59
$\begingroup$

I asked my teacher if a number can have infinitely many digits before the decimal point. He said that this isn't possible, even though there are numbers with infinitely many digits after the decimal point. I asked why and he said that if you keep adding digits to the left of a number it will eventually approach infinity which is not a number and you could no longer distinguish two numbers from one another. Now this is the part of his reasoning I don't understand: why can we distinguish two numbers with infinitely many digits after the point but not before it? Or is there a simpler explanation for this?

$\endgroup$
  • 27
    $\begingroup$ We can have infinitely many digits before the decimal point if the digits are all zeroes. $\endgroup$ – Joel Reyes Noche Sep 24 '14 at 12:50
  • 12
    $\begingroup$ You might be interested in $p$-adic numbers - look at the introduction on the link. Here two numbers are close together if they differ by a high power of the prime $p$ - this means they don't behave as expected - they are rather unlike the familiar real numbers (as well as having important things in common): quite weird things happen, but a sequence of numbers which is increasing according to our normal understanding can converge to a $p$-adic limit and make sense of the kind of expression you are asking about. en.wikipedia.org/wiki/P-adic_number $\endgroup$ – Mark Bennet Sep 24 '14 at 13:01
  • $\begingroup$ If digits 1 to 9 are repeated an infinite number of times to form a number, then that number would be infinite. The decimal point for that number would be irrelevant. The number itself would be irrelevant because it is simply $\infty$ $\endgroup$ – Nick Sep 24 '14 at 14:33
  • 7
    $\begingroup$ Vi Hart has a video on her youtube channel that contains a lot more information about this if you're interested. It's in the context of talking about Cantor's proof that there are more real numbers than integers however, which is something else fascinating you may want to look into. $\endgroup$ – Jack Sep 24 '14 at 15:27
  • 3
    $\begingroup$ This is a job for Archimedes Plutonium! Ah, those halcyon days of sci.math. $\endgroup$ – marty cohen Sep 24 '14 at 19:42

10 Answers 10

46
$\begingroup$

The formal way to understand this is, of course, using the definition of real numbers. A real number is "allowed" to have infinite digits after the decimal point, but only a finite number of digits before. (http://en.wikipedia.org/wiki/Real_number)

(if it interests you, there are numbers that have infinite digits before the decimal point, and only a finite number after. Take a look at http://en.wikipedia.org/wiki/P-adic_number . just to have some fun, know that the $10$-adic expansion of $-1$ is $\color{red}{\dots 99999} = -1$)

If you want to get some intuition about this, first think that, as your teacher said, said number would approach infinity, which is not a real number. This is reason enough.

About the comparing two numbers part: if I give you $$1234.983...$$ and $$1234.981...$$ you know which one is bigger, it does not matter what the other digits are.

But with $$...321.99$$, $$...221.99$$ you don't, because the information relies in the "first" digit. Of course nobody know what the first digit is, since there is no first digit.

But as I said before, this is to gain some intuition; the correct way to think about this is using the definition (which is not trivial)

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ you don't, because the information relies in the "first" digit. Of course nobody know what the first digit is, since there is no first digit.. What if I have $123......321.123$ and $234......432.234$? The second number is larger...? $\endgroup$ – gerrit Sep 24 '14 at 14:43
  • 10
    $\begingroup$ @gerrit be aware about writing with dots. What are the dots supposed to mean? An infinity of numbers? So that $1$ would be the first number? It does not make any sense. It's like asking for the last digit of $\pi$. Having an infinite amount of digits after the decimal point does not mean it is possible to find the last one (precisely because there are infinite) even if we know them all. for example, let's say the the $n$-th digit after the decimal point is $a_n = 3n + 1 \pmod {10}$. What is the last digit? Technically we know them all, but the concept has no meaning. $\endgroup$ – Ant Sep 24 '14 at 14:50
  • 5
    $\begingroup$ For this same reason, I am inclined to say that even with $\frac 13 = 0.3333....$ we do not know what the last digit is, because the concept of last digit is an illusion. It just does not make sense. $\endgroup$ – Ant Sep 24 '14 at 14:52
  • 5
    $\begingroup$ To be pointlessly pedantic, you actually don't know which is bigger if the numbers are 1234.982000... and 1234.9819999.... they are equal. Comparing decimal numbers isn't quite as simple as checking all the digits. $\endgroup$ – cobbal Sep 24 '14 at 15:16
  • 10
    $\begingroup$ @gerrit No you still can't compare and say whether 123......321 and 234......432 is larger, because it depends on how many digits those ellipses represent. And if both are infinitely long, the notion of which is bigger is undefined. $\endgroup$ – Mobius Pizza Sep 25 '14 at 11:13
28
$\begingroup$

