Does a proof or theorem exist for the following.

calcmandan · 1st October 2011, 11:54 PM

A notion crossed my mind the other night about odd and even integers, during dinner actually. It got me curious enough to do some searching on google but came up empty handed. And I spent plenty of time reading through a bunch of existing proofs and theorems before I got tired.

Unfortunately, it's been ages since my proof writing days in college, so here goes.

Proof: In the universe of integers, there are an equal number of odd and even integers.

Is the answer possibly obvious?

I'm just curious.

Daniel

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Vorpal · 1st October 2011, 11:59 PM

"Equal number" just means that you can put them into bijection (one-to-one correspondence, so that every odd number is mapped to exactly one even number and every even number is mapped to by some odd number).
f(n) = n+1 works.

Actually, every infinite subset of the integers has an equal number of members as the integers.

DrDave · 2nd October 2011, 12:01 AM

Yes it's true. See this thread for a discussion on similar proofs http://forums.randi.org/showthread.php?t=219795

In brief, if you can show that for each natural number there is one odd number (or even) you show there are necessarily the same number.

For even numbers the formula is f(x)=2x
For odd numbers the formula is f(x)=2x+1

So we have shown there are infinite of each. To quickly show there are the same number of each.

Drachasor · 2nd October 2011, 12:03 AM

There's an equal number of odd and even integers. this number is equal to the total number of integers as well as the number of primes, perfect squares, etc, etc, etc.

The basic idea is you make a 1-1 ratio with the positive integers.

Code:

1  2  3  4  5  6  7  8  9 10...
1 -1  3 -3  5 -5  7 -7  9 -9...
0  2 -2  4 -4  6 -6  8 -8 10...
2  3  5  7 11 13 17 19 23 29...
1  4  9 16 25 36 49 64 81...

And so forth. It's part of Set Theory.

Tim Thompson · 2nd October 2011, 12:05 AM

I can't give you the formal proof, but I think you will find that the paradoxical answer is this: Yes, there are just as many odd integers as there are even integers, and there are just as many odd or even integers as there are integers total. The reason for this is that the set of all even or odd integers is a countable infinity. In other words, you can put either set of odd or even integers into a one-to-one correspondence with the set of all integers (which is really what you do whenever you count anything).

If you stop short at a finite number, then of course there will always be more integers in your finite set than there are even or odd integers. But if you include the entire infinite sets, then they are the same order of infinity.

I remember learning that when I was a kid in the book One, Two, Three ... Infinity by George Gamow.

calcmandan · 2nd October 2011, 12:10 AM

Quote:

There's an equal number of odd and even integers. this number is equal to the total number of integers as well as the number of primes, perfect squares, etc, etc, etc.

The basic idea is you make a 1-1 ratio with the positive integers.

1 2 3 4 5 6 7 8 9 10...
1 -1 3 -3 5 -5 7 -7 9 -9...
0 2 -2 4 -4 6 -6 8 -8 10...
2 3 5 7 11 13 17 19 23 29...
1 4 9 16 25 36 49 64 81...

And so forth. It's part of Set Theory.

My initial impression was set theory, but I kept second guessing myself. Yall are so FAST at responding!

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calcmandan · 2nd October 2011, 12:15 AM

Quote:

Yes it's true. See this thread for a discussion on similar proofs http://forums.randi.org/showthread.php?t=219795

In brief, if you can show that for each natural number there is one odd number (or even) you show there are necessarily the same number.

For even numbers the formula is f(x)=2x
For odd numbers the formula is f(x)=2x+1

So we have shown there are infinite of each. To quickly show there are the same number of each.

One to one correspondence, phew takes me way back to math 108.

So basically, it's more or less a tautology?

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sol invictus · 2nd October 2011, 01:03 AM

Originally Posted by calcmandan

One to one correspondence, phew takes me way back to math 108.

So basically, it's more or less a tautology?

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I don't think it's a tautology, no. As Tim mentioned, the same techniques can be used to prove that there are as many positive even integers as there are integers, that there are as many primes as integers, that there are as many integers as rational numbers, as many reals between 0 and 1 as all reals, etc. Most people do not find those facts very obvious.

calcmandan · 2nd October 2011, 01:20 AM

Quote:

I don't think it's a tautology, no. As Tim mentioned, the same techniques can be used to prove that there are as many positive even integers as there are integers, that there are as many primes as integers, that there are as many integers as rational numbers, as many reals between 0 and 1 as all reals, etc. Most people do not find those facts very obvious.

Some of those facts are starting to come back to me. I studied math in college but it's been a while, so a lot of it is misfiled in the head. Thanks for your input.

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Acleron · 2nd October 2011, 11:32 PM

Originally Posted by Vorpal

Actually, every infinite subset of the integers has an equal number of members as the integers.

Yes, I particularly like that it can be proved that 100% of integers contain the digit 3.

WhatRoughBeast · 3rd October 2011, 07:13 AM

Originally Posted by Acleron

Yes, I particularly like that it can be proved that 100% of integers contain the digit 3.

No, that's not what he said. Please pay attention.

The number of integers which end in the number 3 is equal to the number of integers. That two sets have the same number of elements does not mean that the two sets are equal. Even if one set is a subset of the other. As long as both sets are infinite.

Acleron · 3rd October 2011, 09:26 AM

Originally Posted by WhatRoughBeast

No, that's not what he said. Please pay attention.

The number of integers which end in the number 3 is equal to the number of integers. That two sets have the same number of elements does not mean that the two sets are equal. Even if one set is a subset of the other. As long as both sets are infinite.

Vorpal said

Quote:

Actually, every infinite subset of the integers has an equal number of members as the integers.

Please tell me where my statement conflicts with either Vorpal's or yours. Except of course where you talk about integers ending in 3, closer reading perhaps.

psionl0 · 3rd October 2011, 09:44 AM

It is pretty pointless to compare the number of elements in the set of even numbers to that in the set of odd numbers because both sets are infinitely large.

Trying to compare one infinitely large number with another is meaningless because infinity is not a number.

The folly of treating infinity like it was a number can be demonstrated in the following example:

What is 1 - 1 + 1 - 1 + 1 - 1 + . . .? It turns out that there are three possible answers depending on how you do the calculation.

The expression could be written this way: (1 - 1) + (1 - 1) + (1 - 1) + . . . = 0
Alternatively, we could write 1 - (1 + 1 - 1 + 1 - 1 + 1 - 1 + . . .) = 1 - 0 = 1
Finally, if the total is X then we could write an algebraic expression: X = 1 - X for which the solution is X = 0.5

We will never know the actual sum because it depends on the last digit in the series and we will never get to that number.

lomiller · 3rd October 2011, 10:43 AM

Originally Posted by DrDave

Yes it's true. See this thread for a discussion on similar proofs http://forums.randi.org/showthread.php?t=219795

In brief, if you can show that for each natural number there is one odd number (or even) you show there are necessarily the same number.

For even numbers the formula is f(x)=2x
For odd numbers the formula is f(x)=2x+1

So we have shown there are infinite of each. To quickly show there are the same number of each.

I don’t like the highlighted part. Just because 2 sets are both infinite doesn’t mean they are equal.

If you go back a step and say that you have shown that for every odd number there exists an even number that can be found by adding 1(and that for every odd number there exists an even number that can be found by subtracting 1). Since even and odd numbers always occur in pairs there must be an equal number of them.

