Principia

You presume to think that people care what you will accept. The arguments that you seek are all there. But, maybe... just maybe it will take someone with more than a college algebra education to get it. You know, In Christ Dougie just might be able to help you out here. No point, after you've shown your true colors, for anyone to entertain you further. And sorry, I'm not playing.

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

wade-w

DNAunion:

I guess I should have been more clear in my answer to your question about the definition of irrational numbers, instead of giving a simple "no". I think the difference here really goes more towards mathematical philosophy, and what constitutes a definition.

The definition you are using is logically equivalent to the usual definition, and as such can be used as a definition if one really wants to. But as Claudia has so clearly explained, it is, at the end of the day, an essentially useless definition.

And it is easy to show that, for example, sqrt (2) is irrational without ever refering to decimal representation.

As for the demonstration you ask for, that requires the calculus you say I can't use in my argument.

Wizardry

DNAunion

If you have your heart set on it, I suppose you can define an irrational number as a number whose decimal representation neither terminates nor repeats, since it can be proven that there is a one to one correpsondence between those numbers whose decimal representation neither terminates nor repeats and those numbers and those numbers that cannot be represented as the ratio of two integers. In other words, it's the same set, so in practical terms it really doesn't matter which definition you use. If you prove that one condition applies, then so does the other. In all cases.

However, in more rigorous contexts, it makes more sense to define a rational number as a number that can be expressed in the form p/q where p and q are integers, and then logically that irrational numbers are those that cannot be represented in that form. There are a couple reasons that mathematicians prefer this definition.

First, because it turns out to be one helluva lot easier to prove that there exists some contradiction in representing a number in the form p/q than to prove that the algorithm necessary to generate the decimal representation of a given number results in a pattern that neither terminates nor repeats.

Secondly, because historically, fractions have been in use far longer than have decimal numbers. The Greeks used fractions, but the decimal representation didn't come into existence until after Fibbonaci AFAIK. Decimals are used mostly for practical purposes, and to get rough estimates. They are rarely used in rigorous contexts.

Thirdly, I don't know about you, but I really can't think of a way to even define a decimal representation without first defining a rational number as a number representable as a fraction. The easiest way I know of to construct the decimal system is to define it as the sum of fractions, where the denominator of said fractions are all integer powers of a common base, which of course presupposes the definition of the rational numbers for all the negative powers. So it makes sense to define irrational numbers as numbers which cannot be expressed as a ratio.

On the first point, you said:

Well, the most famous demonstration of the irrationality of a number would be the proof of the irrationality of the square root of 2, which if I'm not mistaken goes something like this:

Assume that the square root of 2 is a rational number, meaning that it can be expressed in the form p/q.

It follows that p^2/q^2 = 2.

Now, we know from basic fraction math that either p^2 or q^2 must be odd, otherwise we could cancel the common factor 2 ad infinitum until one is odd.

Multiplying by q^2, we get the equation

p^2 = 2q^2

which implies that p^2 is even, which means that p must also be even. If we set p = 2c then we get:

(2c)^2 = 2q^2

which means

4c^2 = 2q^2

dividing by two we get

2c^2 = q^2

which implies that q^2 is even.

So p^2 is even and q^2 is even, which we said before wasn't true. Thus we arrive at a contradiction, proving that sqrt(2) is irrational.

Of course we used the fraction definition rather than the decimal one. The proof of the irrationality of pi given earlier on this thread uses a similar definition. It's just far easier to work with.

The link provided earlier is a proof of the irrationality of pi. <a href="http://www.mathpages.com/home/kmath400.htm" target="_blank">Here's one that demonstrates the irrationality of e</a>, again using the fractional definition.

Fact of the matter is that there is no algorithm, of which I am aware, that calculates pi or e in terms of its decimal representation, in the sense that it calculates the nth digit before the n+1th digit, making it rather difficult to determine whether such an algorithm produces a repeating pattern. Not impossible, but why hurt yourself?

There are certain contexts where using the decimal property is more useful, say if the number is defined in terms of its decimal representation

0.101001000100001000001000000100000001......

where there exists a clear pattern that can be shown not to repeat, but those are rare instances, of course.

wade-w

Excellent post, Wizardry. Thats exactly what I was trying to say in my last.

DNAunion, I suggest you read what Wizardry and Claudia posted carefully.

lpetrich

Furthermore, a decimal representation is specific to base 10, and essentially the same thing is true for other number-system bases. As pointed out earlier, fractional representations are more general.

DNAunion

DNAunion: Thanks Wizardry, I was able to easily follow all of that (see, y'all didn't need no calculus nohows).

So yes, NOW I (finally!) see that the fractional definition can in fact be useful in cases where the decimal definition is not, making the former a more useful definition than the latter.

Shake

I was always of the mind that n/0 was undefined, not equal to infinity. But you can talk about n/x approaching infinity as x gets closer to 0. This, of course, is the idea of limits.

lim n/x = inf
x=&gt;0

for any non-zero n.

Since someone brought up pi, I am reminded that pi/4 can be expressed as the infinite sum:

inf
sum ((-1)^n)/(2n+1)
n=0

wade-w

I debated for a while on whether to answer this, but I can't leave it alone.

Sorry Shake, but I have to disagree. Only a technicality, but that's math for you

lim n/x is undefined as well.
x-&gt;0

To see why, consider the left and right limits. From the left, this limit approaches - inf, and from the right, it approaches + inf. Unless of course n is negative, in which case the signs are simply reversed. Either way, the left and right limits do not agree, and thus the limit does not exist.

That said, if you only consider the magnitude of the values you are conceptually correct. i.e.
lim |n/x| does become arbitrarily large.
x-&gt;0

I dislike saying that this limit is equal to infinity, since infinity is a concept, not a number. A better why to interpret the symbols in this case is that as x becomes arbitrarily close to 0, |n/x| becomes arbitrarily large.

Edited to add: I am aware that many textbooks and websites will have equations of the form:

lim f(x) = inf.
x-&gt;a

What this really means is that f has an asymptote at a.

I am talking about how to read these symbols, so please, lets not start any more flame wars!

[ December 06, 2002: Message edited by: wade-w ]</p>

Thomas Metcalf

This is not exactly relevant, but it's interesting.

A consequence of infinity theory is that there are some irrational numbers that are forever inaccessible to us. The number of real numbers is greater than the number of sentences that we can have in our language, so some numbers must be indescribable to us. We'll never be able to represent them with a sentence, and I think a case can thereby be made that we'll never be able to know of them.

I can explain more if anyone's interested.

Jesse

Principia:
BTW, we all know that you are trying to have people lose track of your original (laughable) question, which has already been answered to great detail (and your great embarrassment, if I might add)

DNAunion:
No, only a stupid person would think that (so I hope you are speaking only for yourself, and not for the others that you include in your statement).

Speaking as a moderator: please try to hold back on these sorts of personal comments, guys.