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This is a very silly question since nobody will actually do this because it makes very little sense in the real world but I just want to know how would it actually work if possible.

For example let us take an amount of 2 dollars and 2 cents and multiply that buy 2 dollars and 2 cents. Would the result be 4 dollars and 4 cents? The only way to make sense out of this for me is to treat them as vectors and multiply components.

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    $\begingroup$ As an analogy... would you multiply a length 2 feet, 2 inches by itself "component-wise" to get 4 feet and 4 inches? $\endgroup$ – pjs36 Apr 16 '15 at 0:14
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    $\begingroup$ Most of the effort in the real world has been in multiplying money by real numbers, especially those larger than one. $\endgroup$ – Jorge Fernández Hidalgo Apr 16 '15 at 0:53
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    $\begingroup$ I wonder what $\$^2$ would mean… $\endgroup$ – Akiva Weinberger Apr 16 '15 at 1:52
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    $\begingroup$ @columbus8myhw, the product of two dollars is a en.wikipedia.org/wiki/Polydisc :-). $\endgroup$ – Martín-Blas Pérez Pinilla Apr 16 '15 at 7:04
  • $\begingroup$ 2 dollars and 2 cents isn't a vector: it's the single number \$2.02. The dot there is a genuine decimal point -- "cent" means "hundredth". $\endgroup$ – David Richerby Apr 17 '15 at 9:01
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Consider the notion of calculating the variance of an array of prices. The result would be some amount of square dollars. Even if this measure is simply an intermediate step toward determining a standard deviation, two currency values are being multiplied together before their square root is taken, so some definition of dollar multiplication must exist. The most natural notion would be as drowdemon suggested; simply multiply the fixed point values together like real/rational numbers.

The idea of a square dollar may not seem to make much sense, but the idea of a square second doesn't really make much sense either until put in the context of acceleration. Just as $m/s^2$ is really just meters per second per second, there is no reason we cannot consider the notion of how the rate of shares one gets per dollar changes per dollar invested: ie shares per dollar per dollar or shares per dollar squared.

More humorous examples abound. Try googling "pound pound dollar rate".

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    $\begingroup$ Someone pointed out something similar to "pound pound dollar" a while back - here in NZ, rather than miles/gallon, we liters/100km (more akin to "gallons per 100 miles", I guess). A liter (or gallon) is a volume, and dividing a volume by a length gets you a surface area - my car runs at a fuel efficiency of about 0.05 square millimeters! This means that if the fuel line is 0.05 square millimeters, the fuel would flow at my driving speed. $\endgroup$ – AMADANON Inc. Apr 16 '15 at 2:07
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    $\begingroup$ @AMADANONInc: or that you could drive along the route of a 0.05 square millimeter pipe full of stationary fuel, using the fuel as you go. At least until you hit traffic lights. $\endgroup$ – Steve Jessop Apr 16 '15 at 12:07
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    $\begingroup$ @AMADANONInc See the end of this wonderful xkcd post for fuel efficiency in units (length)^2. what-if.xkcd.com/11 $\endgroup$ – Ethan Bolker Apr 16 '15 at 16:50
  • $\begingroup$ @EthanBolker - yes, that's who pointed it out. I knew I saw it somewhere. Credit to what-if.xkcd.com/11 ! $\endgroup$ – AMADANON Inc. Apr 17 '15 at 1:29
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No, it would not make sense. And in any case, there is no standard multiplication of vectors anyhow.

The key is that currency has a unit of measurement, e.g. dollars. Just like multiplying one length with another length gives you length-squared, multiplying 2 dollars by 2 dollars would give you 4 dollars squared.

Now, we could treat currency as an algebraic object and define multiplication on it, and there's nothing wrong with that.

But just because we can do it mathematically in an abstract sense doesn't mean that it's useful in any way in the real world. Squared-dollars can't buy you lunch.

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    $\begingroup$ They might buy you a square meal. $\endgroup$ – Robert Israel Apr 16 '15 at 0:15
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    $\begingroup$ But if that meal has a negative nutritional value "imagine" the complexities ;-) $\endgroup$ – skullpatrol Apr 16 '15 at 8:30
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To explain a bit deeper why your definition does not quite work, let's think about unit conversions. Specifically, let's think about using unit conversions to try to get rich quick.

