $v$. Now say we change the scale of U, i.e. we instead use a different unit U' such that $1U = 10U'$. For $V$ we do the same, only that there we choose $V'$ such that $1V = 5V'$. If we compute the sum $s$ of $u$ and $v$ in units U,V we get $$ s = u + v $$ If, instead, we compute the sum in units $U'$ and $V'$, however, we get $$ s' = 10\cdot u + 5\cdot v $$ Note that $s$ and $s'$ don't just differ by a factor, i.e. we can't convert $s$ from unit $U+V$ to $s'$ in unit $U'+V'$ without knowing the original values of $u$ and $v$.

Compare this to the situation of a product. If we compute the product $p$ of $u$ and $v$ in units $U$ and $V$, we get $$ p = u\cdot v $$ If, instead, we compute it in units $U'$ and $V'$, we get $$ p' = (10\cdot u) \cdot (5\cdot v ) = 50\cdot p \text{.} $$ So $p'$ is simply $p$, expressed in a different unit P', with $1P = 1UV = 50P' = 1U'V'$.


So why do you want scale invariance? We want that, because the scale of physical units is usually completely arbitrary. There's nothing fundamental about 1 meter, or 1 inch, or 1 Volt - we just picked some reference value. But since the reference value is arbitrary, the actual physics must not change if we replace it by a different one. Which it doesn't, so long as we only multiply and divide, but not add or subtract values with different units, as the example above shows.

And this is also why apples and oranges are a bad example. We don't expect scale invariance for these, because apples and oranges are discrete objects, so there's a canonical definition of what "1 apple" means. So adding apples and oranges makes perfect sense, and we may e.g. assign the result the unit fruits.

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Addition is "putting one after another". You can put one meter stick after a foot stick and get a stick of length $1.3048$ meter. Putting a meter stick after a second doesn't make sense. Multiplying is like orthogonal coordinates. I can multiply two lengths to get an area. I can multiply a mass and an acceleration to get a force. There is no requirement that each axis is measured in the same units. –  Ross Millikan May 13 at 4:03
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I like to think of units as unknown quantities, like x and y in $12x+2y$. Usually, units stay mutually distinct, so even if we add them, the terms can't be joined. If we get an additional equation to help us relate x to y, then we have a method to combine them. This can happen with meters and seconds, for instance, if you feel like making your velocities dimensionless values between 0 and 1. And as a further consequence of this meters-to-seconds relation, you wind up measuring energy and momentum in grams. Fun stuff. –  Cory May 13 at 14:27
    
Does this have connections with Linear Algebra and Vector Spaces? –  Johann Franklin May 13 at 22:24
    
This question got a lot of up votes, but i see multiplicative invariance as a consequence, more than as a reason, and i will cite a part of your answer to sustain my idea: "the actual physics must not change if we replace it by a different one. Which it doesn't, so long as we only multiply and divide, but not add or subtract values with different units" using this as an argument constitutes a tautology in my opinion. invariance is not absolute, but relative to some operations. why we would not want invariance relatively to sum? the position of zero is arbitrary as well –  danza May 14 at 19:50
    
@danza It's not a tautology - it tells you that you may add quantities with the same unit, and multiply arbitrary quantities without violating scale invariance. Which answer the question of why we put that restriction on physical theories. You're right that it assumes that we don't expect translation invariance, i.e. the quantity zero is unambiguous defined independent of the unit. That's true for very many physical quanties, but you're right that it's no true for all of them. I left that out because I wanted to keep it simple... –  fgp May 14 at 22:54

I suppose you can add apples and oranges. Just take the external direct sum of the fruit spaces. Moreover, you may recall Feynman's theory of everything: take the equations of physics and write them homogeneously $E_1=0, E_2=0, \dots$ then the theory of everything is simply: $$ 0=E_1=E_2= \cdots $$ you might note this is dimensionally inconsistent.

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Fruit spaces?! +1. –  Aidan F. Pierce May 12 at 21:46
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Do you regard apples and oranges as orthogonal fruit spaces? Isn't the OP rather asking for fruit rings? –  String May 12 at 22:11
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@String well, I suppose it depends on the geometry. I mean, are these abstract fruit spaces, or do they inherit geometry from the induced produce metric? –  James S. Cook May 12 at 22:49
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@String: Mmm, fruit rings... –  Rahul May 12 at 22:49
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@JamesS.Cook The reason Feyman's TOE get away with being dimensionally inconsistent is exactly because it equates every to zero. That doesn't break scale invariance because zero times anything is zero again... –  fgp May 13 at 11:22

The addition question is related to the distributive property of numbers: $ax+bx=(a+b)x$

