lpetrich

If you have a number like

0.273240653682303929829016482931...

there is no way to tell whether it is rational or irrational unless you have some procedure for finding the rest of the digits. However, what DNAunion might be thinking of is what is the minimum amount of algorithm complexity necessary for generating a number that looks like this.

The algorithm would use integers, elementary arithmetic operations, and control statements (if-then-else, looping). This will get all rational numbers; to get all irrational ones, infinite limits must also be introduced.

There will be a complexity penalty for large integers, because using sufficiently large ones can generate any rational number with at most one additional step.

Infinite limits are necessary for doing infinite series, sum-to-integral limits, successive-approximation equation-solving limits, and other ways of generating irrational numbers.

Thus, if one has a number

0.11111111111....

the number 1/9 is a low-complexity representation of it.

So what is the lowest-complexity representation of DNAunion's Number? At least relative to some weighting of algorithm elements.

lpetrich

As to the connection between rational numbers and infinite repeats, an infinitely-repeating number representation in base b is essentially:

x = n0*b^(-p0) + n1*b^(-p0-p1) + n1*b^(-p0-2*p1) + n1*b^(-p0-3*p1) + ...

where n0 and n1 are integers and p0 and p1 are nonnegative integers.

This can be collapsed to

x = n/(b^(p0-p1)*(b^p1 - 1))

where n is an integer.

(I hope all this ASCII math notation is not too difficult to follow)

But does the converse hold? Does every rational number have infinitely-repeating trailing digits?

This is equvalent to successfully anwering the following question: if one has an integer d that is relatively prime to b, are there integers c and p such that

c*d = b^p - 1

?

If so, one can express a rational number as

x = n0/d0 [n0 and d0 both integers]

= n/(b^p01*d) [d relatively prime to b, p01 an integer]

= n*c/(b^p01*c*d) [so c*d = b^p1 - 1]

= n*c/(b^p01*(b^p1-1))

I will try to come up with a proof in my next posting.

(Edited to correct a malapropism)

[ December 03, 2002: Message edited by: lpetrich ]</p>

Principia

Good contribution, lpetrich! In fact, the argument hinges on the well known result from number theory known as <a href="http://mathworld.wolfram.com/EulersTheorem.html" target="_blank">Euler's Theorem</a>, of which one way to state it is: given b, d relatively prime, there always exists a positive integer p such that b^p == 1 mod d. But, I won't steal lpetrich's thunder here. A good proof is always fun to read.

[ December 03, 2002: Message edited by: Principia ]</p>

lpetrich

Thanx, Principia, that was the theorem I was looking for.

Also, it can easily be shown that there is no number-system base for which an irrational number's representation terminates. If it terminates, then that number can be represented in the form

n/b^p

where n and p are integers and b is the base, necessarily a positive integer. Meaning that it would then be rational.

And if it has an infinite repeat of some of its trailing digits, then that also implies rationality.

Euler's theorem states that, for b and n relatively prime to each other, b^p - 1 is divisible by n if p is Euler's Totient Function of n, the count of all positive integers relatively prime to n.

This can be proved using some simple Group Theory.

Let's call the defining condition C: a number b must satisfy b^p = 1 mod n

Consider two numbers b1 and b2 satisfying C. It is easy to show that their product also satisfies C. Thus, the set of all b's, B, is closed under multiplication mod n: B*B = B

The set B has a multiplicative identity, 1, which trivially satisfies C, and an inverse for each member. Multiplying the members of B by one of its members yields B again, since there are no distinct numbers b1 and b2 such that b*b1 = b*b2 for some b. And these members include 1, thus implying the existence of an inverse.

Thus, B and multiplication mod n form a group.

Every member b of B will have a certain order, the smallest power r of b where b^r = 1 mod n. The set of {1, b, b^2, ..., b^r-1} and this multiplication form a "cyclic group" of order (number of set members) r. This will be a subgroup of B's group, and by Lagrange's Theorem, the number of set members in that cyclic group will evenly divide the number of set members in B's group.

This means that b^(size of B) = 1 mod n -- and since p = size of B, that proves Euler's Theorem.

lpetrich

As to whether pi is irrational, here is <a href="http://pi314.at/math/irrational.html" target="_blank">an outline of Ivan Niven's proof that pi is irrational</a>. It is a proof by contradiction that uses some of the properties of the trigonometric functions; it should not be difficult for those with some understanding of calculus.

DNAunion

DNAunion: Well, I guess that's about as close as a confirmation as I am going to get from this group: I'll take it (along with the confirmation from the several math sources I quoted earlier, of course).

DNAunion: No, non-rational isn't a definition of irrational - it's just a tautology.

That is, unless you are implying that the only definition of a rational number is a number that can be expressed as the ratio of two integers. But that's not the case. A rational number can also be defined as one whose decimal representation either terminates or repeats.

1) A number is rational if and only if its decimal representation either terminates or repeats.

That is just as strong and just as valid as:

2) A number is rational if and only if it can be represented as a ratio of two integers (with the denominator not being equal to 0)

Anyone who believes differently is free to present a counterexample: a single real number for which (2) holds but (1) does not.