We can see that $\$2 + 2\text{ cents} = 202\text{ cents}$. So, we might think that $(\$2 + 2\text{ cents}) \times (\$2 + 2\text{ cents}) = (202\text{ cents}) \times (202\text{ cents})$. But, using your definition,

$$(\$2 + 2\text{ cents}) \times (\$2 + 2\text{ cents}) = \$4 + 4\text{ cents}$$ $$(202\text{ cents}) \times (202\text{ cents}) = 40804\text{ cents} = \$408 + 4\text{ cents}$$

This doesn't seem like how things should work! Just by converting to cents before we multiplied our money, we became over $100$ times richer than we would have been had we instead left our currency in bills. There are two problems: first, we should really be multiplying term by term, so

$$(\$2 + 2\text{ cents}) \times (\$2 + 2\text{ cents}) = (\$2)(\$2) + (\$2)(2\text{ cents}) + (2\text{ cents})(\$2) + (2\text{ cents})(2\text{ cents})$$ $$ = \$^24 + \$8\text{ cents} + 4\text{ cents}^2$$

and second, we should be multiplying the units together, so

$$(202\text{ cents}) \times (202\text{ cents}) = 40804\text{ cents}^2$$

Now, these two things look different, but you have to keep in mind the $\$1 = 100\text{ cents}$, so

$$\$^24 + \$8\text{ cents} + 4\text{ cents}^2 = (100\text{ cents})^2(4) + (100 \text{ cents})8\text{ cents} + 4\text{ cents}^2 = 40804\text{ cents}^2$$

exactly what we got by converting to cents first.

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    $\begingroup$ Remember that there are 10,000 square cents to the square dollar! $\endgroup$ – AMADANON Inc. Apr 16 '15 at 2:09
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    $\begingroup$ I love $8 cents. Going to start measuring areas in kilometre miles. $\endgroup$ – Ben Millwood Apr 16 '15 at 22:20
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You'd just combine the cents and dollars into one dollar value: $\$2.02*\$2.02 = 2.02*2.02=4.0804$. I guess the units would be square dollars...

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    $\begingroup$ My dollars are rectangular, does that mean it won't work? 8-) $\endgroup$ – Avraham Apr 16 '15 at 0:15
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    $\begingroup$ Technically, if you think of concepts like investment gains, taxes etc. where it's possible to measure dollar per dollar (more commonly uttered as some number of cents to the dollar), it's quite possible to imagine a formula that ends up calculating or using dollar dollar ($^2). But let's not give economists even more ideas on how to mess with the economy... $\endgroup$ – slebetman Apr 16 '15 at 2:53
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Squaring an amount of money could reflect a long-term investment plan. Assume that you invest an amount $x$ and you wonder "what should be the yearly return so that I will have $x^2$ in $n$ years?", if I leave principal and interest intact for the duration?

You're young, say 20, and you think, "let's forget USD 100 for the next 40 years. What would it take to make them USD 10,000?"

Solving

$$100\cdot (1+r)^{40} = (100)^2 \implies r \approx 12.20 \%$$

Hmmm, rather large for a yearly real interest rate (inflation subtracted), averaged over 40 years-this is probably why squaring a monetary sum is not frequently encountered. You could get something like $r= 6\%$, and you will need $72$ years to reach $10,000$. So start when you're born.

But it may be the case that you are a risk-loving (or risk-seeking) individual, in which case your utility function with respect to wealth would be convex - and

$$u(W) = W^2$$ is as good as any.

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    $\begingroup$ I don't like this notion - because you don't want (100$)² you want (100²) $ in 40 years so you are not operating with square dollars! Don't ignore the units, make sense of them $\endgroup$ – Falco Apr 17 '15 at 8:11
  • $\begingroup$ @Falco Good point -but it goes with the discipline. And in Economics we don't have much use for composite units of measurement. $\endgroup$ – Alecos Papadopoulos Apr 17 '15 at 8:54

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