pengkuan Posted December 5, 2015 Author Posted December 5, 2015 When? When does it "pass through" 1/3? If we label the "binary numbers" as xi (for example, x1 = 1/2, x2 = 1/4, x3 = 3/4), for which i is xi equal to 1/3? You have not explained it whatsoever. You've simply denied it. Actually, it does not pass through 1/3, but it will at infinity, where 1/3 is 0.010101010101010101010101010101010101010101010101010101..... In fact, in order to pass through 1/3, we have to use 3 base numeral system. Why are infinitely many binary numbers countable? In fact, they don't. They cannot exist. You can never express the expansion of a irrational number completely.
studiot Posted December 5, 2015 Posted December 5, 2015 Why are infinitely many binary numbers countable? To answer this you need to go back to your definition of countable and countably infinite.
uncool Posted December 6, 2015 Posted December 6, 2015 Actually, it does not pass through 1/3, but it will at infinity, where 1/3 is 0.010101010101010101010101010101010101010101010101010101.....Exactly. That means that your "path" does not "pass through" every "binary number", or even every rational number expressed as a "binary number", and therefore your "proof" that they are not countable fails. In fact, in order to pass through 1/3, we have to use 3 base numeral system. Why are infinitely many binary numbers countable? In fact, they don't. They cannot exist. You can never express the expansion of a irrational number completely. You are moving the goalposts here. You claimed that your path would "pass through" every "binary number". As there is no i such that xi = 1/3, that is false - and the "binary numbers" do not have to be countable.
pengkuan Posted December 6, 2015 Author Posted December 6, 2015 Exactly. That means that your "path" does not "pass through" every "binary number", or even every rational number expressed as a "binary number", and therefore your "proof" that they are not countable fails. I agree that I was wrong in saying that my path passes through 1/3. Actually, it does not, nor through 1/5, /7, 2/5 3/5 ...... So, it is clear that binary numbers are less numerous than rational numbers. You are moving the goalposts here. You claimed that your path would "pass through" every "binary number". As there is no i such that xi = 1/3, that is false - and the "binary numbers" do not have to be countable. So, binary numbers are in some sort "less countable" than rationals rather than uncountable.
uncool Posted December 7, 2015 Posted December 7, 2015 (edited) I agree that I was wrong in saying that my path passes through 1/3. Actually, it does not, nor through 1/5, /7, 2/5 3/5 ...... So, it is clear that binary numbers are less numerous than rational numbers. No. It is clear that "binary numbers" of finite length are "less numerous" than rational numbers, in the sense that they form a subset. Your path only "passes through" "binary numbers" of finite length. That does not imply that the sets are not still in bijection. So, binary numbers are in some sort "less countable" than rationals rather than uncountable.It turns out they are exactly as countable. Your "path" shows that there is a bijection between natural numbers and "binary numbers of finite length"; the original path (along with some set theory) shows that there is a bijection between natural numbers and rational numbers - so there is a bijection between "binary numbers of finite length" and rational numbers. All countable sets are "equal in size", that is, there is a bijection between them. Edited December 7, 2015 by uncool
Keen Posted December 7, 2015 Posted December 7, 2015 Well honestly I am kind of lost already in your first paragraph. When it mentions "We let n go to the cardinal number of the set of natural numbers", what does that thing even mean? From what I understand, what is called here a binary fractional number, is pretty much a binary number with finite decimals, which actually is simply a rational number. Claiming, that this set has the cardinal of the powerset of natural numbers is a very bold assertion, since it challenges already well established results, so this magical use of limit "We let n go to the cardinal number of the set of natural numbers." had better be very well explained. The problem is that with limits you can make pretty much anything if you don't use them carefully and here, I do not think there is a sufficient justification behind that use of limit. If on the other hand the set of binary fractional numbers is the set of all numbers in [0,1[ written in binary, then it seems to me there is a misrepresentation, because the process that is used to list those numbers only deals with numbers with finite decimals. So yeah it's all quite messy and needs rewriting and I am almost 100% sure, that if those arguments were written correctly a flaw would quickly appear, because one does not simply contradict Cantor's diagonal argument.