The short answer to your question is that by definition we only allow real numbers to have finitely many digits before the decimal point. There are very good reasons for this:

Formally, we can think of a number as a finite sequence of digits $x_0,\ x_1, \ \ldots , x_N$, where the number $x$ is equal to $$x=\sum_{n=0}^Nx_n10^n$$

For example, the number $126 = 6 + 2\times 10 + 1 \times 100$.

Inherently there is no reason to restrict ourselves to finite sequences of digits. Given an infinite sequence, we could define a "number" as above and we could differentiate between two numbers by saying that they are different if they differ in at least one digit.

The problems start coming in when try to do arithmetic. Consider the "number" $\ldots 99999$ made up of infinitely many $9$'s. If this is a number, we should be able to add and subtract as normal. But what happens when we try to add $1$ to this number?

$$\begin{array}[t]{r} \ldots 9\ 9\ 9\ 9\ 9\ 9\ 9\\ + \qquad\qquad \ \ \ \ 1 \\ \hline \ldots0\ 0 \ 0\ 0\ 0\ 0\ 0 \end{array}$$

At each stage, we get $10$ and carry a $1$, but this happens at every stage, and never stops. In other words, adding $1$ to this number gives us $0$ - not especially coherent with what we would expect from our number system.

Or consider the "number" $\ldots 11111.1111\ldots$ made up of infinitely many $1$'s before and after the decimal point. What happens when we multiply this number by $10$? We get exactly the same number - meaning that this number and others like it would become solutions to the equation $10x = x$. But in our number system, we would like $0$ to be the only solution to this equation.

It is possible to create a coherent number system where we allow arithmetic like this, but it would be completely different from our own number system.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Integers have infinitely many digits to the left, right? Wouldn't numbers in $\mathbb {Z } $ satisfy $10x=x$, by what you're saying? I get that you can create a number system that satisfys that property, but the integers don't. $\endgroup$ – dylan7 Sep 25 '14 at 12:36
  • $\begingroup$ Two's complement arithmetic essentially says that numbers may have an infinite number of 0's or 1's to the left, but only a finite number of digits may follow the last 0-1 or 1-0 transition. Subtract 1 from 0, and one gets an infinite string of ones. Note that if one uses the power-series formula to compute the value of 1+2+4+8+16... one gets -1 (which makes sense, since it's the result of subtracting one from zero).l $\endgroup$ – supercat Sep 25 '14 at 15:06
8
$\begingroup$

For ordinary integers, there can only be infinitely many digits to the left of the decimal point if the are only finitely many nonzero digits among them, in other words if there is a point to the left of which there are only zero digits. Usually one does not write those digits for reasons that are easy to imagine.

The reason this is so is because numbers are not strings of digits, they are just denoted by strings of digits. This is a lot like my name not being the same thing as I am, but is just used to designate me. It is a bit more difficult to imagine, since while you can see me (or you could if you were in the same room as I am) and see the difference with my name, you cannot see the number designated by $421.94$, as it is an abstract quantity. You might image however two line segments of which one is exactly $421.94$ times as long as the other, and say that $421.94$ is the (abstract) ratio of those two lengths, and you might agree that this has a meaning quite independent and different from the glyphs in "421.94". (By the way such ratios is how the ancient Greeks thought about the nature of numbers.) Numbers are things that exist in our minds independently of how we represent them (or of the question whether we can represent them at all, if you are sufficiently imaginative).

One can very well imagine digits going on forever to the left as well as to the right (and for that matter I can imagine them being stacked in infinitely many rows in the plane as well) but we only say that a string of digits represents a number if it corresponds according to a definite convention to a number. As long as there are only finitely many digits, that convention (for the decimal number system) is that according to its position each digit is multiplied by a certain power of ten, and all those values are added together to get our number. Adding finitely many numbers together is something we imagine always possible, and always returning some number.

