[Guest]

Infinity-infinity=???

Tricky.......

[Guest]

I dont know i guess i equate never ending to the left or the infinity-esque counting numbers to never ending to the right or irrational numbers, but i guess that part is hard to prove

[bonanova]

I would argue if you exclude 1 then map every number to its decimal

or rather the other way around (for example .123759869 would be mapped to 123759869 )

shows the numbers between 0 and 1 only have one more item.

Decimal representation is one way to see that the reals are more dense than the rationals.

First, note that finite decimals [1/2 = 0.5] are a special case of infinite repeating decimals: 0.5000...

Then the decimal real/rational distinction is just this:

• Real numbers are represented by infinite decimals.
  
• Rational numbers are represented by infinite decimals that eventually repeat.
  

  • 2 = 2.000... [repeating: rational]
    
  • 1/2 = 0.5000... [repeating: rational]
    
  • 1/3 = 0.333... [repeating: rational]
    
  • pi = 3.1415926536... [not repeating: irrational]
    
  • e = 2.71828459045... [not repeating: irrational]
    

  Why does this difference make the reals [rationals + irrationals] more dense than the rationals?

  For every rational number there is a value of N such that the decimal digits after the N^th place, since they repeat, are determined.

  For each N, there is a "next" rational number. There are holes between them; they are discrete.

  Now since N is arbitrarily large, the "holes" are arbitrarily small.

  So between any two rational numbers there are an infinity of other rational numbers.

  Still, the rationals are countable but the reals are not.

  Why not?

  Consider the rationals:

  Specify a number N. There are a countable number of rational numbers with the first N decimal digits freely specified, but the remaining infinite digits fixed [since they repeat.] Precisely, there are 10^N such rationals. As N increases without bound, the number of rationals also increases without bound. But the rationals remain countable as this happens. At any granularity level [value of N] the rationals are countable, and they remain countable, even as this granularity becomes arbitrarily fine [N becomes arbitrarily large].

  Now consider the reals:

  Again specify a number N, and consider the 10^N values represented by a string of N decimal digits. Recall that for each of these strings there is a unique associated rational number. In contrast, since for reals there is no constraint on the remaining digits, associated with a string of N decimal digits there are an infinite number of reals.

  This shows that you can't even begin to count [map a single counting number to] the reals. Each individual rational number corresponds to an infinity of real numbers. The rationals are infinitely dense, but the reals in some sense are twice-infinitely dense. In any finite interval [a, b] there are an infinite number of rationals, but there are an infinite number of infinite numbers of reals.

  To express this difference, infinities are sorted into cardinality classes, the first two of which are Aleph₀ [the countable cardinality of the integers and the rationals] and Aleph₁ [the uncountable cardinality of the reals]. Aleph is the first letter of the Hebrew alphabet.

[Guest]

Ah what the hell it's such a lovely proof I think I'll write it out, it's one of my favourites

Real numbers can all be expressed as an unending decimal sequence (proof of that would make this a bit tedious). This isn't necessarily unique, two decimal strings can signify the same number, like 0.345999999... = 0.346000000... (9's and 0's repeat indefinitely). But that problem only occurs with infinite strings of 9's or 0's so in the next bit we'll avoid 0's and 9's for that reason. Apart from that, each different decimal sequence signifies a different real number.

Suppose there were a 1 to 1 mapping between counting numbers and real numbers between 0 and 1.

So for example lets say

1 maps to 0.435986596439...

2 maps to 0.988673253523...

3 maps to 0.101982029164...

4 maps to 0.908350786553...

(or something like that)

In order to span the set of real numbers from 0 to 1 we need to have every possible decimal string covered by this mapping.

To prove that such a mapping won't work, let's construct a number which isn't covered. Take the first decimal place of the first number, the second decimal place of the second number, and so on...

1 maps to 0.435986596439...

2 maps to 0.988673253523...

3 maps to 0.101982029164...