With: $$\begin{align}3\,\mathrm{oranges}+5\,\mathrm{oranges}&=(3+5)\,\mathrm{oranges}\\&=8\,\mathrm{oranges}\end{align}$$

And the lack of any such property for an expression like: $$\begin{align}3\,\mathrm{oranges}+5\,\mathrm{apples}\end{align}$$

Meanwhile with multiplication, the associative and commutative properties permit this: $$\begin{align}(3\,\mathrm{oranges})\cdot(5\,\mathrm{apples}) &=(\mathrm{oranges}\,3)\cdot(5\,\mathrm{apples})\\ &=\mathrm{oranges}\,(3\cdot(5\,\mathrm{apples}))\\ &=\mathrm{oranges}\,((3\cdot5)\,\mathrm{apples})\\ &=\mathrm{oranges}\,(15\,\mathrm{apples})\\ &=(\mathrm{oranges}\,15)\,\mathrm{apples}\\ &=(15\,\mathrm{oranges})\,\mathrm{apples}\\ &=15\,(\mathrm{oranges}\,\mathrm{apples})\\ &=15\,\mbox{orange-apples}\\ \end{align}$$

So from one perspective, the question is about which properties of arithmetic we use, and why isn't there a property like $ax+by=(a+b)(x+y)$. And there is no such property because it doesn't even hold with integers, with integer multiplication defined as repeated addition.

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It may be worth noting that it is possible to add units, but the results aren't terribly useful. If one feeds (1 horsepower + 10 watts) worth of power into a system for one hour and 15 minutes, the total energy will be one horsepower hour plus 10 watt hours plus 15 horsepower minuts plus 150 watt minutes. Not the most convenient bunch of figures to work with. –  supercat May 14 at 4:17
    
@supercat But horsepower and watts are the same unit (except for a scale factor). Your example is as reasonable as adding 1m to 1km and multiplying with 1hour + 1sec. There's a perfectly determined result for both. –  Jens May 15 at 13:04
    
@Jens: My point was that one could do math even without being able to reduce things, but multiplication by "compond" factors would cause expressions to quickly grow out of hand if one couldn't factor them. Perhaps a better example might have been an inductor in series with a capacitor. At any given frequency, conversions will exist from both henry and farad to reactive ohms, but without a specified frequency the addition must be processed using separate units. –  supercat May 15 at 17:51

I think you should look at this symbolically. The units are just unevaluated "dummy" variables that stay symbolic (because they represent a physical quantity they can't be given a numerical value). This is nothing special, we also like to keep $i$ and $\pi$ unevaluated even when they have a bit more mathematical properties defined than our units do.

When you are computing with units, you are leaving them unevaluated - both in summation and multiplication! Think about it:

$$3\text{orange}\times 4\text{apple}=12(\text{orange}\times \text{apple})$$

The multiplicative group that units belong to may define an alias for this particular product, but that's just a substitution rule (like joule=newton*meter). So, in essence, you are not multiplying oranges and apples, you are leaving the product unevaluated. The same goes for summation: $$3\text{orange}+4\text{apple}=3\text{orange}+4\text{apple}$$ It's just that you can't simplify into a product of value*unit. Because we expect final expressions to be in the form value*unit, we say we can't do that, but above expression by itself is mathematically valid (although physically nonsensical, because there is no physical quantity with units 3orange+4apple, or, 0.75orange+apple if you want).

The same goes for evaluating pure mathematical functions on values with units! For instance, $\sin(40\text{apple})$ is perfectly valid expression, but it's irreducible. It has to be left in this form, because there is no numerical value that we can substitute into $\text{apple}$. However, this is somewhat alleviated with logarithms. It's common in physics to get intermediate expressions of the form $$\log V_1-\log V_2=\beta t$$ or something along these lines. $\log (\rm m^3)$ of course doesn't have a numerical value, it's just an irreducible symbolic entity. However, logarithms have a nice property of converting product into summation, so the problematic symbolic entity cancels out (producing $\log\frac{V_1}{V_2}$ which is a pure function evaluated on a pure number).

As soon as a unit can evaluate to numerical value, the "problem" disappears. For instance, the degree is simply ${}^\circ=\frac{\pi}{180}$ and percent is $\%=\frac{1}{100}$ and radians are just $\text{rad}=1$, so you can write $\sin(45^\circ+50\%+2\text{rad}+5)$ and there are no problems with summation of different units whatsoever.

To sum up: units are quantities that by definition don't need to evaluate to numerical values (they are "handles" that point to the physical world). We treat the unevaluated product of value and unit as valid but not sums of mismatched units simply because the first can be reinterpreted back into the physical world, while the second usually have no reasonable meaning. Mathematically, there's no difference.