Since both are exact definitions of a rational number, then each one is equal to the other (basically, if A = C and B = C, then A = B).

So...

1) If a number can be written as the ratio of two integers, then its decimal representation will either terminate or repeat.

2) If a number has a decimal representation that either terminates or repeats, then it can be written as the ratio of two integers.

or combining those:

3) A number can be written as the ratio of two integers if and only if its decimal representation either terminates or repeats.

Okay people, do we agree on that much, or not?

DNAunion: To show not rational seems to me to be the same problem either way - definitively demonstrating a negative.

Here's my take on the counters.

Although I can show that a decimal representation that does repeat DOES repeat (for example, 0.3333333.... or 0.666666.....), I cannot actually demonstrate that a decimal representation that doesn't repeat does NOT repeat.

I don't see a material difference here. That is, someone can show that a number that can be represented as a fraction of two integers (for example, 1/3 or 2/3) CAN be represented as such, but can they actually show that a number that cannot be represented as such, CAN'T be?

I don't see how definitively demonstrating either negative is possible.

I guess that's my hang up. Now, if someone could explain how a single given number can be known for sure to NOT be representable as a fraction of two integers, while also NOT being able to tell that it has a decimal representation that neither terminates nor repeats, I would appreciate it.

Oh yeah, and I didn't major or minor in math, so if you are going to communicate with me, you need to keep it at or below the college algebra level - no calculus, for example.

Oh, Principia, you're not invited to play. You're too rude and beligerant.

[ December 03, 2002: Message edited by: DNAunion ]</p>

Principia

Fine by me. Only so far I can take a crystal clear explanation. I can't help it if you don't try to understand.
Pot. Kettle. Black.

[ December 03, 2002: Message edited by: Principia ]</p>

DNAunion

DNAunion: Talk about logical fallacies. No wait, what Principia actually put forward was an underhanded tactic frowned upon by all sides.

DNAunion: Uhm, where did I claim I could show that? Nowhere. So why should I, according to your implications, be forced to try to demonstrate that I could? In fact, I myself made it clear that NEITHER method could be used to determine if the number was irrational. So again, why should I, according to your implications, be forced to try to demonstrate that I could? And, uhm, where did I even claim that number was irrational? Nowhere.

So what in the world is Principia really refuting? NOTHING I SAID! He's just beating up on the strawman version of my position that he concocted.

Now, what is really funny is that Principia himself was unable to use "his" preferred definition to show what he wants me to show about that number - that that number is irrational. That is, he was unable to show whether or not it could be written as a fraction of two integers. He missed the very point that was being made. NEITHER DEFINITION COULD BE USED TO ACTUALLY DETERMINE WHETHER OR NOT THE NUMBER WAS IRRATIONAL. How could he have missed that? Especially after I spelled it out clearly just a few posts ago.

[ December 03, 2002: Message edited by: DNAunion ]</p>

Principia

Yawn. That was the 14th edit, DNAunion made in his attempt to save face above. I wonder when he'll get it right?

DNAunion

DNAunion: And let’s look at more of Principia’s childish antics.

DNAunion: So show us legitimate material from multiple, mainstream mathematics books that give that definition for that term. Unless you can, that definition is NOTHING LIKE the definition for an irrational number I gave.

DNAunion: Your asinine example is NOT analogous to what I stated. That makes your totally worthless (except as an insult, which is obviously what you were angling for).

First, somehow, you seem to still miss the specific point that was being made: my thought experiment showed that NEITHER “my definition” NOR “your definition” could be used for that number. That countered (at least in that example and other similar ones) your dismissal of my definition because non-repeating couldn’t actually be demonstrated. I did make a point, you’re just too ____ to get it, even after I explained it multiple times.

Second, let’s see just how silly your little game is.

******************************
Principia: A stupidonumber is one which does not contain 1234 anywhere in its fractional representation.

I am thinking of a stupidonumber, and I tell you that the denominator does not contain 1234 any in it.

Now, you all say -- so what, prove to us that it is in fact a stupidonumber using only Principia's definition... See, there is a reason why Principia-type 'definitions' aren't really valid definitions at all.
*******************************

DNAunion: So I guess I just refuted your definition of an irrational number, heh?!?!?!
It's all really silly, isn’t it.

DNAunion: So show us a dictionary and multiple other sources that state that definition for that term. If you can’t, then your example is NOTHING LIKE the definition I gave.

Especially since yours is a self-contradiction: something that mine definitely was not.

So two valid reasons for rejecting your above “refutation” of me as being nothing more than a _____ showing his true colors once again.

**************************************

DNAunion: Finally, something that MIGHT be correct.

However, you or someone else first needs to demonstrate what I asked you people to demonstrate just two or so posts above: basically, demonstrate that “your” definition doesn’t suffer from the same problem of trying to definitively demonstrate a negative in order to determine that a number is irrational.

As soon as someone does that, then I will accept that “my definition” is not useful in determining that a number is irrational, but that the other is.

Until that time, I don't see any real limitation "my definition" faces that the other definition doesn't face as well.

[ December 03, 2002: Message edited by: DNAunion ]</p>