4 maps to 0.908350786553...

to create a new number 0.4813...

Then change each digit of that number to a different digit, which is not 0 or 9 (you've got at least 7 choices for each digit)

So we could go for the lowest possible digits, we'd get 0.1121... or increment the digit (avoiding 0 and 9) to get 0.5124..., it doesn't matter

The number we end up with is a real number. Where does that number appear in our mapping? Well, it's not the 1st item, since the 1st digit is different from the 1st item. It's not the 2nd item, since the 2nd digit is different. And so on. The number is not mapped to at all, and since we could construct such a number regardless of what the mapping is, no such mapping exists which spans all the real numbers from 0 to 1.

Another interesting bit of trivia. If you think of the real numbers as points on a line from 0 to 1, the cardinality of those points is exactly the same as the cardinality of points in a square of side 1, or even a cube. Thinking about that in physical terms it's a little counterintuitive...

[Guest]

Consider the rationals:

Specify a number N. There are a countable number of rational numbers with the first N decimal digits freely specified, but the remaining infinite digits fixed [since they repeat.] Precisely, there are 10^N such rationals. As N increases without bound, the number of rationals also increases without bound. But the rationals remain countable as this happens. At any granularity level [value of N] the rationals are countable, and they remain countable, even as this granularity becomes arbitrarily fine [N becomes arbitrarily large].

Now consider the reals:

Again specify a number N, and consider the 10^N values represented by a string of N decimal digits. Recall that for each of these strings there is a unique associated rational number. In contrast, since for reals there is no constraint on the remaining digits, associated with a string of N decimal digits there are an infinite number of reals.

This shows that you can't even begin to count [map a single counting number to] the reals. Each individual rational number corresponds to an infinity of real numbers. The rationals are infinitely dense, but the reals in some sense are twice-infinitely dense. In any finite interval [a, b] there are an infinite number of rationals, but there are an infinite number of infinite numbers of reals.

I disagree with your logic there. If you were to consider rational numbers as fractions you might want to first order them by denominator, then numerator. Since there are infinitely many denominators for any numerator, you'd never get past the first numerator, thereby "proving" them uncountable. But we know they are countable, it's just that we have proposed a counting order which won't work. In order to prove them uncountable we must demonstrate that no counting order will work (hence the proof by contradiction I posted). Same applies to any countably infinite set, obviously. Say if you were to count the integers by doing the even numbers first, you'd never get around to the odds.

[Guest]

anyway to map the or count rather the rational fractions i think you can arrange it in a 2d array each column a denominator, row a numerator. and start in the top left corner and move to the right one then diagonally down to the left. Then return to the top one over and go diagonal... so on Im not sure this is what you were talking about tho.

(i think you might know this and be trying to make a point but...)

yah i realize my previous statement was retarded but left it because every time i explain my last mistake i seem to make another one. I have this amazing ability to learn things really quickly but it goes out the back door about as fast. Anyway Im going to try to explain my mistake in simple terms that i can understand straight forward and you guys can once again correct me.

So we need to map this set to natural numbers. now first you can see that no matter how big a natural number is it cannot be never ending as irrationals are so were doomed to fail from the start.

anyway assume T:N->[0,1] is one to one and more over that (i mean a through B i think this is the right annotation) T(A,B)->all irrationals between (0,1). Take C the largest irrational and D the smallest irrational. As irrationals can be formed by subtracting an irrational from a rational 1-D must be the biggest rational. Then C is such that C+D is rational and C must equal 1-D and the D must equal 1-C.

(i abandon this proof here as i dont think its going anywhere and think i have another)

assume T:N->[0,1] is one to one and more over that T(A,B)->all irrationals between (0,1). Take C the largest irrational and D the smallest irrational.

So D/2 is either less then 0 (impossible) or rational. so a rational is expressed as an Integer over an Integer number which would require an Integer E such that D*E ->R.