On the other hand you wanted to compare numbers divisible by 2 to numbers dividable by 3 again both sets are infinite but for every 3 numbers divisible by 2 there are only 2 that are divisible by 3 so numbers divisible by 2 will always be a larger value as you move towards infinity.

Ferguson · 3rd October 2011, 01:13 PM

Originally Posted by lomiller

I don’t like the highlighted part. Just because 2 sets are both infinite doesn’t mean they are equal.

Sure does.

Originally Posted by lomiller

On the other hand you wanted to compare numbers divisible by 2 to numbers dividable by 3 again both sets are infinite but for every 3 numbers divisible by 2 there are only 2 that are divisible by 3 so numbers divisible by 2 will always be a larger value as you move towards infinity.

For each finite set you pick, yes. But once we're talking infinity, there are as many. You will never "run out" of numbers divisible by 3.

Modified · 3rd October 2011, 01:40 PM

Originally Posted by Acleron

Vorpal said
Please tell me where my statement conflicts with either Vorpal's or yours. Except of course where you talk about integers ending in 3, closer reading perhaps.

You seem to have assumed that "sets A and B have the same number of members and B is a subset of A" implies that every property shared by all members of B is also shared by all members of A. That is only true if A and B are of finite length, in which case they must be identical.

lomiller · 3rd October 2011, 02:42 PM

Originally Posted by Ferguson

Sure does.

For each finite set you pick, yes. But once we're talking infinity, there are as many. You will never "run out" of numbers divisible by 3.

No. Infinity isn’t a number. This means you can and do get cases where two things that both equal infinity are not equal to each other.
x=z
y=2*z
y/x will still equal 2 even if you allow z to go to infinity (making both x and y infinity)

Acleron · 3rd October 2011, 03:00 PM

Originally Posted by Modified

You seem to have assumed that "sets A and B have the same number of members and B is a subset of A" implies that every property shared by all members of B is also shared by all members of A. That is only true if A and B are of finite length, in which case they must be identical.

If I did, I didn't mean to, but I don't think I did. I was more amused that the calculation of the percentage of the numbers containing a 3 is 100% of all numbers. The amusement is because every property shared by A ie containing the digit 3 is not shared by the set of all numbers.

Drachasor · 3rd October 2011, 03:04 PM

Originally Posted by lomiller

No. Infinity isn’t a number. This means you can and do get cases where two things that both equal infinity are not equal to each other.
x=z
y=2*z
y/x will still equal 2 even if you allow z to go to infinity (making both x and y infinity)

Infinity is like a number in several ways, but of course there are a lot of different types of infinity. How it might show up in a given problem can vary a lot.

There's a significant difference between limits (what you have above) and an infinity like "the number of integers". X = Z could be "the number of positive integers equal or less than Z". y = 2z could be "the number of positive even integers equal or less than 4z." You have, by your argument, shown that the latter set of numbers is twice as big as the former, which is decidedly not the case.

It's much more useful to consider the fact you can match up any countable and infinity set as a 1-1 ratio with the positive integers. You can also do it at a 2-1, 1-2, 4-1, or any other sort of ratio that you like. You'll never reach a point where you can't find numbers to fill in for this matching process. It's on this basis they are said to be of equal size, because we are comparing elements in a set.

Of course, there are infinities much bigger than this, such as the set of Real numbers (this includes fractions and irrational numbers...the set of Rational numbers, which has fractions composed of integers, is countable). This set can't be set up in a 1-1 correspondence with the natural numbers, because they simply can't be placed into an ordered list. There are then infinities bigger still than this, which can't be placed into a 1-1 correspondence with the Real numbers. In fact, there are an infinite number of infinity sizes.

W.D.Clinger · 3rd October 2011, 03:17 PM

Originally Posted by lomiller

No. Infinity isn’t a number. This means you can and do get cases where two things that both equal infinity are not equal to each other.

Since infinity isn't a real number, no real numbers equal infinity.

Besides, this thread is about bijections between sets and the notion of cardinal numbers, which don't have much in common with the real numbers you seem to have in mind.

Originally Posted by lomiller

x=z
y=2*z
y/x will still equal 2 even if you allow z to go to infinity (making both x and y infinity)

You appear to be thinking about the limit of a function as its argument increases without bound ("goes to infinity" in the vernacular). The use of the infinity symbol in

$ \begin{align} x(z) &= z \\ y(z) &= 2 z \\ f(z) &= \frac{y(z)}{x(z)} \\ \lim_{z \rightarrow \infty} f(z) &= 2 \end{align} $

doesn't mean there's any such thing as a real infinity. Line (4) above actually means

$ \[ (\forall \epsilon > 0)(\exists z_0)(\forall z > z_0) \left| {f(z) - 2} \right| < \epsilon \] $

where the variables in that last line range over real numbers (so there's no actual infinity anywhere).

As I said, that doesn't have anything to do with the topic of this thread.

Modified · 3rd October 2011, 03:24 PM

Originally Posted by Acleron

If I did, I didn't mean to, but I don't think I did. I was more amused that the calculation of the percentage of the numbers containing a 3 is 100% of all numbers. The amusement is because every property shared by A ie containing the digit 3 is not shared by the set of all numbers.

If you have a proof of that, then there is a mistake somewhere, since the counterargument is obvious. I'd guess the percentage of numbers less than n containing the digit 3 approaches 100 as n approaches infinity, but that is not the same thing unless your definition of "100% of all numbers" is different from the norm.

lomiller · 4th October 2011, 08:07 AM

Originally Posted by Drachasor

Infinity is like a number in several ways, but of course there are a lot of different types of infinity. How it might show up in a given problem can vary a lot.

There's a significant difference between limits (what you have above) and an infinity like "the number of integers". X = Z could be "the number of positive integers equal or less than Z". y = 2z could be "the number of positive even integers equal or less than 4z." You have, by your argument, shown that the latter set of numbers is twice as big as the former, which is decidedly not the case.

It's much more useful to consider the fact you can match up any countable and infinity set as a 1-1 ratio with the positive integers. You can also do it at a 2-1, 1-2, 4-1, or any other sort of ratio that you like. You'll never reach a point where you can't find numbers to fill in for this matching process. It's on this basis they are said to be of equal size, because we are comparing elements in a set.

Of course, there are infinities much bigger than this, such as the set of Real numbers (this includes fractions and irrational numbers...the set of Rational numbers, which has fractions composed of integers, is countable). This set can't be set up in a 1-1 correspondence with the natural numbers, because they simply can't be placed into an ordered list. There are then infinities bigger still than this, which can't be placed into a 1-1 correspondence with the Real numbers. In fact, there are an infinite number of infinity sizes.

You are pretty much saying the same thing as I did above to explain why the number if even and odd numbers must be equal. I suspect you missed part of the discussion leading up to my post so I’ll backtrack a bit

The question being addressed in the post you were quoting is a little different, however. The argument is being made that the number of even and odd numbers are equal because they are both infinity, while the number of even and odd numbers are equal it is not because both sets are infinite so I went on to give an example of two infinite sets where the number of members is not equal
The first set is the numbers evenly divisible by 2 (even numbers)
The second set is the numbers evenly divisible by 3

In this case every time we count to a multiple of 6 w get 3 numbers divisible by 2 and only 2 that are divisible by 3. No matter how high we count this relationship does not change so even though both sets are infinite the set of numbers divisible by 2 is larger.

