Rational numbers == infinitely repeating sequences of digits

[lpetrich]

A feature of rational numbers is that their decimal representations always have infinitely repeating sequences of digits. This is true not only for base 10, but also for every possible base of a place system.

Examples:

1 = 1.000000... = 0.999999...
1/3 = 0.333333...
1/6 = 0.166666...
1/7 = 0.142857142857142857...
1/9 = 0.111111...

In base 2:
1/3 (11) = 0.01010101...
1/5 (101) = 0.001100110011...

In base 3:
1/2 = 0.11111111...

That is why decimal representations are usually not exactly translated into floating-point binary representations. One only has a finite number of digits available, and one has to cut off an infinite sequence of them.


It is easy to prove that an infinite repeating sequence of digits gives a rational number.

It is more difficult to prove that every rational number can be represented with an infinite repeating sequence of digits, but it can be done.

 

[SLD]

A feature of rational numbers is that their decimal representations always have infinitely repeating sequences of digits. This is true not only for base 10, but also for every possible base of a place system.

Examples:

1 = 1.000000... = 0.999999...
1/3 = 0.333333...
1/6 = 0.166666...
1/7 = 0.142857142857142857...
1/9 = 0.111111...

In base 2:
1/3 (11) = 0.01010101...
1/5 (101) = 0.001100110011...

In base 3:
1/2 = 0.11111111...

That is why decimal representations are usually not exactly translated into floating-point binary representations. One only has a finite number of digits available, and one has to cut off an infinite sequence of them.


It is easy to prove that an infinite repeating sequence of digits gives a rational number.

It is more difficult to prove that every rational number can be represented with an infinite repeating sequence of digits, but it can be done.

Why is .99999999... always 1? Is there any proof? I would think that it is actually the real number closest to 1. But would it be rational?

In broader terms, if any two real numbers are next to each other, wouldn’t the decimal expansion of the smaller one repeat forever with 9's? At least at some point?

Which brings me to even a larger point. If numbers are infinite, why can’t we come up with two arbitrarily large numbers and divide them to come up with pi?

SLD

 

[untermensche]

Why does any of this matter?

Numbers are arbitrary constructs and so are operations.

The things that come out of these arbitrary schemes are just random noise.

 

[SLD]

One more point about Pi. If Pi has infinite digits, then it should have an infinite number combinations of digits. At some point wouldn’t it then be repeating the entire sequence up to that point? And for say two trillion times in a row? Granted that’s not infinite, but for all practical purposes if we discovered that point, we’d be mightily confused.

 

[lpetrich]

Why is .99999999... always 1? Is there any proof? I would think that it is actually the real number closest to 1. But would it be rational?

Yes there is a proof. Let us consider n digits past the decimal point. Its value is 9/10 + 9/10^2 + ... + 9/10^n. It is rather obvious that we have a geometric series, and the sum of a geometric series is
\(\sum_{k=0}^n a^k = \frac{1 - a^{n+1}}{1 - a}\)
This result is easy to prove by mathematical induction.

Applying it to this problem, I find 9/10 * (1 - 1/10^(n+1)) / (1 - 1/10) = 1 - 1/10^(n+1)

As n -> infinity, this sum tends to 1, and thus, 0.999999... = 1.

This proof is easily generalized to other number bases. Base-2 0.111111... = 1, base-3 0.222222... = 1, etc.

In broader terms, if any two real numbers are next to each other, wouldn’t the decimal expansion of the smaller one repeat forever with 9's? At least at some point?

There is no such thing as two real numbers being next to each other, because if they are different, then one can easily find a number between them, their mean.

Which brings me to even a larger point. If numbers are infinite, why can’t we come up with two arbitrarily large numbers and divide them to come up with pi?

No, because that it is not possible. In fact, that is what makes an irrational number irrational. Proving Pi is Irrational: a step-by-step guide to a “simple proof” requiring only high school calculus – Mind Your Decisions  Proof that π is irrational

One can get arbitrarily close to pi by using rational numbers, but one cannot get pi itself.

It is not just pi that is irrational. The square root of 2 was discovered to be irrational almost 2,500 years ago, and there is a simple proof of it being irrational.

 

[lpetrich]

Why does any of this matter?

Numbers are arbitrary constructs and so are operations.

The things that come out of these arbitrary schemes are just random noise.

untermensche, why do you think that numbers are arbitrary? Do you think that 2 + 2 = 5 is just as valid as 2 + 2 = 4? The truth of 2 + 2 = 4 and the falsehood of 2 + 2 = 5 arise from the definitions of 2, 4, 5, addition, and equality of numbers.

 

[untermensche]

Why does any of this matter?

Numbers are arbitrary constructs and so are operations.

The things that come out of these arbitrary schemes are just random noise.

untermensche, why do you think that numbers are arbitrary? Do you think that 2 + 2 = 5 is just as valid as 2 + 2 = 4? The truth of 2 + 2 = 4 and the falsehood of 2 + 2 = 5 arise from the definitions of 2, 4, 5, addition, and equality of numbers.

What is arbitrary is saying 2 = 2.

You have to look at it abstractly and totally ignore that they are not really the same thing. They are in different places. They are separate entities.

Once 2 is defined an 4 is defined and '+' is defined and '=' is defined then 2 + 2 = 4 exists due to definition. The definitions do not allow 2 + 2 to = 5.