pengkuan Posted December 7, 2015 Author Posted December 7, 2015 No. It is clear that "binary numbers" of finite length are "less numerous" than rational numbers, in the sense that they form a subset. Your path only "passes through" "binary numbers" of finite length. That does not imply that the sets are not still in bijection. It turns out they are exactly as countable. Your "path" shows that there is a bijection between natural numbers and "binary numbers of finite length"; the original path (along with some set theory) shows that there is a bijection between natural numbers and rational numbers - so there is a bijection between "binary numbers of finite length" and rational numbers. All countable sets are "equal in size", that is, there is a bijection between them. Of cause "binary numbers" of finite length are a set of the same size than the natural numbers. When I said "in some sort "less countable" " it was to just to emphasize that binary numbers are countable, not as real numbers which are uncountable. - and the "binary numbers" do not have to be countable. What does this mean? Well honestly I am kind of lost already in your first paragraph. When it mentions "We let n go to the cardinal number of the set of natural numbers", what does that thing even mean? This is to get the transfinite number aleph 0 directly for 2n=2À0 without saying "n goes to infinity and n=aleph 0" . Claiming, that this set has the cardinal of the powerset of natural numbers is a very bold assertion, since it challenges already well established results, so this magical use of limit "We let n go to the cardinal number of the set of natural numbers." had better be very well explained. The problem is that with limits you can make pretty much anything if you don't use them carefully and here, I do not think there is a sufficient justification behind that use of limit. Binary numbers are a subset of the rationals, so they have the cardinality of the natural numbers. What I'm saying is not that binarys have the cardinality of the reals 2À0 , but the power set of the naturals has not the cardinality of the reals, but that of the naturals. If on the other hand the set of binary fractional numbers is the set of all numbers in [0,1[ written in binary, then it seems to me there is a misrepresentation, because the process that is used to list those numbers only deals with numbers with finite decimals. So yeah it's all quite messy and needs rewriting and I am almost 100% sure, that if those arguments were written correctly a flaw would quickly appear, because one does not simply contradict Cantor's diagonal argument. In fact, fractional binaries cannot write all the reals in [1,0[ because they cannot have infinitely many digits. They are only rationals. 1
studiot Posted December 7, 2015 Posted December 7, 2015 (edited) In fact, fractional binaries cannot write all the reals in [1,0[ because they cannot have infinitely many digits. They are only rationals. Why can't they have infinitely many digits? Incidentally the conventional way to state this is to say they neither repeat nor terminate. It avoids the use to the term infinity. Edited December 7, 2015 by studiot
pengkuan Posted December 7, 2015 Author Posted December 7, 2015 Why can't they have infinitely many digits? Because infinitely many digits have no sense. They cannot be used to perform mathematics. Incidentally the conventional way to state this is to say they neither repeat nor terminate. It avoids the use to the term infinity. Yes, but this term does not emphasize infinity that is the crux of the problem.
Carrock Posted December 7, 2015 Posted December 7, 2015 In "2" you say "Because the intervals are always empty when n→∞, the set of all fractional binary numbers is not continuous but discrete" I suspect you implicitly mean by n→∞ that n is or becomes larger than any given finite number but n is still finite. Yes. When n→∞, it has to start from 1, then +1 then +1 ...... At each step, n is a finite number. Only if n=aleph-null are all the real numbers (from 0 to 1) represented and there are are then no intervals between numbers. This is THE problem. How can n become aleph_0 without first has been a finite number? One cannot leap to aleph_0, one just go the old way, 1+1, then +1, then +1 . During the process, the intervals are all empty. At the end, since there is no end, the intervals will never be filled. There will be empty intervals between numbers, binary or decimal or in any base. In fact, n is not kept finite, but cannot be infinite, that is, never n=aleph_0. So, real number cannot be mirrored by digital numbers. I will try another approach. I assume you accept that there are [latex]{\aleph_{0}}[/latex] elements in the infinite integer series 1,2,3..... ie the series does not end at some arbitrarily large finite number. If not, I'm done. Rather than count n by starting from 1, avoid that objection by simply constructing figure 3 in the same way as you construct the infinite integer series, with each distinct n set equal to a distinct positive integer. You don't have to count or iterate or step through every or any value of n; constructing either the infinite integer series or figure 3 does not require you to 'count' every member. If you consider the number of rows generated by different values of n to be finite, you have to explain how n cannot be set equal to some members of the infinite integer series 1,2,3..... If this proof I've outlined here and earlier in thread, that there are [latex]{2}^{\aleph_{0}}[/latex] ie [latex]{\aleph_{1}}[/latex] distinct binary fractions is valid, your contradicting proofs must be invalid.