Ferguson tried to argue that this isn’t the case by saying the number of integers in each set were both equal to infinity and therefore equal. While this would be true for equality to a real number infinity is not a real number and you can apply equality in this way.

lomiller · 4th October 2011, 08:14 AM

Originally Posted by W.D.Clinger

As I said, that doesn't have anything to do with the topic of this thread.

Follow read my post above or follow my posts though to the one you quoted.

Several people in this thread are attempting to argue that the number of even and odd numbers are equal because both equal infinity. This is not a valid proof because infinity is not a number and you can’t apply equality in that way with it.

Even though they get lucky and get the correct answer in this case, there are other cases where their logic fails.

Acleron · 4th October 2011, 08:54 AM

Originally Posted by Modified

If you have a proof of that, then there is a mistake somewhere, since the counterargument is obvious. I'd guess the percentage of numbers less than n containing the digit 3 approaches 100 as n approaches infinity, but that is not the same thing unless your definition of "100% of all numbers" is different from the norm.

Hmm. what I'm saying is that it is 100% when you have reached infinity.

Look here for the proof.

sol invictus · 4th October 2011, 08:55 AM

Originally Posted by lomiller

I went on to give an example of two infinite sets where the number of members is not equal
The first set is the numbers evenly divisible by 2 (even numbers)
The second set is the numbers evenly divisible by 3

Your "example" is not an example. Those two sets have exactly the same number of elements. The proof is identical in structure to the proof for evens and odds: construct a bijection between the two sets. In the example you gave, that is trivial: map each even number 2n to the divisible-by-3 number 3n, and vice versa (where n is any integer). That's a 1-to-1 map, so it proves that the number of elements in those two sets are equal.

An example of two infinite sets with different numbers of elements are the set of all real numbers between 0 and 1, and the integers. The set of reals is larger.

W.D.Clinger · 4th October 2011, 09:09 AM

Originally Posted by lomiller

Originally Posted by W.D.Clinger

As I said, that doesn't have anything to do with the topic of this thread.

Follow read my post above or follow my posts though to the one you quoted.

Several people in this thread are attempting to argue that the number of even and odd numbers are equal because both equal infinity. This is not a valid proof because infinity is not a number and you can’t apply equality in that way with it.

No, their proofs are valid because

The integers are countable (obviously).
The set of even numbers is a subset of the set of integers (obviously).
The set of odd numbers is a subset of the set of integers (obviously).
There are infinitely many even numbers (obviously).
There are infinitely many odd numbers (obviously).
So there's a countable infinity of even numbers (from facts 1, 2, 3), and there's a countable infinity of odd numbers (from facts 1, 3, 5).
All countably infinite sets have the same size; to put it more precisely, given any two countably infinite sets, there's a bijection between them (this is trivial, as it follows immediately from the definition of countable infinity).
The set of even numbers has the same size as the set of odd numbers (from facts 6 and 7).

It would be easier just to describe a one-to-one correspondence (bijection) between the even and odd integers, but it's important to recognize the validity of the above proof because that kind of reasoning is easier to apply in some less obvious cases, and generalizes to more powerful proof techniques for truly difficult cases.

lomiller · 4th October 2011, 09:26 AM

Originally Posted by sol invictus

Your "example" is not an example. Those two sets have exactly the same number of elements. The proof is identical in structure to the proof for evens and odds: construct a bijection between the two sets. In the example you gave, that is trivial: map each even number 2n to the divisible-by-3 number 3n, and vice versa (where n is any integer). That's a 1-to-1 map, so it proves that the number of elements in those two sets are equal.r.

You haven’t correctly characterised the example.

Take x as the set of integers from 4 to n
Count the number of times x is evenly divisible by 2
Count the number of times x is evenly divisible by 3
If you allow n to go to infinity both counts are infinite but for any value of n the first will always be greater.

WhatRoughBeast · 4th October 2011, 09:28 AM

Originally Posted by Acleron

If I did, I didn't mean to, but I don't think I did. I was more amused that the calculation of the percentage of the numbers containing a 3 is 100% of all numbers. The amusement is because every property shared by A ie containing the digit 3 is not shared by the set of all numbers.

That's what I thought what was going on. And yes, it's sufficiently counter-intuitive (one might even say absurd) to be amusing.

But that's infinity for you.

Drachasor · 4th October 2011, 10:11 AM

Originally Posted by lomiller

You are pretty much saying the same thing as I did above to explain why the number if even and odd numbers must be equal. I suspect you missed part of the discussion leading up to my post so I’ll backtrack a bit

No, I'm saying you are wrong. I suspect you missed my first post on this thread, because I explained how all countable sets are the same size.

Originally Posted by lomiller

The question being addressed in the post you were quoting is a little different, however. The argument is being made that the number of even and odd numbers are equal because they are both infinity, while the number of even and odd numbers are equal it is not because both sets are infinite so I went on to give an example of two infinite sets where the number of members is not equal
The first set is the numbers evenly divisible by 2 (even numbers)
The second set is the numbers evenly divisible by 3

Those two sets are NOT different in size. This is part of Set Theory.

Originally Posted by lomiller

In this case every time we count to a multiple of 6 w get 3 numbers divisible by 2 and only 2 that are divisible by 3. No matter how high we count this relationship does not change so even though both sets are infinite the set of numbers divisible by 2 is larger.

You can line up the members of these sets so that they are in a 1-1 relationship with each other. For any member of one set, you can produce a member of the other in that manner. They are the same size.

This is how you count things. You line up the members of one and compare them to the members of another. You don't go "ok, how many members are there less than 5? How about 6? How about 7?"

Originally Posted by lomiller

Ferguson tried to argue that this isn’t the case by saying the number of integers in each set were both equal to infinity and therefore equal. While this would be true for equality to a real number infinity is not a real number and you can apply equality in this way.

Sometimes I wonder if anyone reads my posts. I mean, I'm the third reply, but you act like I haven't said anything in this thread.

As you say, infinity isn't quite a number. Certainly not an ordinary number. Yet you are treating it like one. You are taking a limit as two numbers go to infinity and divide each other and acting like this is a solid comparison of the size of the two infinities. It isn't. As I demonstrated you can have that ratio come out to be ANY number you wish by adjusting how you count and still counting ALL members of both sets. That's how we know your counting system is messed up.

Instead, we say that when things are like that, the two sets are equal in size. If the ratio must be infinity (e.g. you can't make a 1-1 correspondence between two sets), then we say one is bigger than the other.

Solitaire · 4th October 2011, 08:49 PM

Originally Posted by psionl0

It is pretty pointless to compare the number of elements in the set of even numbers to that in the set of odd numbers because both sets are infinitely large.

Trying to compare one infinitely large number with another is meaningless because infinity is not a number.

I must disagree.

Ordinality And Cardinality

An ordinal is number you can reach by counting the elements finite set. For example, the set {0, 1, 2, 3}, has an ordinality of four – the same ordinality as the set {-1, 1, -2, 2}.

A cardinal is number you cannot simply count them by you can compare the size of them by lining up the members – see illustration.

Thus when I say infinity is a number I mean it is a cardinal number not an ordinal number.

The Function Trap

Functions with an infinite number terms in series often get confused with the issue of the size of infinity. Infinity has a definite size, not an arbitrary one. Hence when you create a term series such as the 1 - 1 + 1 - 1 summing series depends upon how you evaluate it.

(1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) . . . = 0

Move the parenthesis and extract the first term.

1 - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) . . . = 1

Rotate the terms so that the -1 is in front.