However adding infinitely many numbers together is not something we can imagine always possible, at least not if we care about preserving the beautiful properties that numbers have. Imagine adding together infinitely many copies of the number $1$. If that were to give a number as a result (you might want to call it "$\infty$") then one could also ask to compute $1+\infty$. On hand that would still be adding up infinitely many copies of the number $1$, so it should give $\infty$, but on the other hand adding $1$ to a number should always produce a different number (the difference between the new and the old number should be$~1$, not$~0$). As a consequence of such simple observations, one cannot consider that in general adding up infinitely many numbers gives a result that is a number. (Saying the result "is infinitely large" is just a manner of speach, and is not a number.)

Now in certain special cases one can agree that adding up infinitely many terms gives a definite number as result. A somewhat boring instance of this is when all terms except finitely many ones are zero; in this case one can agree that the (finite) sum of the nonzero terms also gives the sum of all the terms (by definition of the latter). This is why I said one can allow infinitely many digits $0$ going off to the left (or to the right for that matter).

Another case that is usually considered to give a well defined result is when the terms are all positive (or zero) but get smaller and smaller, in such a way that there is a threshold that none of the results of adding finitely many terms passes this threshold. In this case we imagine there is a smallest possible such threshold, one that is never passed, but such that any smaller threshold will be ultimately passed by some finite sum of terms. I said "imagine" here, because one needs to invent the real numbers specifically for this purpose: if for instance all our terms are rational numbers, one for which we have fractions to express them precisely, then all finite sums will also be rational numbers, but the smallest threshold might not be equal to any rational number. However the real numbers are, by design, such that whenever any threshold exists for the finite sums, there exists a (unique) real number that is the smallest threshold; this number is then defined to be the sum of the infinite sequence of terms.

If negative terms are allowed things get a bit more complicated, but still there are certain cases where the finite sums ultimately get confined in smaller and smaller intervals, and are said to tend to a limit; this condition defines "convergent" infinite sums of terms, and these are the only infinite summations that conventionally designate a number (namely that limit). All other summations are said to be "divergent", and no value at all is attributed to adding up their terms. $1+1+1+\cdots$ is a typical divergent summation.

Now you can imagine that adding up ever higher power of $10$, each multiplied by a nonzero digit, is even worse than doing $1+1+1+\cdots$; the summation diverges and produces no number. However when taking ever smaller powers of$~10$, it is easy to find a threshold for the finite sums; indeed stopping at any term, and taking the finite sum obtained after adding $1$ to the final digit, always produces threshold valid for all finite sums of the original terms. This is why infinite progressions of decimals after the decimal point always designate a real number. (They do not define a real number, as it existed independently, and even less is it true that they are real numbers; in particular it can happen that two different progressions of decimals designate the same real number, which confuses people who think of numbers as being progressions of decimals a lot.)

I might add, if this interests you, that other number systems than the real numbers exist, and that other notions of converging sums can be proposed (not based on getting "ever closer" to a limit in the usual sense). In one such number systems, the $10$-adic integers, infinite progressions of decimals to the left (but not to the right!) do define a convergent sum, and therefore designate a number. These numbers however do not have all the usual propreties of numbers; notably there is no notion of "less than" and "more than" among the $10$-adic integers. Can you see why that would be a difficult thing to define for numbers whose digits run off to the left?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ If one thinks about how arithmetic is performed, it would make sense that (in decimal) subtracting 1 from 0 would yield an infinite string of nines. The meaning of an infinite string of digits other than zeroes or nines is a bit dodgy, but saying that an infinite string of nines is -1 can be quite useful (the binary equivalent is the way two's complement math works). $\endgroup$ – supercat Sep 25 '14 at 15:08
  • $\begingroup$ @supercat: That is what happens in $10$-adic integers, which some other answers say a bit more about. But note that then there is no longer an order relation, and ($10$ not being prime) there are some other curious facts to be dealt with. $\endgroup$ – Marc van Leeuwen Sep 25 '14 at 15:43
  • $\begingroup$ 10-adic integers allow for the possibility that the leading infinite string is something other than all zeroes or all nines, and those possibilities get really weird, but the scenario of all nines (or, more generally, base-1) can be useful even if one doesn't allow for other leading infinite sequences. Since every number of that form has an additive inverse with only a finite number of non-zero leading digits, ordering relationships, as well as things like division, remain well defined. $\endgroup$ – supercat Sep 25 '14 at 15:52
6
$\begingroup$