pengkuan Posted December 7, 2015 Author Posted December 7, 2015 I will try another approach. I assume you accept that there are [latex]{\aleph_{0}}[/latex] elements in the infinite integer series 1,2,3..... ie the series does not end at some arbitrarily large finite number. If not, I'm done. Rather than count n by starting from 1, avoid that objection by simply constructing figure 3 in the same way as you construct the infinite integer series, with each distinct n set equal to a distinct positive integer. You don't have to count or iterate or step through every or any value of n; constructing either the infinite integer series or figure 3 does not require you to 'count' every member. If you consider the number of rows generated by different values of n to be finite, you have to explain how n cannot be set equal to some members of the infinite integer series 1,2,3..... If this proof I've outlined here and earlier in thread, that there are [latex]{2}^{\aleph_{0}}[/latex] ie [latex]{\aleph_{1}}[/latex] distinct binary fractions is valid, your contradicting proofs must be invalid. Sorry. I do not understand your reasoning. What are "objection ", "rows " or "does not require you to 'count' every member"? What is my "contradicting proofs must be invalid"? What are you trying to prove? Binary numbers are countable? Uncountable?
uncool Posted December 8, 2015 Posted December 8, 2015 (edited) Pengkuan: When you say "binary numbers", do you mean numbers that have finitely many digits in binary, or can they have infinitely many digits? I think much of your confusion comes from mixing up the two. The former are countable. The latter are uncountable, with as many elements as the reals, and equally, as many elements as the powerset of the integers. You've said things that apply to the former but not the latter at some times, and the latter but not the former at others. Edited December 8, 2015 by uncool 2
pengkuan Posted December 8, 2015 Author Posted December 8, 2015 Pengkuan: When you say "binary numbers", do you mean numbers that have finitely many digits in binary, or can they have infinitely many digits? I think much of your confusion comes from mixing up the two. The former are countable. The latter are uncountable, with as many elements as the reals, and equally, as many elements as the powerset of the integers. You've said things that apply to the former but not the latter at some times, and the latter but not the former at others. For me binary numbers are 0+.+sequence of n digits with n going to infinity. n is finite at any stage. In fact, a number with infinitely many digits does not have sense. When one says irrational have infinitely many digits one does not means it have actually infinitely many, but it does have a last one, which means that the binary number does not have a definite value.
imatfaal Posted December 8, 2015 Posted December 8, 2015 For me binary numbers are 0+.+sequence of n digits with n going to infinity. n is finite at any stage. In fact, a number with infinitely many digits does not have sense. When one says irrational have infinitely many digits one does not means it have actually infinitely many, but it does have a last one, which means that the binary number does not have a definite value. Surely the whole point of an irrational is that it cannot be the quotient of two integers - and if it has a last digit then it has a fractional representation and is rational
pengkuan Posted December 8, 2015 Author Posted December 8, 2015 Surely the whole point of an irrational is that it cannot be the quotient of two integers - and if it has a last digit then it has a fractional representation and is rational Infinitely many digits number do not equal irrational number because without the last digit, it cannot have definite value.
imatfaal Posted December 8, 2015 Posted December 8, 2015 Irrational numbers do not terminate nor become periodical - they do not have a final digit in contradiction to your assertion that they do ... When one says irrational have infinitely many digits one does not means it have actually infinitely many, but it does have a last one, ...