(-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) . . . = 0

Move the parenthesis and extract the first term.

-1 + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) . . . = -1

None of these series expresses the true size of infinity. They only show the possible values of a function with an infinite number of terms.

Infinity As A Constant

Infinity acts like a constant similar to Pi, e, or phi. You can add, divide, multiply, and subtract a number from it. Such operations do not change its size, for example: ∞ + 1 = ∞ - 1. But, if you subtract two functions with these values you won't find ∞ + 1 ≡ ∞ - 1 because the two functions subtracted will give you ∞ + 1 - (∞ - 1) = 2.

This subtraction of two infinite functions that differ shows up in Quantum Mechanics as the Casimir Force between two plates. The infinite wave modes outside the plates generates an infinite force inward. The infinite wave modes inside the plages generate an infinite force outward. Yet when you do the math and the experiment in the lab you find the two don't cancel out perfectly at zero, but produce a predictable inward force on the plates depending upon the distance.

Does this clear things up?
Or did I make the subject more confusing?

psionl0 · 4th October 2011, 09:24 PM

Originally Posted by lomiller

No. Infinity isn’t a number. This means you can and do get cases where two things that both equal infinity are not equal to each other.
x=z
y=2*z
y/x will still equal 2 even if you allow z to go to infinity (making both x and y infinity)

You accidentally made the mistake of treating infinity as a number in this example after all. This becomes clear if we add a couple of words to the example:

For any finite number z NO MATTER HOW LARGE
if x = z
and y = 2 * z
then y/x = 2

Although this is described in calculus as a "limit to infinity", it is still dealing with finite numbers. The expression "∞ / ∞" is known in calculus as "indeterminate".

psionl0 · 4th October 2011, 09:45 PM

Originally Posted by Solitaire

Does this clear things up?
Or did I make the subject more confusing?

Definitely more confusing.

It is nonsense to talk of such things as ∞ + 1 or ∞ - 1 but let us assume that your claim that ∞ + 1 = ∞ - 1 is true.

You also claim that (∞ + 1) - (∞ - 1) = 2 but because of your first postulate we can write (∞ + 1) - (∞ - 1) = (∞ - 1) - (∞ - 1) = 0.

So, does 0 = 2?

Vorpal · 4th October 2011, 10:28 PM

There, Solitaire is probably talking about the extended reals. Affinely extended reals (with +∞ and -∞) are standard in real analysis, while projectively extended reals (with just ∞) are the analogue of the Riemann sphere, the one-point compatification of the complex numbers, commonly used in complex analysis. For neither one are the expression ∞ - ∞ (or +∞ - +∞) defined, but many artihmetic operations, such as ∞+1 = ∞-1 = ∞, are meaningful and useful.

What happens in Casimir force or physics in general is a bit more subtle, though. More relevant are the Cauchy principal value of integrals and Euler's zeta function trick for series (or exponential regulator for physics-folk), or dark sorcery.

Drachasor · 4th October 2011, 11:08 PM

Originally Posted by psionl0

Definitely more confusing.

It is nonsense to talk of such things as ∞ + 1 or ∞ - 1 but let us assume that your claim that ∞ + 1 = ∞ - 1 is true.

You also claim that (∞ + 1) - (∞ - 1) = 2 but because of your first postulate we can write (∞ + 1) - (∞ - 1) = (∞ - 1) - (∞ - 1) = 0.

So, does 0 = 2?

Quite. It becomes unclear what you get when you subtract one infinity from another. When it happens in mathematics dealing with reality, you can take some sort of limit to figure out the answer (and indeed it can be pretty much anything, and depends on the physical constraints).

sol invictus · 5th October 2011, 03:04 PM

Originally Posted by lomiller

You haven’t correctly characterised the example.

I think I have.

Quote:

Take x as the set of integers from 4 to n
Count the number of times x is evenly divisible by 2
Count the number of times x is evenly divisible by 3
If you allow n to go to infinity both counts are infinite but for any value of n the first will always be greater.

That's true. Nevertheless, in the limit, those two sets are the same size.

psionl0 · 5th October 2011, 05:15 PM

Originally Posted by sol invictus

That's true. Nevertheless, in the limit, those two sets are the same size.

Wrong!

If the number of elements in the first set is n/2 and the number of elements in the second set is n/3 then the ratio between the two is 3:2 and in the limit the ratio tends to 3:2.

Again this shows the absurdity of treating infinity as a number. You can't count up to infinity nor can you add up finite numbers and get an infinite result.

Modified · 5th October 2011, 05:19 PM

Originally Posted by Acleron

Hmm. what I'm saying is that it is 100% when you have reached infinity.

What does "reached infinity" mean?

Solitaire · 5th October 2011, 05:31 PM

Originally Posted by psionl0

Definitely more confusing.

It is nonsense to talk of such things as ∞ + 1 or ∞ - 1 but let us assume that your claim that ∞ + 1 = ∞ - 1 is true.

You also claim that (∞ + 1) - (∞ - 1) = 2 but because of your first postulate we can write (∞ + 1) - (∞ - 1) = (∞ - 1) - (∞ - 1) = 0.

So, does 0 = 2?

Clearly not. My approach isn't working, try this.

You can have a function f(x) whose series equals infinity and a function g(x) whose series equals infinity, but when you subtract f(x) from g(x), term by term, you get a finite nonzero number. The size of f(x) and g(x) remain the same, but something remains. Hence the desire to include this in some sort of notation.

Perhaps even the equation, ∞_g = ∞_f + 3, is overly ambitious in trying to express this.

Originally Posted by Vorpal

There, Solitaire is probably talking about the extended reals. Affinely extended reals (with +∞ and -∞) are standard in real analysis, while projectively extended reals (with just ∞) are the analogue of the Riemann sphere, the one-point compatification of the complex numbers, commonly used in complex analysis. For neither one are the expression ∞ - ∞ (or +∞ - +∞) defined, but many artihmetic operations, such as ∞+1 = ∞-1 = ∞, are meaningful and useful.

What happens in Casimir force or physics in general is a bit more subtle, though. More relevant are the Cauchy principal value of integrals and Euler's zeta function trick for series (or exponential regulator for physics-folk), or dark sorcery.

Yes. I couldn't think of Cauchy at the moment.

Acleron · 5th October 2011, 05:45 PM

Originally Posted by Modified

What does "reached infinity" mean?

Look at the link I gave

sol invictus · 5th October 2011, 05:47 PM

Originally Posted by psionl0

Wrong!

No, it really isn't.

Quote:

If the number of elements in the first set is n/2 and the number of elements in the second set is n/3 then the ratio between the two is 3:2 and in the limit the ratio tends to 3:2.

You're assuming the value in the limit is equal to the value the ratio tends to under your chosen procedure. That assumption is unwarranted, and false in this instance.

Quote:

Again this shows the absurdity of treating infinity as a number. You can't count up to infinity nor can you add up finite numbers and get an infinite result.

Not at all - it simply shows the absurdity of making incorrect assumptions about infinite sets.

There's a very simple proof that shows that those two sets have exactly the same number of elements. It was explained at length in the thread on primes; I recommend you read that.

