I've seen a number of people ask why, if you can have arbitrary-sized integers that do everything exactly, you can't do the same thing with floats, avoiding all the rounding problems that they keep running into.

If we compare the two, it should become pretty obvious at some point within the comparison.

Integers

Let's start with two integers, a=42 and b=2323, and add them. How many digits do I need? Think about how you add numbers: you line up the columns, and at worst carry one extra column. So, the answer can be as long as the bigger one, plus one more digit for carry. In other words, max(len(a), len(b)) + 1.

What if I multiply them? Again, think about how you multiply numbers: you line up the smaller number with each column of the larger number, then there's that extra digit of carry again. So, the answer can be as long as the sum of the two lengths, plus one more. In other words, len(a) + len(b) + 1.

What if I exponentiate them? Here, things get a bit tricky, but there's still a well-defined, easy-to-compute answer if you think about it, asking how many digits are in a**b is just solving for x in 10**(x-1) = a**b and then rounding up. So, log10 both sides and add one, and x = log10(a**b) + 1 = log10(a) * b + 1. Fit in your variables, and it's log10(42) * 2323 ~= 3770.808, which rounds up to 3771. Try len(str(42**2323)) and you'll get 3771.

You can come up with other fancy operations to apply to integers--factorials, gcds, whatever--but the number of digits required for the answer is always a simple, easy-to-compute function of the number of digits in the operands.^*

_{* Except when the answer is infinite, of course. In that case, you easily compute that the answer can't be stored in any finite number of digits and use that fact appropriately--raise an exception, return a special infinity value, whatever.}

Decimals

Now, let's start with two decimals, a=40 and b=.2323, and add them. How many digits do I need? Well, how many digits do the originals have? It kind of depends on how you count. But the naive way of counting says 2 and 4, and the result, 42.2323 has 6 digits. As you'd suspect, len(a) + len(b) + 1 is the answer here.

What if I multiply them? At first glance, it seems like it should be easy--our example gives us 9.7566, which has 5 digits; multiplying a by itself is the same as integers, and b by itself for 0.05396329 is just adding 4 decimal digits to 4 decimal digits, so it's still len(a) + len(b) + 1.

What if I exponentiate them? Well, now things get not tricky, but impossible. 42**.2323 is an irrational number. That means it has an infinite number of digits (in binary, or decimal, or any other integer base) to store. (It also takes an infinite amount of time to compute, unless you have an infinite-sized lookup table to help you.) In fact, most fractional powers of most numbers are irrational--2**0.5, the square root of 2, is the most famous irrational number. This means it takes an infinite number of digits to store.

And it's not just exponentiation; most of the things you want to do with real numbers--take the sine, multiply by pi, etc.--give you irrational answers. Unless you stick to nothing but addition, subtraction, multiplication, and division, you can't have exact math.

Even if all you want is addition, subtraction, multiplication, and division: a=1, b=3. How many digits do I need to divide them? Start doing some long division: 1 is smaller than 3, so that's 0. 10 has three 3s in it, so that's 0.3, with 1 left over. 10 has three 3s in it, so that's 0.3 with 1 left over. That's obviously going to continue on forever: there is no way to represent 1 / 3 in decimal without an infinite number of digits. Of course you could switch bases. For example, in base 9, 1 / 3 is 0.3. But then you need infinite digits for all kinds of things that are simple in base 10.

Fractions

If all you want actually is addition, subtraction, multiplication, and division, you're dealing with fractions, not decimals. Python's fractions.Fraction type does all of these operations with infinite precision. Of course when you go to print our the results as decimals, they may have to get truncated (otherwise, 1/3 or 1/7 would take forever to print), but that's the only limitation.

Of course if you try to throw exponentiation or sine at a Fraction, or multiply it by a float, you lose that exactness and just end up with a float.

Aren't decimals just a kind of fraction, where the denominator is 10^d, where d is the number of digits after the decimal point? Yes, they are. But as soon as you, say, divide by 3, the decimal result is a fraction with an infinite denominator, as we saw above so that doesn't do you any good. If you need exact rational arithmetic, you need fractions with arbitrary denominators.

Accuracy

In real life, very few values are exact in the first place.^* Your table isn't exactly 2 meters long, it's 2.00 +/- 0.005 meters.^** Doing "exact" math on that 2 isn't going to do you any good. Doing error-propagating math on that 2.00, however, might.

Also, notice that a bigger number isn't necessarily more accurate than a smaller one (in fact, usually the opposite), but the simple decimal notation means it has more precision: 1300000000 has 10 digits in it, and if we want to let people know that only the first 3 are accurate, we have to write something like 1300000000 +/- 5000000. And even with commas, like 1,300,000,000 +/- 5,000,000, it's still pretty hard to see how many digits are accurate. In words, we solve that by decoupling the precision from the magnitude: 1300 million, plus or minus 5 million, puts most of the magnitude into the word "million", and lets us see the precision reasonably clearly in "1300 +/- 5". Of course at 13 billion plus or minus 5 million it falls down a bit, but it's still better than staring at the decimal representation and counting up commas and zeroes.