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this is similar to the answer by @alex-jordan , it is an interesting and smart way to tackle the problem –  danza May 14 at 19:10

In principle $a x^2+b x+c$ would be regarded as adding a 3D,2D and 1D object from a 17th century mathematics perspective. They asserted a law of dimensional homogeneity and would rather write $a x^2+b^2 x+c^3$.

So it has a lot to do with how we model the physical world and whether we regard the ontological status of mathematical objects as either purely abstract or ideals of physics or even to be in 1-1 correspondance with the physical world.

From a purely abstract perspective we could define addition, subtraction, multiplication and division on almost any collection of abstract objects as long as we can make it well defined and consistent. So multiplying apples and oranges ... Why not? Sounds like fun!


Another funny example is that (at least in my country) teachers tend to visualize fractions as pizzas. But that makes multiplication and division rather mysterious. Also negative fractions and irrational proportions seem very strange from this perspective.

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Multiplying apples and oranges? Easy, just do $\mbox{Apples}\otimes\mbox{Oranges}$. –  Neal May 12 at 22:33
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@Neal: Yes! What about the field of pizzas above? What would be a natural definition of multiplication consistent with real life pizzas? –  String May 12 at 22:40
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Well, to visualize $(a/b)(c/d)$ with pizzas ... $d$ is the size of the initial slice of pizza. You have $c$ of them. Now take each of them and cut them into $b$ pieces. Now make $a$ copies of what you have, and that's the result. –  Neal May 12 at 23:30
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@Neal: Thanks! Could you do a video of $\frac{355}{113}\cdot\frac{7}{22}$ for me as a promotion of this new branch of mathematics and upload it on youtube? And then we'll rename \pi as \pizza ... –  String May 12 at 23:41

I'm sure there are many different ways of conceptualizing this. Here's one that is a little different from the ways discussed in the answers so far.

Historically, addition and multiplication were defined in geometrical terms. This approach is developed in Euclid (including a bunch of number theory) in a style that is very alien to the modern math student. Basically multiplication is thought of as a computation of the area of a rectangle.

But this approach includes a notion of rotation, which is not meaningful when all you want to do is arithmetic. Suppose you're multiplying 2 oranges by 2 apples. You have a square that has a width of 2 oranges, a height of 2 apples, and an area of 4 orange-apples. If rotations made sense, then you could rotate this square so that the 2 apples were brought on top of the 2 oranges. But that doesn't make sense, because apples are not comparable to oranges.

So we can take Euclidean geometry and throw away the notions of rotation and angular measure. What we get then is affine geometry. In affine geometry, lengths are only comparable if they're along parallel lines. There is no notion of angle, but there is a notion of area.

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I would like to compliment the answers already posted, especially fgp's excellent answer.

The reason we don't want to add quantities like meters and feet is indeed scale invariance. But this isn't the reason why we don't add meters and seconds--we aren't worried about issues of scaling, it just doesn't make any sense to add the underlying physical quantities. Yet it makes perfect sense to multiply any two quantities. Why? It turns out there is some fairly sophisticated mathematics going on here.

Generally, physical quantities are elements of an abelian group, or if they are continuous a vector space. This means we can add and subtract values of the same quantity, and if they are continuous we can scale them. To go beyond this, we need to construct new vector spaces. It turns out that there are natural constructions which allow us to multiply elements (even in different spaces!), and in some cases to invert elements, but not to add elements in different spaces.

Multiplying elements: Given two vector spaces $V$ and $W$, we can form the tensor product $V\otimes W$ which consists of linear combinations of elements of the form $v\otimes w$ with $v\in V,w\in W$, and obeys the following rules: $$v\otimes w+v'\otimes w=(v+v')\otimes w$$ $$v\otimes w + v\otimes w'=v\otimes (w+w')$$ $$\lambda(v\otimes w)=(\lambda v)\otimes w=v\otimes (\lambda w)$$ The product of $v$ and $w$ is then just $v\otimes w$. One important thing to note is that if $\dim V=1$ or $\dim W=1$, then all elements of $V\otimes W$ are of the form $v\otimes w$ (which is called an elementary tensor), but if either space has dimension greater than $1$ there are elements of $V\otimes W$ which are not elementary.