You can't have infinitely many digits before the decimal point because the geometric series $$\sum_{n=0}^\infty ar^n$$ only converges when $|r| < 1$. So you can have infinitely many digits after the decimal point, because that's adding something of the form $$\sum_{n=1}^\infty a_n(\frac{1}{10})^n \le \sum_{n=1}^\infty 9(\frac{1}{10})^n$$ which converges because $|\frac{1}{10}| < 1$, but having infinitely many terms before the decimal point is of the form $$\sum_{n=0}^\infty a_n(10)^n \ge \sum_{n=0}^\infty 10^n$$ which doesn't converge (even if some $a_n = 0$ you can compare $10^n$ with the next nonzero term which is bigger).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ If e.g. r is an integer greater than 1 and a=r-1, the series yields -1, a result consistent with the effect of subtracting one from zero in base r. If one has a number whose last k digits are zeroes, then after subtracting one the last k digits will be r-1. If there was an infinite string of zeroes before the number prior to the subtraction, there should be an infinite string of (r-1) after. What's weird is what happens if there's an infinite string of digits that aren't all zero or r-1. $\endgroup$ – supercat Sep 25 '14 at 15:12
  • $\begingroup$ I'm not sure what your first sentence means, if r is an integer greater than 1 then the series diverges right? Because the partial sums don't converge $\endgroup$ – Jay Sep 25 '14 at 18:26
  • $\begingroup$ The normal formula for computing 1+n+n^2+n^3... is 1/(1-n). The fact that the formula yields -1 when n=2 might be viewed as nonsensical, but if one regards integers as "wrapping around" after value which, while infinite, is precisely one greater than the sum of that series, such a definition will yield an algebraic ring in which most of the normal properties of integer arithmetic hold, but some some others do as well. $\endgroup$ – supercat Sep 25 '14 at 19:06
  • $\begingroup$ +1. Even though I think it could use some more explanation, this is the only answer that gets to the heart of the issue: "decimal places" are not the definition of a [real] number, but a convenient representation for numbers. Answering OP's question requires looking at what this notation really means - what it's a shorthand for. $\endgroup$ – R.. GitHub STOP HELPING ICE Sep 26 '14 at 0:13
4
$\begingroup$

A partial answer only:

  1. Of course you may write a number with infinitely many zeros to the left.

  2. Your teacher is completely correct, for this reason: I’m guessing that you admit that there is no number greater than every $10^n$, by which I mean bigger than a million, bigger than a billion, than a trillion, etc., bigger than all the “illion”s. Yet the representation that you’re thinking of would, if it represented a number, be just so large. If this doesn’t convince you completely, try to imagine how you would multiply to itself a “number” that was represented with infinitely many $1$’s, both to the left and to the right of the point. Think about it and you realize that calculating each digit in the proposed product involves infinitely many additions, not good.

  3. In very advanced mathematics, there are “$n$-adic numbers”, and these can be written in a notation that allows infinitely many digits to the left of the point, but only finitely many to the right. The mathematical properties of these $n$-adic numbers depends strongly on the particular $n$, the base of the expansion. If you use decimal notation, you get $10$-adic numbers, and if you use binary, you get $2$-adic numbers, also called dyadic numbers. If you’re enthusiastic, maybe you can convince yourself that there are two nonzero $10$-adic numbers, $A$ and $B$, with this amazing property: (1) no digits to the right of the point; (2) $A+B=1$; (3) $A^2=A$ and $B^2=B$; (4) $AB=0$. And even more amazingly, for dyadic numbers, there are no such $A$ and $B$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ A+B=1 and AB=0 implies that the two numbers end in 5 and 6. One can compute any number of digits of the one ending in 5 by beginning with 5 and squaring it until the digits stop changing. Is there any nice way to compute the one ending in 6 other than by adding two to the nines' complement of the one ending in 5? $\endgroup$ – supercat Sep 25 '14 at 15:22
  • $\begingroup$ @supercat, I’m sure that that is the most efficient method. $\endgroup$ – Lubin Sep 25 '14 at 17:03
  • $\begingroup$ That property is cool; I would guess it's a result of 10 being the product of two primes? Even $A^2=A$ will fail with a prime base, though I'm not exactly sure what's "going on" in base 10. What would happen in bases 6, 12 [has a duplicate factor], or 30 [smallest number with three factors]? $\endgroup$ – supercat Sep 25 '14 at 18:03
  • 1
    $\begingroup$ This is indeed interesting, though to my knowledge it’s not big in mathematical research. If your number $n$ is divisible only by the primes $p_1,\cdots,p_k$, then the ring of $n$-adic numbers is the “direct sum” of the $\mathbb Q_{p_i}$, for $i=1,\cdots,k$. So, as you guess, if you consider the $120$-adic numbers, since $120=8\cdot3\cdot5$, the ring of $120$-adics is the direct sum of $\mathbb Q_2$, $\mathbb Q_3$, and $\mathbb Q_5$. You can prove this with Chinese Remainder Theorem. $\endgroup$ – Lubin Sep 26 '14 at 3:12
3
$\begingroup$