Which is a contradiction (irrational times rational)

So there is no smallest D

Now you can do this the same for (1-C)/2+C is rational or greater then 1

which would prove there is no greatest C

So irrationals are infinite in both directions (at least)

but you could also replace 0 and 1 with two variables representing real numbers and prove there are an infinite number of irrationals between any two rationals so there are an infinite set of irrationals between an infinite number of rationals. so definitely not aleph0

I dont think in math terms technically proved it but hey

half way through this I realized this, but now it seems trivial

wait the set of numbers for irrational B such that B+C is Real. is a closed subset. This is pretty much logically follows as it is the Real numbers plus and minus B but it was a realization at the time.

anyway I remember seeing your guys proofs in class but thought id take a stab

[EventHorizon]

anyway to map the or count rather the rational fractions i think you can arrange it in a 2d array each column a denominator, row a numerator. and start in the top left corner and move to the right one then diagonally down to the left. Then return to the top one over and go diagonal... so on Im not sure this is what you were talking about tho.

Yup, that's usual way the rationals are shown to be countable. Here's two interesting things regarding this. If you think of it as simply tuples (ordered pairs of numbers) instead of a numerator and denominator, you can replace the column labelling with the tuple mapped to by the number, and get a way to count 3-tuples (ordered triples). Continuing in this way, you can show all N-tuples are countable for any finite N.

So, now that we know all N-tuples are countable, we can turn the tuple into a function (like division for the initial tuples), and still the cardinality of those numbers created will be Aleph₀.

(e.g., for a 5-tuple (a,b,c,d,e) you could map it to the function (a/b)+(c/d)^-e. The countable set created includes all rationals and many irrationals, but still is not "larger" than natural numbers.)

This essentially shows that Aleph₀ * Aleph₀ = Aleph₀. Something pointed out by Octopuppy when he mentioned the square and cube etc with edges being 1 on a number line all have the same density.

but you could also replace 0 and 1 with two variables representing real numbers and prove there are an infinite number of irrationals between any two rationals so there are an infinite set of irrationals between an infinite number of rationals. so definitely not aleph0

I dont think in math terms technically proved it but hey

Yeah, that didn't prove it. Notice that between any two unequal irrational numbers are infinitely many rationals as well. To see this we'll use the repeating decimal definition of rational numbers that Bonanova used. First, find the first difference in the decimal expansions of the two irrational numbers. Move a few decimal places to the right, then either round up the lower or round down the higher from there (I might need to be more specific here for a solid proof, but you get the idea). You can now tack on the decimal expansion for any rational number, and the result will be a rational number between the two given irrational numbers.

Edited June 19, 2009 by EventHorizon

[bonanova]

I disagree with your logic there. If you were to consider rational numbers as fractions you might want to first order them by denominator, then numerator. Since there are infinitely many denominators for any numerator, you'd never get past the first numerator, thereby "proving" them uncountable. But we know they are countable, it's just that we have proposed a counting order which won't work. In order to prove them uncountable we must demonstrate that no counting order will work (hence the proof by contradiction I posted). Same applies to any countably infinite set, obviously. Say if you were to count the integers by doing the even numbers first, you'd never get around to the odds.

You're right of course.

My statement should have read [still falling short of a proof]

Each individual rational number corresponds to an uncountable infinity of real numbers.

A countable infinity of countable infinities is still countable.

My intent was to show how much more dense the reals are than the rationals.

The reals do not have a vanishing granularity.

Your presentation of Cantor's argument provides the proof.

[Guest]

infinity -1= infinity

infinity -2= infinity

and so on

infinity - infinty = infinty

[hugemonkey]

Easy, we covered this in the first day of zen class

Mu

Isn't that what a sacred Chao says

[Guest]

I dont know i guess i equate never ending to the left or the infinity-esque counting numbers to never ending to the right or irrational numbers, but i guess that part is hard to prove

That was an interesting comment, I only just noticed it and it had me thinking for a bit. The difference is that although there is no limit to the number of digits left of the decimal point, there is no number which has infinitely many. But a number can actually have infinitely many decimal places to the right.