1st October 2011, 11:54 PM	#1
calcmandan Scholar Join Date: Apr 2011 Posts: 69	Does a proof or theorem exist for the following. A notion crossed my mind the other night about odd and even integers, during dinner actually. It got me curious enough to do some searching on google but came up empty handed. And I spent plenty of time reading through a bunch of existing proofs and theorems before I got tired. Unfortunately, it's been ages since my proof writing days in college, so here goes. Proof: In the universe of integers, there are an equal number of odd and even integers. Is the answer possibly obvious? I'm just curious. Daniel -- Sent from my HP TouchPad using Communities
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1st October 2011, 11:59 PM	#2
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,005	"Equal number" just means that you can put them into bijection (one-to-one correspondence, so that every odd number is mapped to exactly one even number and every even number is mapped to by some odd number). f(n) = n+1 works. Actually, every infinite subset of the integers has an equal number of members as the integers.
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2nd October 2011, 12:03 AM	#4
Drachasor Graduate Poster Join Date: Nov 2010 Posts: 1,718	There's an equal number of odd and even integers. this number is equal to the total number of integers as well as the number of primes, perfect squares, etc, etc, etc. The basic idea is you make a 1-1 ratio with the positive integers. Code: 1 2 3 4 5 6 7 8 9 10... 1 -1 3 -3 5 -5 7 -7 9 -9... 0 2 -2 4 -4 6 -6 8 -8 10... 2 3 5 7 11 13 17 19 23 29... 1 4 9 16 25 36 49 64 81... And so forth. It's part of Set Theory.
	Last edited by Drachasor; 2nd October 2011 at 12:04 AM.

2nd October 2011, 12:05 AM	#5
Tim Thompson Muse Join Date: Dec 2008 Location: San Gabriel Valley, east of Los Angeles Posts: 963	Integers I can't give you the formal proof, but I think you will find that the paradoxical answer is this: Yes, there are just as many odd integers as there are even integers, and there are just as many odd or even integers as there are integers total. The reason for this is that the set of all even or odd integers is a countable infinity. In other words, you can put either set of odd or even integers into a one-to-one correspondence with the set of all integers (which is really what you do whenever you count anything). If you stop short at a finite number, then of course there will always be more integers in your finite set than there are even or odd integers. But if you include the entire infinite sets, then they are the same order of infinity. I remember learning that when I was a kid in the book One, Two, Three ... Infinity by George Gamow.
	__________________ The point of philosophy is to start with something so simple as not to seem worth stating, and to end with something so paradoxical that no one will believe it. -- Bertrand Russell

2nd October 2011, 12:10 AM	#6
calcmandan Scholar Join Date: Apr 2011 Posts: 69	Quote: There's an equal number of odd and even integers. this number is equal to the total number of integers as well as the number of primes, perfect squares, etc, etc, etc. The basic idea is you make a 1-1 ratio with the positive integers. 1 2 3 4 5 6 7 8 9 10... 1 -1 3 -3 5 -5 7 -7 9 -9... 0 2 -2 4 -4 6 -6 8 -8 10... 2 3 5 7 11 13 17 19 23 29... 1 4 9 16 25 36 49 64 81... And so forth. It's part of Set Theory. My initial impression was set theory, but I kept second guessing myself. Yall are so FAST at responding! -- Sent from my HP TouchPad using Communities
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2nd October 2011, 12:01 AM	#3
DrDave Guest Join Date: Apr 2005 Location: Geneva Posts: 3,110	Yes it's true. See this thread for a discussion on similar proofs http://forums.randi.org/showthread.php?t=219795 In brief, if you can show that for each natural number there is one odd number (or even) you show there are necessarily the same number. For even numbers the formula is f(x)=2x For odd numbers the formula is f(x)=2x+1 So we have shown there are infinite of each. To quickly show there are the same number of each.

2nd October 2011, 12:15 AM	#7
calcmandan Scholar Join Date: Apr 2011 Posts: 69	Quote: Yes it's true. See this thread for a discussion on similar proofs http://forums.randi.org/showthread.php?t=219795 In brief, if you can show that for each natural number there is one odd number (or even) you show there are necessarily the same number. For even numbers the formula is f(x)=2x For odd numbers the formula is f(x)=2x+1 So we have shown there are infinite of each. To quickly show there are the same number of each. One to one correspondence, phew takes me way back to math 108. So basically, it's more or less a tautology? Sent from my HP TouchPad using Communities
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2nd October 2011, 01:03 AM	#8
sol invictus Philosopher Join Date: Oct 2007 Location: Nova Roma Posts: 8,417	Originally Posted by calcmandan One to one correspondence, phew takes me way back to math 108. So basically, it's more or less a tautology? Sent from my HP TouchPad using Communities I don't think it's a tautology, no. As Tim mentioned, the same techniques can be used to prove that there are as many positive even integers as there are integers, that there are as many primes as integers, that there are as many integers as rational numbers, as many reals between 0 and 1 as all reals, etc. Most people do not find those facts very obvious.

2nd October 2011, 01:20 AM	#9
calcmandan Scholar Join Date: Apr 2011 Posts: 69	Quote: I don't think it's a tautology, no. As Tim mentioned, the same techniques can be used to prove that there are as many positive even integers as there are integers, that there are as many primes as integers, that there are as many integers as rational numbers, as many reals between 0 and 1 as all reals, etc. Most people do not find those facts very obvious. Some of those facts are starting to come back to me. I studied math in college but it's been a while, so a lot of it is misfiled in the head. Thanks for your input. --Sent from my HP TouchPad using Communities
calcmandan Scholar Join Date: Apr 2011 Posts: 69

2nd October 2011, 11:32 PM	#10
Acleron Graduate Poster Join Date: Oct 2007 Location: In a beautifully understandable universe Posts: 1,927	Originally Posted by Vorpal Actually, every infinite subset of the integers has an equal number of members as the integers. Yes, I particularly like that it can be proved that 100% of integers contain the digit 3.

3rd October 2011, 07:13 AM	#11
WhatRoughBeast Muse Join Date: Apr 2011 Posts: 817	Originally Posted by Acleron Yes, I particularly like that it can be proved that 100% of integers contain the digit 3. No, that's not what he said. Please pay attention. The number of integers which end in the number 3 is equal to the number of integers. That two sets have the same number of elements does not mean that the two sets are equal. Even if one set is a subset of the other. As long as both sets are infinite.
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3rd October 2011, 09:26 AM	#12
Acleron Graduate Poster Join Date: Oct 2007 Location: In a beautifully understandable universe Posts: 1,927	Originally Posted by WhatRoughBeast No, that's not what he said. Please pay attention. The number of integers which end in the number 3 is equal to the number of integers. That two sets have the same number of elements does not mean that the two sets are equal. Even if one set is a subset of the other. As long as both sets are infinite. Vorpal said Quote: Actually, every infinite subset of the integers has an equal number of members as the integers. Please tell me where my statement conflicts with either Vorpal's or yours. Except of course where you talk about integers ending in 3, closer reading perhaps.

3rd October 2011, 09:44 AM	#13
psionl0 Illuminator Join Date: Sep 2010 Location: 31°58'S 115°57'E Posts: 4,760	It is pretty pointless to compare the number of elements in the set of even numbers to that in the set of odd numbers because both sets are infinitely large. Trying to compare one infinitely large number with another is meaningless because infinity is not a number. The folly of treating infinity like it was a number can be demonstrated in the following example: What is 1 - 1 + 1 - 1 + 1 - 1 + . . .? It turns out that there are three possible answers depending on how you do the calculation. The expression could be written this way: (1 - 1) + (1 - 1) + (1 - 1) + . . . = 0 Alternatively, we could write 1 - (1 + 1 - 1 + 1 - 1 + 1 - 1 + . . .) = 1 - 0 = 1 Finally, if the total is X then we could write an algebraic expression: X = 1 - X for which the solution is X = 0.5 We will never know the actual sum because it depends on the last digit in the series and we will never get to that number.

