To infinity and beyond

[Qfwfq]

Many of the most important advances in mathematics have been possible due to mathematicians not being overly rigid and revising details in order for a new context and framework to include a familiar one; this goes also for advances with applications in the physical sciences, engineering &c. In some cases it is necessary to keep more separation, in other cases it is just as well to let the same familiar old term have a meaning which is slightly more extended in the wider context and equivalent one by restriction.

[Ben]

OK, I might be willing to open a new thread on geometry, where these things can be discussed. (As it's Friday night here, insobriety may intervene, but let's see)

 

Meanwhile, back at the ranch. I showed you Cantor's proof that there may be "different sizes" of infinity, countable and uncountable. It's called his diagonal argument (that's why I used [math]D[/math] for my key subset). I trust this is obvious.

 

He also used the same argument to show that the set of all real numbers [math]\mathbb{R}[/math] is uncountable. But it gets worse......

 

Consider the real interval [math][0,1][/math] that is all the real numbers between and including 0 and 1. What is the smallest non-zero element in this interval? Say what, we have no way of knowing, since for any eany-weany non-zero real number [math]\epsilon[/math], I can always find, say [math] \frac{\epsilon}{2}[/math] as a non-zero number.

 

Likewise for the largest number less than 1. This may strike you as ugly ugly, but it's true.

 

This forces the conclusion that there are uncountably many real numbers in the interval [math][0,1][/math], exactly the same as there are in the entire real line [math]R^1[/math]. This is counter-intuitive, but also true.

 

It has its uses....

 

So, when Q says, for example, that in the limit that some real [math]x[/math] tends to zero then [math]\frac{1}{x}[/math] tends to infinity, we can breathe a sigh of relief, knowing that [math]x[/math] will never actually get to zero, at least not in Mankind's lifetime.

 

"Bad math" is thereby avoided. So infinity, especially of the uncountable variety, definitely has its uses.

[HydrogenBond]

Say we drew two concentric circles in a plane. We can defines points where the angles will add to 180 degrees. However, since the outer circle is larger it will define more points than the inside circle. Will these still be parallel?

 

The reason I ask this, some theories suggest if we tried to head toward infinity we would curve and return to where we began. This would cause the parallel lines to become concentric circles. On work-around is to have the parallel lines spiral like a helix so they can share the inside and outside paths, maintaining the point to point count.

[Ben]

HB: I thought it had been agreed that we should leave geometry for a separate thread. True, I offered to start one, but I'm begining to feel uneasy about starting up so many new threads.

 

Anyway, geometry aside, here's another example of how weird infinity can be.

 

The set of symbols - letters - that make up an alphabet is finite, and therefore countable. In our case it has cardinality 26.

 

Suppose we make use of this fact to enumerate these, and, for reasons I will explain shortly, designate as follows: A = 10, B = 20, C = 30 etc.

 

Then a word will be of the form 2050140 = BEN. This is clearly a number, i.e. an element in [math]\mathbb{N}[/math]. Now I can stick these words together to make a sentence, sentences to make a paragraph, paragraphs to make a book, etc, and I will still have an element in [math]\mathbb{N}[/math].

 

OK, I simply used zero as a delimiter to avoid under-counting, that is, I know that 1020 = AB and not L = 120, which could happen if 12 = AB or L.

 

So we see there is no word, sentence, paragraph, library, or even collection of libraries (Congress of Libraries??) that cannot be expressed as an element in the countable set [math]\mathbb{N}[/math]

 

But, since the set [math]\mathbb{R}[/math] of real numbers is uncountable, that means there will be uncountably many elements that cannot be given a name using words or sentences or paragraphs, or....

 

Yikes! But it gets worse.

 

Suppose as before that [math]\mathcal{P}(\mathbb{N})[/math] is the powerset on the natural numbers. Recall by my OP this is uncountable.

 

If, for each element of the powerset there is at least one true statement (using words, sentences, paragraphs,....) in some theory of numbers, then there will be uncountably many true statements of number theory. Therefore, there are uncountably many more true statements of number theory than there are words, sentences, paragraphs, books or libraries (since these are countable).

 

It follows that any system that tries to describe number theory as a complete set of true statements is doomed to failure.

 

This is version of an incompleteness theorem.

[C1ay]

True, I offered to start one, but I'm begining to feel uneasy about starting up so many new threads.

 

Feel free to start any threads you like on the topics we discuss here...

[lemit]

If two threads are extended in lines parallel to each other . . . .

 

Sorry.