pengkuan Posted December 8, 2015 Author Posted December 8, 2015 Irrational numbers do not terminate nor become periodical - they do not have a final digit in contradiction to your assertion that they do Irrational number does not have last digit. It was a typo. They do not have definite digital value.
imatfaal Posted December 8, 2015 Posted December 8, 2015 Irrational number does not have last digit. It was a typo. They do not have definite digital value. OK - that makes much more sense.
studiot Posted December 8, 2015 Posted December 8, 2015 (edited) pengkuan Irrational number does not have last digit. It was a typo. They do not have definite digital value. Not having a 'last digit' is not the same as not having a definite value. Not having a definite value is not the same as not being able to be represented by a finite string or ratio of two finite strings. What is true is that in a number system to any base there are numbers that cannot be represented in a finite string in that base. In general these unrepresentable numbers will be different in different number bases but that does not affect their existence. For example in the number system to the base pi, the number we call pi is representable exactly. Edited December 8, 2015 by studiot
uncool Posted December 8, 2015 Posted December 8, 2015 For me binary numbers are 0+.+sequence of n digits with n going to infinity. n is finite at any stage.In other words, any particular "binary number" will have finitely many digits? In fact, a number with infinitely many digits does not have sense. When one says irrational have infinitely many digits one does not means it have actually infinitely many, but it does have a last one, which means that the binary number does not have a definite value.This looks like word salad. What are you trying to prove now?
pengkuan Posted December 8, 2015 Author Posted December 8, 2015 In other words, any particular "binary number" will have finitely many digits? Yes. This looks like word salad. What are you trying to prove now? Back to the origin. There are two sort of binary numbers, finitely many digits and infinitely many digits. The former are rational, the latter are irrational. In the section 1 of my article, I have shown that the cardinality of the former is 2À0 . This cardinality means they seemingly contains irrationals. But as they are rational, the cardinality 2À0 must be that of rational, which I showed afterward. Those with infinitely many digits are said to be irrational numbers. But I find that they do not have last digits, so we do not have their exact values. So, they cannot not be used as numbers. We use pi as point in the line, but not as 3.1415926.................. No operation is done using its infinitely many digits form. So I think binary numbers with infinitely many digits are not numbers at all because they cannot be used.
uncool Posted December 8, 2015 Posted December 8, 2015 Yes. Back to the origin. There are two sort of binary numbers, finitely many digits and infinitely many digits. The former are rational, the latter are irrational. In the section 1 of my article, I have shown that the cardinality of the former is 2À0 . This cardinality means they seemingly contains irrationals. But as they are rational, the cardinality 2À0 must be that of rational, which I showed afterward. Please, prove again why "binary numbers with infinitely many digits" must be rational. Those with infinitely many digits are said to be irrational numbers. But I find that they do not have last digits, so we do not have their exact values. So, they cannot not be used as numbers. We use pi as point in the line, but not as 3.1415926.................. No operation is done using its infinitely many digits form. So I think binary numbers with infinitely many digits are not numbers at all because they cannot be used.This isn't a proof.
pengkuan Posted December 8, 2015 Author Posted December 8, 2015 Please, prove again why "binary numbers with infinitely many digits" must be rational. This isn't a proof. The number of binary number is 2n, when n=infinity, n=À0 , so the cardinality of binary number is 2À0. As this is the cardinality of the power set of the naturals, this set must be the real. But the set of rationals does not contain irrational. So, the power set of the naturals is not the reals. This isn't a proof. I'm not proving. I explain.
studiot Posted December 8, 2015 Posted December 8, 2015 (edited) Back to the origin. There are two sort of binary numbers, finitely many digits and infinitely many digits. The former are rational, the latter are irrational. What rubbish. The number 1/3 is an infinitely long number in decimal. It is also an infinitely long number in binary. Yet it is rational in both systems.ie 1/3 or 1/11 Edited December 8, 2015 by studiot
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