 

[Kharakov]

As n -> infinity, this sum tends to 1, and thus, 0.999999... = 1.

I have some questions.

potato) Why is .999... not called irrational form?

 

[lpetrich]

An irrational number is a number that is not a rational number, a ratio of two integers.

Here is a proof that an infinitely repeating sequence of digits makes a rational number:

A number N can be expressed in base B as

N = N0 + N1*B^-k1 + N2*B^-k1-k2 + N2*B^-k1-2k2 + N2*B^-k1-3k2 + N2*B^-k1-4k2 + ...

where N0 is the integer part, N1 has at at most k1 digits, and N2 has at most k2 digits. Since we have a geometric series,

N = N0 + N1*B^-k1 + N2*B^-k1 / (B^k2 - 1)

Which is, of course, a rational number.


The proof in the opposite direction is more difficult. It involves  Euler's theorem, also called the Fermat-Euler theorem or Euler's totient theorem. It states that, for relatively prime a and n,

a^phi = 1 mod n

where phi is Euler's totient function, the count of all positive integers less than n that are relatively prime to n. phi(1) = 1, phi(p) = p-1 for prime p, phi(p^m) = p^m-1*(p-1), and phi(a*b) = phi(a)*phi(b) when a and b are relatively prime.

Consider a rational number N/D. Decompose D into two parts: D = D0*D1. Of these, D0 evenly divides B^m for some m, and D1 is relatively prime to B. Now use Euler's totient theorem.
B^phi(D1) = 1 mod D1
or
B^d = K*D1 + 1
where d = phi(D1)

N/D thus becomes N/(D0*D1) = N'/D' where N' = N*(B^m/D0)*K is an integer and D' = B^m*(B^d - 1).

By construction, this is a rational number. Furthermore, this construction gives the length of the repeat or some multiple of it.

So (having an infinitely repeating sequence of digits) <-> (being a rational number)

 

[lpetrich]

One more point about Pi. If Pi has infinite digits, then it should have an infinite number combinations of digits. At some point wouldn’t it then be repeating the entire sequence up to that point? And for say two trillion times in a row? Granted that’s not infinite, but for all practical purposes if we discovered that point, we’d be mightily confused.

Yes, it will have repeats, but it won't have form (N0).(N1)(N2)(N2)(N2)(N2)... Only rational numbers can have that form.

 

[untermensche]

One more point about Pi. If Pi has infinite digits, then it should have an infinite number combinations of digits. At some point wouldn’t it then be repeating the entire sequence up to that point? And for say two trillion times in a row? Granted that’s not infinite, but for all practical purposes if we discovered that point, we’d be mightily confused.

Yes, it will have repeats, but it won't have form (N0).(N1)(N2)(N2)(N2)(N2)... Only rational numbers can have that form.

That form is just a contingent happenstance that arises from arbitrarily defining things.

You define numbers and define functions and something will come out of it.

Even something irrational like a digit that repeats without end.

 

[lpetrich]

Even something irrational like a digit that repeats without end.

Why is that supposed to be irrational?

 

[untermensche]

Even something irrational like a digit that repeats without end.

Why is that supposed to be irrational?

Because it is something that both allegedly exists yet never ends.

 

[lpetrich]

Even something irrational like a digit that repeats without end.

Why is that supposed to be irrational?

Because it is something that both allegedly exists yet never ends.

So there is no such thing as an infinitely long sequence?

 

[untermensche]

Because it is something that both allegedly exists yet never ends.

So there is no such thing as an infinitely long sequence?

I've never seen one.

 

[lpetrich]

Because it is something that both allegedly exists yet never ends.

So there is no such thing as an infinitely long sequence?

I've never seen one.

Our finite minds have ways of comprehending mathematical infinities. These ways are extensions of what we do for large finite sets. We usually don't try to list every element of them, but instead find some rule which generates all of them and no others. Limited imagination is not much of an argument.

(ETA: some of what I had posted here I've moved to the "Infinite Sets" thread, where it belonged)

 

[untermensche]

We usually don't try to list every element of them, but instead find some rule which generates all of them and no others.

I can see extremely finite rules but no infinities.

To see something requires at least some time.

To see infinite elements though can't be done even if you have infinite time to do it in.

Infinite elements is by definition an amount of elements that can never be expressed. The end of them can never be observed.

 

[lpetrich]

We usually don't try to list every element of them, but instead find some rule which generates all of them and no others.

I can see extremely finite rules but no infinities.

It's more like

"The first hundred positive integers"
vs.
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100}

The first description is much shorter than the second one, and by untermensche's argument, there is no such thing as a large finite set, since he seems to consider only the second kind of description a valid description of a set. Once one accepts rules for generating set elements, like "The first hundred positive integers", it is a small step to infinite sets: "All positive integers".

 

[untermensche]

In theory there is a large finite set.

And in theory any finite set can be expressed.

 

[lpetrich]

And in theory any finite set can be expressed.

Even a set with more members than there are elementary particles in the observable Universe? A number which is approximately 10⁸⁶.

So I can write "the first 10^100 positive integers" without having to write them all down, because doing so is a physical impossibility. What is the fundamental difference between "the first 100 positive integers" and "the first 10¹⁰⁰ positive integers"? Or between those two descriptions and "all positive integers"?