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I actually have no idea how you could misunderstand my point so hard lmao, unless you just didn't read or something

I actually have no idea how you could misunderstand my point so hard lmao, unless you just didn't read or something

A natural number can be represented in any number encoding; for example, in binary, every natural number is some sequence of 0s and 1s. In decimal, it's a sequence of the decimal digits (0, 1, ... 9). Depending on which encoding you pick, if you try to enumerate every element in the sequence (0, 1, 2...), you get a sequence that looks slightly different, and you'll notice the length of some strings changes depending on the number of digits you use. However, it cannot be disputed that the encoding does not have any impact on the number of elements in the sequence.

You can think of the "rational" numbers as being the natural numbers, but encoded in a number system wherein there are 11 digits: the 10 digits from (0, ... 9) and an 11th digit for representing fractions, '/'. Seeing as we've stated that it does not matter how many digits you use to represent the natural numbers, it is perhaps unsurprising that there are as many rational numbers as natural numbers. You of course have to add the constraint that '/' can never precede a sequence of all 0s, but that's a relatively minor thing. Even without that constraint they're still the same set.

This is [B]not[/B] the case for the irrational numbers, because every irrational number is a [B]countably infinite[/B] sequence of digits.

A natural number can be represented in any number encoding; for example, in binary, every natural number is some sequence of 0s and 1s. In decimal, it's a sequence of the decimal digits (0, 1, ... 9). Depending on which encoding you pick, if you try to enumerate every element in the sequence (0, 1, 2...), you get a sequence that looks slightly different, and you'll notice the length of some strings changes depending on the number of digits you use. However, it cannot be disputed that the encoding does not have any impact on the number of elements in the sequence.

You can think of the "rational" numbers as being the natural numbers, but encoded in a number system wherein there are 11 digits: the 10 digits from (0, ... 9) and an 11th digit for representing fractions, '/'. Seeing as we've stated that it does not matter how many digits you use to represent the natural numbers, it is perhaps unsurprising that there are as many rational numbers as natural numbers. You of course have to add the constraint that '/' can never precede a sequence of all 0s, but that's a relatively minor thing. Even without that constraint they're still the same set.

This is not the case for the irrational numbers, because every irrational number is a countably infinite sequence of digits.

[QUOTE=Plotato;968274]Your definitions are awful, inaccurate and unhelpful.

You can't say it is obvious that there is "more of something". There are no more even numbers than there are natural numbers. There are no more numbers of the Fibonacci sequence than there are natural numbers. Alluded to earlier there is no more fractional numbers than there are natural numbers. Cantor's diagonalization states specifically that such a construction exists where you can generate an entirely new number from an infinite list of numbers that was not previously recorded in an infinite mapping with the natural numbers. [/QUOTE]

[QUOTE=Plotato]
Your third line misrepresents the whole argument because the set of rational numbers are countable and are a certain permutation of the natural numbers.
[/QUOTE]

[B]The set [/B]of rational numbers is a permutation of the naturals. Yes, hence why the set of rationals is countable. But for irrational numbers, it is every irrational number, itself, that is a permutation of the entire set of natural numbers and is, hence, countably infinite.

[QUOTE=Plotato]
You haven't defined what "certain permutation" means and therefore is moot and your definition of irrational number is also completely unhelpful.
[/QUOTE]

A permutation is a certain way of ordering some set.

[QUOTE=Plotato]
Permutation implies a certain order and as far as it goes, the decimal construction of irrational numbers have no order to them (and therefore can't be represented as a fraction).
[/QUOTE]

Yes, indeed. However, I'm talking about the permutations of the set of [B]natural[/B] numbers, not reals. Whether or not the real numbers have a well-ordering is not relevant to my point.

[QUOTE=Plotato]
This isn't an "inbetween argument" like the way you put it. Infinity does not work the way you describe it like it has a certain scale or something.
[/QUOTE]

It's not an in-between argument. You misunderstood my point about the "infinity of infinities"; I'm not talking about all the segments of [0, 1) that you can copy paste onto each other over and over, but I'm talking about the numbers [B]themselves[/B] being infinite; their representation is infinite. If their representation is infinite, you clearly need a "higher" ordinal to index into the list of infinite things: a number that can index into an infinite list of finite things has less information than it, because... you can change a finite number number of digits in a finite number, and get a fully unique, distinct number. For infinite numbers, if you change only a finite number of the first few digits you've written, there are still many numbers you could be referring to.