Scientific notation is an even better way of decoupling the precision from the magnitude. 1.3*10¹⁰ +/- 5*10⁶ obviously has magnitude around 10¹⁰, and precision of 3-4 digits.*** And going to 1.3*10¹¹ +/- 5*10⁶ is just as readable. And floating-point numbers give us the same benefit.

In fact, when the measurement or rounding error is exactly half a digit, it gets even simpler: just write 1.30*10¹⁰, and it's clear that we have 3 digits of precision, and the same for 1.30*10¹¹. And, while the float type doesn't give us this simplification, the decimal.Decimal type does. In addition to being a decimal fraction rather than a binary fraction, so you can think in powers of 10 instead of 2, it also lets you store 1.3e10 and 1.30e10 differently, to directly keep track of how much precision you want to store. It can also give you the most digits you can get out of the operation when possible--so 2*2 is 4, but 2.00*2.00 is 4.0000. That's almost always more than you want (depending on why you were multiplying 2.00 by 2.00, you probably want either 4.0 or 4.00), but you can keep the 4.0000 around as an intermediate value, which guarantees that you aren't adding any further rounding error from intermediate storage. When you perform an operation that doesn't allow that, like 2.00 ** 0.5, you have to work out for yourself how much precision you want to carry around in the intermediate value, which means you need to know how to do error propagation--but if you can work it out, decimal can let you store it.

<sub>* Actually, there are values that can be defined exactly: the counting numbers, e, pi, etc. But notice that most of the ones that aren't integers are irrational, so that doesn't help us here. But look at the symbols section for more...

** If you're going to suggest that maybe it's exactly 2.0001790308123812082 meters long: which molecule is the last molecule of the table? How do you account for the fact that even within a solid, molecules move around slowly? And what's the edge of a molecule? And, given that molecules' arms vibrate, the edge at what point in time? And how do you even pick a specific time that's exactly the same across the entire table, when relativity makes that impossible? And, even if you could pick a specific molecule at a specific time, its edge is a fuzzy cloud of electron position potential that fades out to 0.

*** The powers are 10 and 6, so it's at worst off by 4 digits. But the smaller one has a 5, while the bigger one has a 1, so it's obviously a lot less than 4 digits. To work out exactly how many digits it's off, do the logarithm-and-round trick again.</sub>

Money

Some values inherently have a precision cutoff. For example, with money, you can't have less than one cent.^* In other words, they're fixed-point, rather than floating-point, values.

The decimal module can handle these for you as well. In fact, money is a major reason there's a decimal standard, and implementations of that standard in many languages' standard libraries.<sub>***

* Yes, American gas stations give prices in tenths-of-a-cent per gallon, and banks transact money in fractional cents, but unless you want to end up in Superman 3,** you can ignore that.

** And yes, I referenced Superman 3 instead of Office Space. If you're working in software in 2015 and haven't seen Office Space, I don't know what I could say that can help.

*** For some reason, people are willing to invest money in solving problems that help deal with money.</sub>

Symbols

So, how do mathematicians deal with all of this in real life? They don't. They do math symbolically, rather than numerically. The square root of 2 is just the square root of 2. And you carry it around that way throughout the entire operation. Multiply 3 * sqrt(2) and the answer is 3 * sqrt(2). But multiply sqrt(2) * sqrt(2) and you get 2, and multiply sqrt(2) * sqrt(3) and you get sqrt(6), and so on. There are simplification rules that give you exact equivalents, and you can apply these as you go along, and/or at the end, to try to get things as simple as possible. But in the end, the answer may end up being irrational, and you're just stuck with sqrt(6).

Sometimes you need a rough idea of how big that sqrt(6) is. When that happens, you know how rough you want it, and you can calculate it to that precision. To three digits, more than enough to give you a sense of scale, it's 2.45. If you need a pixel-precise graph, you can calculate it to +/- half a pixel. But the actual answer is sqrt(6), and that's what you're going to keep around (and use for further calculation).

In fact, let's think about that graph in more detail. For something simple and concrete, let's say you're graphing radii vs. circumferences of circles, measured in inches, on a 1:1 scale, to display on a 96 pixels-per-inch screen. So, a circle with radius 3" has a diameter of 18.850" +/- half a pixel. Or, if you prefer, 1810 pixels. But now, let's say your graph is interactive, and the user can zoom in on it. If you just scale that 1810 pixels up at 10:1, you get 18100 pixels. But if you stored 6*pi and recalculate it at the new zoom level, you get 18096 pixels. A difference of 4 pixels may not sound like much, but it's enough to make things look blocky and jagged. Zoom in too much more, and you're looking at the equivalent of face-censored video from Cops.

Python doesn't have anything built-in for symbolic math, but there are some great third-party libraries like SymPy that you can use.