$$1/3=0.33333333333333333333333333...$$ $$1/9=0.11111111111111111111111111...$$ These two numbers are obviously different.

Now, imagine if you will

$$1+10+100+1000+10000...$$

and $$3+30+300+3000+30000...$$

Both of the first ones have infinitely many digits after the decimal, but they can still be represented as 1 number (1/3 and 1/9)

Both of the later ones have infinitely many digits before the decimal point, but cant actually be differentiated. You may say "but the second one is obviously three times as big!", but for every term in that second sequence, there are terms in the first sequence bigger than it. The reason this is that you cant compare the numbers is that they aren't numbers. Numbers can't be infinite, the study of infinities is separate from numbers for that very reason.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Your last phrase destroys the sense of your answer. Digits running off to the right are no more "writable" than digits running off to the left. $\endgroup$ – Marc van Leeuwen Sep 24 '14 at 14:21
  • $\begingroup$ But they can be written in some single number form (a fraction, or a root) $\endgroup$ – Asimov Sep 24 '14 at 15:38
  • 3
    $\begingroup$ No they cannot, at least not for all real numbers. $\endgroup$ – Marc van Leeuwen Sep 24 '14 at 15:46
  • 1
    $\begingroup$ That's a good point. I may need to think about how to adjust my definition in the last line to reflect the truth better. $\endgroup$ – Asimov Sep 24 '14 at 15:46
  • $\begingroup$ A real number can be described by a sum of infinitely many terms. But the sum itself (a single value) must be finite. I don't see where @Asimov relied on anything being "writable". $\endgroup$ – David K Sep 25 '14 at 5:27
1
$\begingroup$

What is the underlying reason for having infinitely many digits following the decimal point, but not infinitely many digits left of the decimal point?

The underlying reason is that real numbers can have infinite precision, but only finite size. You can find larger and larger real numbers, but each of them has a finite size. You can have a real number with a gazillion digits left of the decimal point, which is a very, very, very large number, but it's still a finite number.

To the right of the decimal point, the number of decimals is unlimited because real numbers have infinite precision. Even for rational numbers, you need an infinite number of decimals because a finite number of decimals can only represent a tiny fraction of the rational numbers. For example, 10/7 = 1.428,571,428,571,428,571... needs infinitely many decimals because if you cut of the decimals at any point the result is too small, and if you add 1 to the last digit the result is too large.

But real numbers really need an infinite number of decimals, because after every decimal you can add any other decimal you like and you get different real numbers, and go on forever doing so.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Let me just remark that I find it curious that the question exclusively talks about "numbers", and the answers, while sometimes referring to p-adics as something else to think about, exclusively consider the question to be about "real numbers" and/or measures.

Incidentally, the branch of mathematics called "number theory" is very little concerned with real numbers except as an analytic tool.

When we are talking "n-adics" and take, say, a base of 10, then something like

...666666666666667

can be considered a representation of 1/3 since multiplying it by 3 yields

...000000000000001

It turns out that things work out more systematically if you don't take a base of 10 but rather of a prime number. Either way, one can develop theories around such constructs and rules to work with, and calling "real numbers" numbers but such things not is purely arbitrary.

If you choose to do it that way, "number theory" is barely about numbers any more...