3rd October 2011, 10:43 AM	#14
lomiller Philosopher Join Date: Jul 2007 Posts: 6,861	Originally Posted by DrDave Yes it's true. See this thread for a discussion on similar proofs http://forums.randi.org/showthread.php?t=219795 In brief, if you can show that for each natural number there is one odd number (or even) you show there are necessarily the same number. For even numbers the formula is f(x)=2x For odd numbers the formula is f(x)=2x+1 So we have shown there are infinite of each. To quickly show there are the same number of each. I don’t like the highlighted part. Just because 2 sets are both infinite doesn’t mean they are equal. If you go back a step and say that you have shown that for every odd number there exists an even number that can be found by adding 1(and that for every odd number there exists an even number that can be found by subtracting 1). Since even and odd numbers always occur in pairs there must be an equal number of them. On the other hand you wanted to compare numbers divisible by 2 to numbers dividable by 3 again both sets are infinite but for every 3 numbers divisible by 2 there are only 2 that are divisible by 3 so numbers divisible by 2 will always be a larger value as you move towards infinity.
lomiller Philosopher Join Date: Jul 2007 Posts: 6,861	__________________ "Anything's possible, but only a few things actually happen" Last edited by lomiller; 3rd October 2011 at 10:49 AM.

3rd October 2011, 01:13 PM	#15
Ferguson Muse Join Date: May 2009 Location: Lansing, MI Posts: 830	Originally Posted by lomiller I don’t like the highlighted part. Just because 2 sets are both infinite doesn’t mean they are equal. Sure does. Originally Posted by lomiller On the other hand you wanted to compare numbers divisible by 2 to numbers dividable by 3 again both sets are infinite but for every 3 numbers divisible by 2 there are only 2 that are divisible by 3 so numbers divisible by 2 will always be a larger value as you move towards infinity. For each finite set you pick, yes. But once we're talking infinity, there are as many. You will never "run out" of numbers divisible by 3.

3rd October 2011, 01:40 PM	#16
Modified Illuminator Join Date: Sep 2006 Location: SW Florida Posts: 4,062	Originally Posted by Acleron Vorpal said Please tell me where my statement conflicts with either Vorpal's or yours. Except of course where you talk about integers ending in 3, closer reading perhaps. You seem to have assumed that "sets A and B have the same number of members and B is a subset of A" implies that every property shared by all members of B is also shared by all members of A. That is only true if A and B are of finite length, in which case they must be identical.

3rd October 2011, 02:42 PM	#17
lomiller Philosopher Join Date: Jul 2007 Posts: 6,861	Originally Posted by Ferguson Sure does. For each finite set you pick, yes. But once we're talking infinity, there are as many. You will never "run out" of numbers divisible by 3. No. Infinity isn’t a number. This means you can and do get cases where two things that both equal infinity are not equal to each other. x=z y=2*z y/x will still equal 2 even if you allow z to go to infinity (making both x and y infinity)
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3rd October 2011, 03:00 PM	#18
Acleron Graduate Poster Join Date: Oct 2007 Location: In a beautifully understandable universe Posts: 1,927	Originally Posted by Modified You seem to have assumed that "sets A and B have the same number of members and B is a subset of A" implies that every property shared by all members of B is also shared by all members of A. That is only true if A and B are of finite length, in which case they must be identical. If I did, I didn't mean to, but I don't think I did. I was more amused that the calculation of the percentage of the numbers containing a 3 is 100% of all numbers. The amusement is because every property shared by A ie containing the digit 3 is not shared by the set of all numbers.

3rd October 2011, 03:04 PM	#19
Drachasor Graduate Poster Join Date: Nov 2010 Posts: 1,718	Originally Posted by lomiller No. Infinity isn’t a number. This means you can and do get cases where two things that both equal infinity are not equal to each other. x=z y=2*z y/x will still equal 2 even if you allow z to go to infinity (making both x and y infinity) Infinity is like a number in several ways, but of course there are a lot of different types of infinity. How it might show up in a given problem can vary a lot. There's a significant difference between limits (what you have above) and an infinity like "the number of integers". X = Z could be "the number of positive integers equal or less than Z". y = 2z could be "the number of positive even integers equal or less than 4z." You have, by your argument, shown that the latter set of numbers is twice as big as the former, which is decidedly not the case. It's much more useful to consider the fact you can match up any countable and infinity set as a 1-1 ratio with the positive integers. You can also do it at a 2-1, 1-2, 4-1, or any other sort of ratio that you like. You'll never reach a point where you can't find numbers to fill in for this matching process. It's on this basis they are said to be of equal size, because we are comparing elements in a set. Of course, there are infinities much bigger than this, such as the set of Real numbers (this includes fractions and irrational numbers...the set of Rational numbers, which has fractions composed of integers, is countable). This set can't be set up in a 1-1 correspondence with the natural numbers, because they simply can't be placed into an ordered list. There are then infinities bigger still than this, which can't be placed into a 1-1 correspondence with the Real numbers. In fact, there are an infinite number of infinity sizes.

3rd October 2011, 03:17 PM	#20
W.D.Clinger Master Poster Join Date: Oct 2009 Posts: 2,439	Originally Posted by lomiller No. Infinity isn’t a number. This means you can and do get cases where two things that both equal infinity are not equal to each other. Since infinity isn't a real number, no real numbers equal infinity. Besides, this thread is about bijections between sets and the notion of cardinal numbers, which don't have much in common with the real numbers you seem to have in mind. Originally Posted by lomiller x=z y=2*z y/x will still equal 2 even if you allow z to go to infinity (making both x and y infinity) You appear to be thinking about the limit of a function as its argument increases without bound ("goes to infinity" in the vernacular). The use of the infinity symbol in $<br /> \begin{align}<br /> x(z) &= z \\<br /> y(z) &= 2 z \\<br /> f(z) &= \frac{y(z)}{x(z)} \\<br /> \lim_{z \rightarrow \infty} f(z) &= 2<br /> \end{align}<br />$ doesn't mean there's any such thing as a real infinity. Line (4) above actually means $<br /> \[<br /> (\forall \epsilon > 0)(\exists z_0)(\forall z > z_0) \left\| {f(z) - 2} \right\| < \epsilon<br /> \]<br />$ where the variables in that last line range over real numbers (so there's no actual infinity anywhere). As I said, that doesn't have anything to do with the topic of this thread.

3rd October 2011, 03:24 PM	#21
Modified Illuminator Join Date: Sep 2006 Location: SW Florida Posts: 4,062	Originally Posted by Acleron If I did, I didn't mean to, but I don't think I did. I was more amused that the calculation of the percentage of the numbers containing a 3 is 100% of all numbers. The amusement is because every property shared by A ie containing the digit 3 is not shared by the set of all numbers. If you have a proof of that, then there is a mistake somewhere, since the counterargument is obvious. I'd guess the percentage of numbers less than n containing the digit 3 approaches 100 as n approaches infinity, but that is not the same thing unless your definition of "100% of all numbers" is different from the norm.