Inverting elements: Given a vector space $V$, we can form the dual space $V^*$ consisting of all linear maps $f:V\to \mathbb R$. If $\dim V=1$, for any $v\ne 0$ in $V$ there is a unique linear map $v^{-1}:V\to\mathbb R$ such that $v^{-1}(v)=1$, which we call the inverse of $v$. This behaves much the way we'd expect of an inverse: we have a natural map $V\otimes V^*\to \mathbb R$ defined by $v\otimes f\mapsto f(v)$, and so $v\otimes v^{-1}\mapsto 1$. If $\dim V>1$, we need more structure in order to define an inverse. If $V$ is an inner product space, we can take an orthonormal basis $e_1,\ldots,e_n$ of $V$ and define the inverse of $v=a_1e_1+\cdots+a_ne_n$ to be the function $v^{-1}$ such that $$v^{-1}(e_i)=\frac{a_i}{\|v\|^2}=\frac{a_i}{a_1^2+\cdots+a_n^2}$$ which satisfies $v^{-1}(v)=1$ and turns out to be independent of our choice of orthonormal basis.

Let's bring these together in an example. Let $V$ be the space of displacements, which has units of $(m_x,m_y,m_z)$ where $m_x$ is meters in the $x$-direction, etc. Let $W$ be the space of times, which has units $s$. Then $W^*$ has units $s^{-1}$, and $V\otimes W^*$ has units $(m_x\cdot s^{-1},m_y\cdot s^{-1}, m_z\cdot s^{-1})$, so $V\otimes W^*$ is the space of velocities. If a displacement $d$ occurs over a time $t$, we then get a velocity $d\otimes t^{-1}$. Conversely, given a velocity $v$ and a time $t$, $v\otimes t$ is an element of $V\otimes W^*\otimes W$. The map $W^*\otimes W\to \mathbb R$ gives us a map $V\otimes W^*\otimes W\to V$; this is just cancelling the units $s$ and $s^{-1}$ to get a displacement.

If this formulation seems cumbersome and unnecessary, that's because when one of the quantities we are working with is $1$-dimensional we can ignore most of the details of tensor products and everything becomes much easier. Let's examine what would happen if time were $2$-dimensional. Let $V$ and $W$ be as before. The space of velocities is still $V\otimes W^*$, but this is now $6$-dimensional. Now not every element of $V\otimes W^*$ is elementary, so we can no longer interpret a velocity as a displacement over a period of time! Indeed, two different velocities can have the same displacement over the same (nonzero) time. Things get even worse when we consider acceleration (which is $12$-dimensional) or even higher derivatives. In these cases, the algebra of tensor products is really necessary.

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If you think of it as a word problem, it becomes easier to understand:

Imagine you have a total of $15$ oranges and $9$ apples, and you want to evenly distribute them in packages, then you would get $5$ oranges for every $3$ apples or $5$ oranges$/$$3$ apples.

Now imagine you have $10$ oranges you have to send out and for each orange you need to include $5$ apples, $10$ oranges $\times5$ apples result in $50$ orange apples. Ok, not the best but I think you get the idea.

When you add or especially subtract, you're not actually doing anything to either number if they're not the same unit. $5$ apples + $3$ oranges indicates you have $5$ apples and $3$ oranges (in programming you'd say they're of the same class, fruit, so you can say $5$ apples and $3$ oranges $= 8$ fruit).

Especially weird when you do subtraction. If you have $5$ apples and you subtract $3$ oranges, what does that even mean? You can't do that obviously. But it does tell you 2 things: You have $5$ apples, and you owe someone $3$ oranges.

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This question relies on an assumption that is actually not correct, you actually can add apples and oranges (or any two things of dissimilar units/meaning), and we do it all the time. Consider this statement:

He is five feet and ten inches tall.

I have just added two dissimilar measurements while describing his height, feet and inches, and you understand what that means, likely because you know what a foot is and what an inch is. Now you can't add them and get down to only one number unless there is a conversion between the two, or you are going to change the unit/meaning of the answer that you want. Consider the following elementary word problem:

I have two apples and three oranges, how many fruits do I have? Answer: three (fruits) It is still two apples and three oranges.

Multiplication inherently involves two numbers of different units/meaning. You can't actually multiply two numbers of the same units/meaning together without a third underlying assumption. Consider the following elementary word problem:

I have two bags each containing three oranges, how many oranges do I have overall?

Answer: 6 (oranges). This would not work if both numbers referred to a collection of oranges because then multiplication wouldn't make any sense without a third underlying assumption (i.e. each orange in one collection actually represents a set of oranges).

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Quite simply, you can simplify $ax \cdot bx$ to $ab \space x^2$, but you can't simplify $ax + by$.

Imagine we have two values (with units), $7\ \text{m}$ and $5\ \text{s}$. Why would you try to add $7$ meters to $5$ seconds? You can't because they aren't the same unit. However, we can divide them to get a velocity: $\frac{7\ \text{m}}{5\ \text{s}} = \frac{7}{5} \text{m s}^{-1}$.

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