[bonanova]

Previous post can be touched up a bit to form a proof.

Binary representation of the reals can be thought of as selecting subsets of a countably infinite sequence of binary digits: where the"1"'s are the selectors. Cantor's diagonal proof assumes the reals [subsets] are countable and associates integers with the [assumed complete] list. Cantor's diagonals [2's complemented] are seen not to be in the list, so the list is not complete. The assumption of countability is thus negated.

Briefly, countable infinite sets have cardinality Aleph₀.

The subsets of a countably infinite set [in this case the reals] have cardinality 2^Aleph0 = Aleph₁ and are not countable.

e.g. there are 2³ subsets of a set of 3 objects.

In my distinction of the countability of the rationals and reals I showed a 1-1 correspondence of the rationals [having digits that terminate] with the reals that start with the binary representation of a rational and append Aleph₀ binary digits that are freely specified as 0 or 1. Appending those binary digits defines a set of reals with cardinality 2^_Aleph0 = Aleph₁.

Thus each rational can be paired with an uncountable set of reals.

The set of all reals is thus uncountable.

[Guest]

First of all, one must define "infinity", eg- there is an infinte amount of rational numbers, and an infinite amount of complex numbers, but there are more complex numbers than there are rational numbers.

In a strictly mathematical sense, there are many types of infinity.

In another strictly mathematical sense, infinity is not a number. A "number" is a mathematical phenomenon that you can place on a number line. Although any given number line is theoretically infiniyely long, you cannot place infinity on it. And since the definition of subtraction is: lx-zl=distince from "x" to "z" on a number line, you cannot mathematicaly define infinity-infinity. Since infinity doesn't have a value, neither does infinity-infinity. This question is pointless...

[Guest]

Mathematically,

infinity - infinity = indefinite number

The answer can be 0, 1, 2, or even any real number like -9871.2345, for example.

See the link on Infinity Minus Infinity for the proof.

But, the author of this puzzle is asking -

Infinity-infinity=???

Not ∞ - ∞ = ?

So, what is the difference between "Infinity" and "infinity"?

[Guest]

infinity -1= infinity

infinity -2= infinity

and so on

infinity - infinty = infinty

If you are indeed right,

infinity -1= infinity implies infinity - infinity = 1

infinity -2= infinity implies infinity - infinity = 2

and so on

infinity - infinty = infinty implies infinity - infinity = infinity as well.

[Guest]

some one else can explain the more complicated parts of this but infinity isnt a number or variable, its a concept so such formulas do not work. the only thing you can do with infinties is multiplication and division (or orders).

infinity squared /infinity = infinity

infinity/ infinity squared=0

n^infinity>infinity (n>1)and so on

the infinity that is growing the fastest will always overtake the other infinity when you look at a number sufficiently large

[Guest]

infinity,,,,

[bonanova]

some one else can explain the more complicated parts of this but infinity isnt a number or variable, its a concept so such formulas do not work. the only thing you can do with infinties is multiplication and division (or orders).

infinity squared /infinity = infinity

infinity/ infinity squared=0

n^infinity>infinity (n>1)and so on

the infinity that is growing the fastest will always overtake the other infinity when you look at a number sufficiently large

You're probably thinking of L'Hopital's rule where limits of functions are in view.

The OP does not refer to functions or limits, just two references to "infinity."

Since infinity comes in different flavors [cardinalities] it's no so simple as saying infinity²/infinity = infinity.

If the infinity in the denominator is C and in the numerator is [Aleph₀]², then the result is zero.

[dyalDragon]

one way to represent infinity is with the harmonic series, or sum of 1/n for n=0 --> n=infinity

so maybe harmonic series - harmonic series = disharmony??

this seems to fit, especially since disharmony is the main result that i can see from all the differing opinions

Edited July 12, 2009 by dyalDragon