#### mathstudent1234

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mathonline.wikidot.com

If you were counting on infinity being absolute, your number's up

www.scientificamerican.com

This maps the non-negative rationals onto the non-negative integers.

Having done that the negative rationals are just mapped onto the negative integers using the obvious relation.

A quick wiki reveals no lack of proofs that the reals are uncountable. Have you tried reading any of them?

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Here is one that is unbelievable.I have now thanks. Yes it's clear. I just don't like the fact that it is true but that doesnt matter.

A bit string is a sequence the entries of which are all either a zero or a one.

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Alright I've done research I get it still unbelievable.Here is one that is unbelievable.

A bit string is a sequence the entries of which are all either a zero or a one.

The collection of all bit strings is uncountable.

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Yes, the set of all algebraic numbers is countable. Yes, there is a proof that "such a bijection" (from the natural numbers to all real numbers) can't exist. Here it is:

We prove that there exist no bijection from the set of all natural numbers to the set of all real numbers between 0 and 1. If that is true then there certainly cannot be a bijection to the set of

Suppose such a bijection exists. That is, suppose that we can assign one real; number to 1, another to 2, etc such that every natural number is assigned a unique real number and every real number, between 0 and 1, is assigned to a unique natural number, so we have a list of real numbers, between 0 and 1, r1, r2, r3, … that contains all such real numbers.

Construct a real number, x, thusly: The first digit of x is anything other than the first digit of r1, the second digit is anything other than the second digit of r2, etc. x is a real number between 0 and 1 and is not on that list since it differs from each number on the list at at least one decimal place. That contradicts the assumption that every real number is on the list.

(There are some technical points such as the fact that 0.4999... with the "9" infinitely repeating is the same as 0.5 but that is easily fixed by requiring that the choice of new digit is not always the same.)

Once we accept that the real numbers are a purely mental construction (as of course are the transfinite numbers), we no longer get confused about what seems to be physically impossible. The diagonal proof is a proof by contradiction involving an infinite list of items each of which is itself an infinite list. The human imagination cannot truly visualize such a thing, but human reason can think about it without visualization.

Oh I don't know. Imagine the plane covered in tiles. Each row is an infinite list, and there are infinite rows.The human imagination cannot truly visualize such a thing

Nevertheless I can pick a tile and proceed in a one tile per step spiral path that will, taken in the limit, include every tile.

That's pretty easy to visualize.

Except after you counted a tile it splits into a bunch of smaller tiles. You go back, count them and continue along your spiral. But the smaller tiles split yet again. Good luck.Oh I don't know. Imagine the plane covered in tiles. Each row is an infinite list, and there are infinite rows.

Nevertheless I can pick a tile and proceed in a one tile per step spiral path that will, taken in the limit, include every tile.

That's pretty easy to visualize.

My point exactly. You cannot stretch any physical analogy about the real numbers very far, particularly not with the infinite, because no one has any tangible experience with an infinite physical object. It is better to follow the logic without attempting a bijection onto a physical object.Except after you counted a tile it splits into a bunch of smaller tiles. You go back, count them and continue along your spiral. But the smaller tiles split yet again. Good luck.

Well that's not what JeffM stated. He stated an infinite list of lists.Except after you counted a tile it splits into a bunch of smaller tiles. You go back, count them and continue along your spiral. But the smaller tiles split yet again. Good luck.

Of course the analogy breaks down for the reals otherwise they'd be countable... which they aren't.

I don't find that unbelievable, because that includes all infinite length bit strings.Here is one that is unbelievable.

A bit string is a sequence the entries of which are all either a zero or a one.

The collection of all bit strings is uncountable.

However, the collection of all and only finite length bit strings is countable.

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To: lookagain, thank you for your comments. However, you failed to note that I defined a bit string as a sequence and because a sequence is a function from a countablely infinite set then there are infinitely long.I don't find that unbelievable, because that includes all infinite length bit strings.

However, the collection of all and only finite length bit strings is countable.

But you are quite correct the collection of bit strings of finite length is a countable collection.

There are \(\displaystyle \ 2^1 \ \ 1-bit \ strings, \ \ 2^2 \ \ 2-bit \ \ strings, \ \ and \ \ in \ \

general, \ \ 2^n \ \ n-bit \ \ \ \ \ \ \ \ \ \ strings. \ \ \) Correspond the first two bit strings

with 1 and 2 in a count. Correspond the next four bit strings with 3 through 6.

Correspond the next eight bit strings after those with 7 through 14. And so on.

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I don't see how that works. It may, but I don't see it.There are \(\displaystyle / 2^1 / 1-bit / / strings, / / 2^2 / 2-bit / / strings, / / and / / in / /

general, / / 2^n / / n-bit / / strings. / / \) Correspond the first two bit strings

with 1 and 2 in a count. Correspond the next four bit strings with 3 through 6.

Correspond the next eight bit strings after those with 7 through 14. And so on.

Please help me. To what does the string \(\displaystyle 00010111001100001111111001010111111\) correspond?

It's one of the 35-bit length sequences. It will correspond to somewhereI don't see how that works. It may, but I don't see it.

Please help me. To what does the string \(\displaystyle 00010111001100001111111001010111111\) correspond?

between location number \(\displaystyle \ 2^{35} - 1 \ \ and \ \ 2^{36} - 2, \ \ inclusive.\)

It's not required to be able to state what it corresponds to exactly.

Locations 1 --> 2, 3 --> 6, 7 --> 14, . . . , (2^n -- 1) --> [2^(n+1) - 2], . . .

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We start with a list that can be put into 1-1 correspondance with the natural numbers. Each element of the list consists of an infinite string of two symbols, say plus and minus, but there is no infinite repetition of either symbol.I don't see how that works. It may, but I don't see it.

Please help me. To what does the string \(\displaystyle 00010111001100001111111001010111111\) correspond?

We now assume that the list exhausts the possible configuration of such strings. But Cantor's diagonal argument shows that we can construct an infinite string that meets the criteria and is not in the list. Therefore there is no such list, and there are transfinite numbers greater than \(\displaystyle \aleph_0.\)

Moreover, we can put the binary representations of every irrational number > 0 and < 1 into 1-1 correspondence with the set of infinite strings of plus or minus with no infinite repetitions. This proves that the number of irrationals exceeds the number of positive integers. Therefore the number of reals exceeds the number of positive integers, and the number of irrational numbers exceeds the number of rational numbers.

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Oh my friend yes indeedIt's not required to be able to state what it corresponds to exactly.[/tex]

Lookagain, I have long objected to your correcting posts on minor point about which I suspected you know nothing. Now I admit that when you replied to this thread the way you did, I knew that I had you. With this postings you have shown that you do not even have the level of mathematics that I expect of a third year mathematics major. In other words, you my friend are an amateur at mathematics. I happen to know that others here think that too. Your snide corrections are not well received.

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