 

--lemit

[Qfwfq]

If two threads are extended in lines parallel to each other . . . .

Yup that beer was strong stuff, wasn't it? And the bill, of course, is on Ben!!! :(

 

Thread proceeding well now, as planned by Author. :hihi: Popcorn... popcorn... get your popcorn here........

[Ben]

Ya know what? I loathe popcorn, salted or sugared......

 

Anyway, some of you be surprised to learn I don't posses a calculator, so can someone confirm (or otherwise) the following, which I shall need to proceed:

 

For any [math]n,\,\,k \in \mathbb{N}[/math], there is NO rational number that cannot resolved as [math]\frac{k}{2^n}[/math]?

 

I can do a few of these in my head, and a proof would obviously be ideal, but going to a few values of n and k would be a great help.

 

Stuck for now

 

*blush*

[Qfwfq]

For any [math]n,\,\,k \in \mathbb{N}[/math], there is NO rational number that cannot resolved as [math]\frac{k}{2^n}[/math]?

If you meant exactly that, I somewhat doubt it on rational... er, formal grounds.

 

If you meant that, for any rational, there will be some pair [math]n,\,\,k \in \mathbb{N}[/math], I still doubt it because not every natural denominator is a power of two. Perhaps you meant some different lookalike? :confused:

[lemit]

So, is infinity an invention to make formulae that don't seem to be working come out all right, or is it, like zero, almost in the nature of a discovery?

 

And, is that as dumb a question as it looks like to me?

 

Thanks.

 

--lemit

[Turtle]

Ya know what? I loathe popcorn, salted or sugared......

 

Anyway, some of you be surprised to learn I don't posses a calculator, so can someone confirm (or otherwise) the following, which I shall need to proceed:

 

For any [math]n,\,\,k \in \mathbb{N}[/math], there is NO rational number that cannot resolved as [math]\frac{k}{2^n}[/math]?

 

I can do a few of these in my head, and a proof would obviously be ideal, but going to a few values of n and k would be a great help.

 

Stuck for now

 

*blush*

 

if you use the definition of [math] \mathbb{N}[/math], a Natural number, that includes 0 (zero), then when n=0, [math]\frac{k}{2^n}[/math] is rational for all values of k because 2⁰ = 1 and the definition of rational numbers only excludes 0 as a denominator.

 

there is NO rational number that cannot [be] resolved as [math]\frac{k}{2^n}[/math]

do you mean by that, "cannot have a solution in"? , or "cannot be re-written as" ?

[Turtle]

So, is infinity an invention to make formulae that don't seem to be working come out all right, or is it, like zero, almost in the nature of a discovery?

 

And, is that as dumb a question as it looks like to me?

 

Thanks.

 

--lemit

 

we have a whole thread arguing whether math is discovered or invented, so let's just accept that "it is" & get on our way. :hihi:

 

zero is a number, infinity is not. the title of this thread is wrong on one account because it implies infinity is a place/destination/value, and wronger again in that it is redundant inasmuch as infinity is defined/understood by mathematicians as "growth without bound" and you can't go beyond a bound that isn't there.

 

infinity is a useful concept-tool for shortening-the-time-required/speeding-up certain mathematical calculations, much as a saw is a useful tool-concept for a carpenter to use to speed-up/shorten-the-time-required cutting some wood. which is to say that if you're not a mathematician or a carpenter you don't need either infinity or a saw. :D :D

[Jay-qu]

I wonder in the opposite is true: ie k/(2^n) is never irrational

 

Intuitively this makes sense, for you can aways express this as

[math]k \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} ... [/math]

 

And for any terminating decimal number (rational) you can always continually double it until it becomes an integer. If you dont believe me grab a calculator and try it, punch in an arbitrary decimal and keep hitting x2x2x2x2.... and you will get an integer (providing you have enough decimals on your calc..).

 

So perhaps there is an analytical proof of this?

[Qfwfq]

I wonder in the opposite is true: ie k/(2^n) is never irrational

This is obvious by construction, it's the ratio of two naturals.

 

And for any terminating decimal number (rational) you can always continually double it until it becomes an integer. If you dont believe me grab a calculator and try it, punch in an arbitrary decimal and keep hitting x2x2x2x2.... and you will get an integer (providing you have enough decimals on your calc..).

 

So perhaps there is an analytical proof of this?

:umno:

 

If [imath]d\in\mathbb{N}[/imath] has prime factors other than 2 then doubling will not make the ratio integer. If you don't believe me grab a calculator and try it, but punch in 0.2 and keep on doubling it as long as you like.