4th October 2011, 08:07 AM	#22
lomiller Philosopher Join Date: Jul 2007 Posts: 6,861	Originally Posted by Drachasor Infinity is like a number in several ways, but of course there are a lot of different types of infinity. How it might show up in a given problem can vary a lot. There's a significant difference between limits (what you have above) and an infinity like "the number of integers". X = Z could be "the number of positive integers equal or less than Z". y = 2z could be "the number of positive even integers equal or less than 4z." You have, by your argument, shown that the latter set of numbers is twice as big as the former, which is decidedly not the case. It's much more useful to consider the fact you can match up any countable and infinity set as a 1-1 ratio with the positive integers. You can also do it at a 2-1, 1-2, 4-1, or any other sort of ratio that you like. You'll never reach a point where you can't find numbers to fill in for this matching process. It's on this basis they are said to be of equal size, because we are comparing elements in a set. Of course, there are infinities much bigger than this, such as the set of Real numbers (this includes fractions and irrational numbers...the set of Rational numbers, which has fractions composed of integers, is countable). This set can't be set up in a 1-1 correspondence with the natural numbers, because they simply can't be placed into an ordered list. There are then infinities bigger still than this, which can't be placed into a 1-1 correspondence with the Real numbers. In fact, there are an infinite number of infinity sizes. You are pretty much saying the same thing as I did above to explain why the number if even and odd numbers must be equal. I suspect you missed part of the discussion leading up to my post so I’ll backtrack a bit The question being addressed in the post you were quoting is a little different, however. The argument is being made that the number of even and odd numbers are equal because they are both infinity, while the number of even and odd numbers are equal it is not because both sets are infinite so I went on to give an example of two infinite sets where the number of members is not equal The first set is the numbers evenly divisible by 2 (even numbers) The second set is the numbers evenly divisible by 3 In this case every time we count to a multiple of 6 w get 3 numbers divisible by 2 and only 2 that are divisible by 3. No matter how high we count this relationship does not change so even though both sets are infinite the set of numbers divisible by 2 is larger. Ferguson tried to argue that this isn’t the case by saying the number of integers in each set were both equal to infinity and therefore equal. While this would be true for equality to a real number infinity is not a real number and you can apply equality in this way.
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4th October 2011, 08:14 AM	#23
lomiller Philosopher Join Date: Jul 2007 Posts: 6,861	Originally Posted by W.D.Clinger As I said, that doesn't have anything to do with the topic of this thread. Follow read my post above or follow my posts though to the one you quoted. Several people in this thread are attempting to argue that the number of even and odd numbers are equal because both equal infinity. This is not a valid proof because infinity is not a number and you can’t apply equality in that way with it. Even though they get lucky and get the correct answer in this case, there are other cases where their logic fails.
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4th October 2011, 08:54 AM	#24
Acleron Graduate Poster Join Date: Oct 2007 Location: In a beautifully understandable universe Posts: 1,927	Originally Posted by Modified If you have a proof of that, then there is a mistake somewhere, since the counterargument is obvious. I'd guess the percentage of numbers less than n containing the digit 3 approaches 100 as n approaches infinity, but that is not the same thing unless your definition of "100% of all numbers" is different from the norm. Hmm. what I'm saying is that it is 100% when you have reached infinity. Look here for the proof.

4th October 2011, 08:55 AM	#25
sol invictus Philosopher Join Date: Oct 2007 Location: Nova Roma Posts: 8,417	Originally Posted by lomiller I went on to give an example of two infinite sets where the number of members is not equal The first set is the numbers evenly divisible by 2 (even numbers) The second set is the numbers evenly divisible by 3 Your "example" is not an example. Those two sets have exactly the same number of elements. The proof is identical in structure to the proof for evens and odds: construct a bijection between the two sets. In the example you gave, that is trivial: map each even number 2n to the divisible-by-3 number 3n, and vice versa (where n is any integer). That's a 1-to-1 map, so it proves that the number of elements in those two sets are equal. An example of two infinite sets with different numbers of elements are the set of all real numbers between 0 and 1, and the integers. The set of reals is larger.

4th October 2011, 09:09 AM	#26
W.D.Clinger Master Poster Join Date: Oct 2009 Posts: 2,439	Originally Posted by lomiller Originally Posted by W.D.Clinger As I said, that doesn't have anything to do with the topic of this thread. Follow read my post above or follow my posts though to the one you quoted. Several people in this thread are attempting to argue that the number of even and odd numbers are equal because both equal infinity. This is not a valid proof because infinity is not a number and you can’t apply equality in that way with it. No, their proofs are valid because The integers are countable (obviously). The set of even numbers is a subset of the set of integers (obviously). The set of odd numbers is a subset of the set of integers (obviously). There are infinitely many even numbers (obviously). There are infinitely many odd numbers (obviously). So there's a countable infinity of even numbers (from facts 1, 2, 3), and there's a countable infinity of odd numbers (from facts 1, 3, 5). All countably infinite sets have the same size; to put it more precisely, given any two countably infinite sets, there's a bijection between them (this is trivial, as it follows immediately from the definition of countable infinity). The set of even numbers has the same size as the set of odd numbers (from facts 6 and 7). It would be easier just to describe a one-to-one correspondence (bijection) between the even and odd integers, but it's important to recognize the validity of the above proof because that kind of reasoning is easier to apply in some less obvious cases, and generalizes to more powerful proof techniques for truly difficult cases.

4th October 2011, 09:26 AM	#27
lomiller Philosopher Join Date: Jul 2007 Posts: 6,861	Originally Posted by sol invictus Your "example" is not an example. Those two sets have exactly the same number of elements. The proof is identical in structure to the proof for evens and odds: construct a bijection between the two sets. In the example you gave, that is trivial: map each even number 2n to the divisible-by-3 number 3n, and vice versa (where n is any integer). That's a 1-to-1 map, so it proves that the number of elements in those two sets are equal.r. You haven’t correctly characterised the example. Take x as the set of integers from 4 to n Count the number of times x is evenly divisible by 2 Count the number of times x is evenly divisible by 3 If you allow n to go to infinity both counts are infinite but for any value of n the first will always be greater.
lomiller Philosopher Join Date: Jul 2007 Posts: 6,861	__________________ "Anything's possible, but only a few things actually happen"

4th October 2011, 09:28 AM	#28
WhatRoughBeast Muse Join Date: Apr 2011 Posts: 817	Originally Posted by Acleron If I did, I didn't mean to, but I don't think I did. I was more amused that the calculation of the percentage of the numbers containing a 3 is 100% of all numbers. The amusement is because every property shared by A ie containing the digit 3 is not shared by the set of all numbers. That's what I thought what was going on. And yes, it's sufficiently counter-intuitive (one might even say absurd) to be amusing. But that's infinity for you.
WhatRoughBeast Muse Join Date: Apr 2011 Posts: 817	Last edited by WhatRoughBeast; 4th October 2011 at 09:31 AM.