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ A curious feature of the n-adics with prime bases is that they can represent any fraction except those whose denominator would be a multiple of the base. Thus, dynadic numbers can represent any fraction whose denominator is odd, but can't represent any fraction whose denominator is even. $\endgroup$ – supercat Sep 25 '14 at 15:24
  • $\begingroup$ @supercat You can write those numbers with (a finite amount of) digits to the right of the decimal point though. (I think. Is this correct?) $\endgroup$ – Oscar Cunningham Sep 26 '14 at 14:41
  • 1
    $\begingroup$ @OscarCunningham: In binary, the value -1/3 would be "10...101010101.0"; the value 1/3 would be twice that, plus one, or "10...010101011.0". A fraction like 1/6 would be half of that, or "10...1010101.10" (which may be viewed as -1/3 +1/2). I think this approach will allow for all rational numbers, though those whose denominator is a multiple of 2 without being a power of 2 will have an infinite number of digits to the left as well as a non-zero finite number of digits to the right. $\endgroup$ – supercat Sep 26 '14 at 19:27
0
$\begingroup$

Just because a number has an infinite number of digits doesn't mean it isn't well defined...

With a great deal of effort you may even be able to redefine a number with an infinite number of digits into being a finite number...

consider 1+2+3+4+5+6+...=-1/12 which is the solution to riemannZeta(-1)

proof

1-1+1-1+1-......=A
1-A=1-(1-1+1-1+1-1+1-....)=1-1+1-1+1-1+1-....=A
1-A=A
A=1/2

1-2+3-4+5-6+7-....=B
B+B= 1-2+3-4+5-6+7-....
      +1-2+3-4+5-6+7-....
   = 1-1+1-1+1-1+1-......
B+B=A
B=1/4

1+2+3+4+5+6+7+8=C
C-B=1+2+3+4+5+6+7+8
   -1+2-3+4-5+6-7+8-...
   =  4  +8 +12 +16 +...
   =4*(1 +2 +3  +4  +...)
   =4*C
C-B=4*C
C=-B/3
 =-1/4/3
 =-1/12

boya

and as Mathmo123 pointed out numbers with repeating digits may all be considered equal to zero. since ...111111.11111.... *10 = ...11111111.11111.... and since the only solution to x*10=x is zero then all numbers with repeating digits left and right of the decimal place can be considered equal to zero.

there is precedent for this as the riemannZeta function is zero for all negative even integers... i.e. 0=1^2+2^2+3^2+...=1^4+2^4+3^4+...=1^6+2^6+3^6+...

and as supercat has mentioned in several comments, the power-series formula yields -1 for the geometric series 1+2^2+2^3+2^4+2^5+2^6+2^7+... even though it is out side the traditionally defined domain of that formula.

consider what 1+2^1+2^2+2^3+2^4+... is in binary... it happens to be equal to ...11111111111.0 in binary and if ...1111111.1111111.... (all repeating digits) is equal to zero and 0.11111111111 which is equal to one... then

 ...1111111111.111111111...(binary)  0
-...0000000000.111111111...(binary) -1
=...1111111111.000000000...(binary)=-1

which satisfies the power series formula!

Everything works out perfectly!

So in summary: a number with an infinite number of digits to the left of the decimal place can be well defined.

However this does not mean that ALL such numbers are well defined... (although they may all be well defined, I can't prove that all infinite numbers are well defined... it's beyond my current level of skill and time available to prove)

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ What do you mean by "well defined?" I don't think that your definition works. As a side note: By referring to the zeta function, it seems as if you're arguing that divergent infinite series are equal to $0$. But, then you reference the power series formula which is also a divergent series. You must be very careful with divergent series in general, and you can't make many assumptions on what this means with a "standard" interpretation of the sum. $\endgroup$ – apnorton Sep 26 '14 at 12:04
  • $\begingroup$ Your second highlighted line should say that "0000.1111" is +1 rather than -1. Note that Boolean operators could be defined on real numbers using the above principles if one asserts that all negative numbers should be regarded as having an infinite string of "1"'s on both sides and all positive numbers with an infinite string of zeroes [under such rules, -1 would be regarded not as ...1111.0000... but ...1110.1111....]. $\endgroup$ – supercat Sep 26 '14 at 19:32
-4
$\begingroup$

It's a matter of semantics and definition. Technically, since integers are a subset of real numbers and there are an infinite number of them, you can have an infinite number of digits; however, for the sake of defining properties, and defining numeric fields, you limit the number of digits to n... which can be as large as you like.

So, yes, you CAN have an infinite number of digits - because you can always make your number larger by multiplying it by 10, say - but, for the sake of definition and math theory, it is NOT possible, by nature of its definition.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ There could be indefinite many numbers before he decimal point but infinite numbers only if all but a finite number of them are zero. $\endgroup$ – Lehs Sep 24 '14 at 19:24

Not the answer you're looking for? Browse other questions tagged or ask your own question.