 

So, is infinity an invention to make formulae that don't seem to be working come out all right, or is it, like zero, almost in the nature of a discovery?

A couple of formal definitions have made some great advances in modern mathematics possible. The one Ben is discussing (infinite cardinality) pertains to set theory. The other one pertains to calculus (more complicated) and is what Turtle hints at, being sometimes described as increase without bound. In modern theoretical physics there are some inevitable infinities, which it took a humongous effort to make head or tail of (renormalization).

[Ben]

Well, I'll leave the philosophizing to others; I am just a simple mechanic.

 

So. I must have had my head on upside down last night. As Q points out, [math]\frac{k}{2^n}[/math] is essentially by definition rational. But any arbitrary rational can only rarely be reduced to this form using the cancellation law.

 

For, if [math]\frac{a}{b}[/math] is rational, then it is reducible iff there exists [math]c[/math] such that [math]a = xc[/math] AND [math] b = yc \Rightarrow \frac{a}{b} \mapsto \frac{x}{y}[/math].

 

So if either numerator or denominator are prime, say, this cannot be.

 

But, even though I don't like his arithmetic approach, Jay gave me an idea on how to proceed.

 

I said earlier that the real interval [math][0,1][/math] is exactly the same size (same cardinality) as the entire real line [math]R^1[/math]. Our intuition rebels (at least mine does); surely [math]R^1[/math] is "longer" than [math][0,1][/math]? Of course, length makes no sense here, but we can find an analogue, but you will see I fail spectacularly to reassure you....

 

Suppose [math]S[/math] is a set. One says that a measure [math]m[/math] is defined on this set iff [math]m(S) \ge 0[/math] where we assume that [math]m(S)[/math] is a non-negative integer. I will also insist that for the empty set [math] m(\O) = 0[/math], but perhaps not conversely.

 

Since [math]m(S)[/math] is integer, then I may have that [math]m(S) + m(T) = m(S \cup T[/math]. (Oh... notice that both S and T are subsets of [math]S \cup T[/math]).

 

I further notice that strict additivity fails when [math] S \cap T \ne \O[/math], so write [math]m(S \cup T) = m(S) + m(T)- m(S \cap T)[/math]. This is the crucial point, so don't go forgetting it!

 

So consider those rationals in the set [math][0,1][/math]; let's call these as the set [math]A[/math]. Notice that this set is countable (since the quotient of a countable set by a countable set is of necessity countable - proof, anyone?)

 

Now for any [math]a_n \in A[/math] and for some [math]\epsilon >0[/math], we will insist it is contained in an interval of measure [math] \frac{\epsilon}{2^n}[/math]. So 1 is contained in an interval of measure [math]\epsilon[/math], and [imath] \frac{1}{2}[/imath] in an interval of measure [imath] \frac{1}{2}\epsilon[/imath] etc.

 

By a theorem of Dedekind (I think), we know the rationals are dense in the reals, which means that most of these intervals overlap i.e. intersect

 

Now, the sum [math]\sum \frac{1}{2^n} \mapsto 2[/math] as n [math] n \mapsto \infty[/math]. Recalling that we have [math]m(S \cup T) = m(S)

+ m(T)- m(S \cap T)[/math], then [math] m(A) = \sum \frac{\epsilon}{2^n} < 2\epsilon[/math].

 

But since I can make [math]\epsilon [/math] as small as I like and then smaller still, the only number definitely smaller is zero.

 

Eek! The rationals are a set of measure ("length") zero.

[coldcreation]

zero is a number, infinity is not. the title of this thread is wrong on one account because it implies infinity is a place/destination/value, and wronger again in that it is redundant inasmuch as infinity is defined/understood by mathematicians as "growth without bound" and you can't go beyond a bound that isn't there.

 

Good point. Ben, though, mentioned above somewhere that he chose the title to attract attention to be slightly light-hearted.:)

 

It's also, of course, a famous quote form Buzz Lightyear (Toy Story, 1995, Walt Disney Pictures, Pixar Animation Studios).

 

 

CC

[Jay-qu]

This is obvious by construction, it's the ratio of two naturals.

If [imath]d\in\mathbb{N}[/imath] has prime factors other than 2 then doubling will not make the ratio integer. If you don't believe me grab a calculator and try it, but punch in 0.2 and keep on doubling it as long as you like.

 

:doh: :doh:

 

So its always rational, but cant make all rationals. Thanks for putting it straight Q