4th October 2011, 10:11 AM	#29
Drachasor Graduate Poster Join Date: Nov 2010 Posts: 1,718	Originally Posted by lomiller You are pretty much saying the same thing as I did above to explain why the number if even and odd numbers must be equal. I suspect you missed part of the discussion leading up to my post so I’ll backtrack a bit No, I'm saying you are wrong. I suspect you missed my first post on this thread, because I explained how all countable sets are the same size. Originally Posted by lomiller The question being addressed in the post you were quoting is a little different, however. The argument is being made that the number of even and odd numbers are equal because they are both infinity, while the number of even and odd numbers are equal it is not because both sets are infinite so I went on to give an example of two infinite sets where the number of members is not equal The first set is the numbers evenly divisible by 2 (even numbers) The second set is the numbers evenly divisible by 3 Those two sets are NOT different in size. This is part of Set Theory. Originally Posted by lomiller In this case every time we count to a multiple of 6 w get 3 numbers divisible by 2 and only 2 that are divisible by 3. No matter how high we count this relationship does not change so even though both sets are infinite the set of numbers divisible by 2 is larger. You can line up the members of these sets so that they are in a 1-1 relationship with each other. For any member of one set, you can produce a member of the other in that manner. They are the same size. This is how you count things. You line up the members of one and compare them to the members of another. You don't go "ok, how many members are there less than 5? How about 6? How about 7?" Originally Posted by lomiller Ferguson tried to argue that this isn’t the case by saying the number of integers in each set were both equal to infinity and therefore equal. While this would be true for equality to a real number infinity is not a real number and you can apply equality in this way. Sometimes I wonder if anyone reads my posts. I mean, I'm the third reply, but you act like I haven't said anything in this thread. As you say, infinity isn't quite a number. Certainly not an ordinary number. Yet you are treating it like one. You are taking a limit as two numbers go to infinity and divide each other and acting like this is a solid comparison of the size of the two infinities. It isn't. As I demonstrated you can have that ratio come out to be ANY number you wish by adjusting how you count and still counting ALL members of both sets. That's how we know your counting system is messed up. Instead, we say that when things are like that, the two sets are equal in size. If the ratio must be infinity (e.g. you can't make a 1-1 correspondence between two sets), then we say one is bigger than the other.
	Last edited by Drachasor; 4th October 2011 at 10:14 AM.

4th October 2011, 08:49 PM	#30
Solitaire Neoclinus blanchardi Join Date: Jul 2001 Location: Johnson City, Tennessee Posts: 1,735	Originally Posted by psionl0 It is pretty pointless to compare the number of elements in the set of even numbers to that in the set of odd numbers because both sets are infinitely large. Trying to compare one infinitely large number with another is meaningless because infinity is not a number. I must disagree. Ordinality And Cardinality An ordinal is number you can reach by counting the elements finite set. For example, the set {0, 1, 2, 3}, has an ordinality of four – the same ordinality as the set {-1, 1, -2, 2}. A cardinal is number you cannot simply count them by you can compare the size of them by lining up the members – see illustration. Thus when I say infinity is a number I mean it is a cardinal number not an ordinal number. The Function Trap Functions with an infinite number terms in series often get confused with the issue of the size of infinity. Infinity has a definite size, not an arbitrary one. Hence when you create a term series such as the 1 - 1 + 1 - 1 summing series depends upon how you evaluate it. (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) . . . = 0 Move the parenthesis and extract the first term. 1 - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) . . . = 1 Rotate the terms so that the -1 is in front. (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) . . . = 0 Move the parenthesis and extract the first term. -1 + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) . . . = -1 None of these series expresses the true size of infinity. They only show the possible values of a function with an infinite number of terms. Infinity As A Constant Infinity acts like a constant similar to Pi, e, or phi. You can add, divide, multiply, and subtract a number from it. Such operations do not change its size, for example: ∞ + 1 = ∞ - 1. But, if you subtract two functions with these values you won't find ∞ + 1 ≡ ∞ - 1 because the two functions subtracted will give you ∞ + 1 - (∞ - 1) = 2. This subtraction of two infinite functions that differ shows up in Quantum Mechanics as the Casimir Force between two plates. The infinite wave modes outside the plates generates an infinite force inward. The infinite wave modes inside the plages generate an infinite force outward. Yet when you do the math and the experiment in the lab you find the two don't cancel out perfectly at zero, but produce a predictable inward force on the plates depending upon the distance. Does this clear things up? Or did I make the subject more confusing?
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4th October 2011, 09:24 PM	#31
psionl0 Illuminator Join Date: Sep 2010 Location: 31°58'S 115°57'E Posts: 4,760	Originally Posted by lomiller No. Infinity isn’t a number. This means you can and do get cases where two things that both equal infinity are not equal to each other. x=z y=2z y/x will still equal 2 even if you allow z to go to infinity (making both x and y infinity) You accidentally made the mistake of treating infinity as a number in this example after all. This becomes clear if we add a couple of words to the example: For any finite* number z NO MATTER HOW LARGE if x = z and y = 2 * z then y/x = 2 Although this is described in calculus as a "limit to infinity", it is still dealing with finite numbers. The expression "∞ / ∞" is known in calculus as "indeterminate".
	Last edited by psionl0; 4th October 2011 at 09:47 PM.

4th October 2011, 09:45 PM	#32
psionl0 Illuminator Join Date: Sep 2010 Location: 31°58'S 115°57'E Posts: 4,760	Originally Posted by Solitaire Does this clear things up? Or did I make the subject more confusing? Definitely more confusing. It is nonsense to talk of such things as ∞ + 1 or ∞ - 1 but let us assume that your claim that ∞ + 1 = ∞ - 1 is true. You also claim that (∞ + 1) - (∞ - 1) = 2 but because of your first postulate we can write (∞ + 1) - (∞ - 1) = (∞ - 1) - (∞ - 1) = 0. So, does 0 = 2?

4th October 2011, 10:28 PM	#33
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,005	There, Solitaire is probably talking about the extended reals. Affinely extended reals (with +∞ and -∞) are standard in real analysis, while projectively extended reals (with just ∞) are the analogue of the Riemann sphere, the one-point compatification of the complex numbers, commonly used in complex analysis. For neither one are the expression ∞ - ∞ (or +∞ - +∞) defined, but many artihmetic operations, such as ∞+1 = ∞-1 = ∞, are meaningful and useful. What happens in Casimir force or physics in general is a bit more subtle, though. More relevant are the Cauchy principal value of integrals and Euler's zeta function trick for series (or exponential regulator for physics-folk), or dark sorcery.
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,005	__________________ For every philosopher, there exists an equal and opposite philosopher. They're both wrong. Last edited by Vorpal; 4th October 2011 at 10:48 PM. Reason: Misattributed which Euler trick was used; the guy has so many!

4th October 2011, 11:08 PM	#34
Drachasor Graduate Poster Join Date: Nov 2010 Posts: 1,718	Originally Posted by psionl0 Definitely more confusing. It is nonsense to talk of such things as ∞ + 1 or ∞ - 1 but let us assume that your claim that ∞ + 1 = ∞ - 1 is true. You also claim that (∞ + 1) - (∞ - 1) = 2 but because of your first postulate we can write (∞ + 1) - (∞ - 1) = (∞ - 1) - (∞ - 1) = 0. So, does 0 = 2? Quite. It becomes unclear what you get when you subtract one infinity from another. When it happens in mathematics dealing with reality, you can take some sort of limit to figure out the answer (and indeed it can be pretty much anything, and depends on the physical constraints).

5th October 2011, 03:04 PM	#35
sol invictus Philosopher Join Date: Oct 2007 Location: Nova Roma Posts: 8,417	Originally Posted by lomiller You haven’t correctly characterised the example. I think I have. Quote: Take x as the set of integers from 4 to n Count the number of times x is evenly divisible by 2 Count the number of times x is evenly divisible by 3 If you allow n to go to infinity both counts are infinite but for any value of n the first will always be greater. That's true. Nevertheless, in the limit, those two sets are the same size.