So perhaps it is unsurprising that if you have an infinite list A of infinite things, that you have more information in A than in the infinite list B of finite things.

Originally Posted by Plotato

Your definitions are awful, inaccurate and unhelpful.

You can't say it is obvious that there is "more of something". There are no more even numbers than there are natural numbers. There are no more numbers of the Fibonacci sequence than there are natural numbers. Alluded to earlier there is no more fractional numbers than there are natural numbers. Cantor's diagonalization states specifically that such a construction exists where you can generate an entirely new number from an infinite list of numbers that was not previously recorded in an infinite mapping with the natural numbers.

Originally Posted by Plotato

Your third line misrepresents the whole argument because the set of rational numbers are countable and are a certain permutation of the natural numbers.

The set of rational numbers is a permutation of the naturals. Yes, hence why the set of rationals is countable. But for irrational numbers, it is every irrational number, itself, that is a permutation of the entire set of natural numbers and is, hence, countably infinite.

Originally Posted by Plotato

You haven't defined what "certain permutation" means and therefore is moot and your definition of irrational number is also completely unhelpful.

A permutation is a certain way of ordering some set.

Originally Posted by Plotato

Permutation implies a certain order and as far as it goes, the decimal construction of irrational numbers have no order to them (and therefore can't be represented as a fraction).

Yes, indeed. However, I'm talking about the permutations of the set of natural numbers, not reals. Whether or not the real numbers have a well-ordering is not relevant to my point.

Originally Posted by Plotato

This isn't an "inbetween argument" like the way you put it. Infinity does not work the way you describe it like it has a certain scale or something.

It's not an in-between argument. You misunderstood my point about the "infinity of infinities"; I'm not talking about all the segments of [0, 1) that you can copy paste onto each other over and over, but I'm talking about the numbers themselves being infinite; their representation is infinite. If their representation is infinite, you clearly need a "higher" ordinal to index into the list of infinite things: a number that can index into an infinite list of finite things has less information than it, because... you can change a finite number number of digits in a finite number, and get a fully unique, distinct number. For infinite numbers, if you change only a finite number of the first few digits you've written, there are still many numbers you could be referring to.

So perhaps it is unsurprising that if you have an infinite list A of infinite things, that you have more information in A than in the infinite list B of finite things.

[QUOTE=Plotato;942837]i knew you would still be confused about this, oh, how the germans have stumped the french once again

mathematically, countability is defined of whether you can assign 1, 2, etc. to whatever you want to map to on a [B]one-to-one[/B] basis, no element left unjerked. if the list is infinitely long, then, as unintuitive as it sounds, then a countably infinite list is full... matched with infinite elements, on a one-to-one basis.

Cantor's argument comes with the above presupposition that you can build such a list where you can map every integer (1 -> 0.1232, etc) with a real number [I]once[/I] (therefore making it countable). the order of the list doesn't matter. now why wouldn't such a list already have the whatever number produced by diagonalization? no, i can, through diagonalization, come up with a new number that [B]won't have an sole integer assigned to[/B]. i know it sounds unintuitive, but i will make bold the affirmative. the case is: i have produced a new number [B]of which there is not an assignment of an integer[/B], therefore proving my idea that such a list is countable wrong. guess its not countable. this idea also implies the varying sizes to infinity, as we can't match one-to-one every element from one set of numbers to another without leaving something behind...

the thing with math is that you have to explicitly define, prove and state everything in a sheer logical manner. the definition of "countable" is mentioned earlier, and if i, for some reason, can't make that one-to-one complete map from the integers to whatever, the definition is now uncountable by contradiction. intuition doesn't matter. somethings in math are just unintuitively the case. yes, every number that would exist exists between 0 and 1... can you "count" them all, given the definition of "countable"? if you're still thinking that you could always generate a new integer for a new real number produced, well that would just break the definition of "countable" that we have, which is a pretty good one, based off of mathematical reasoning. if we hold the definition of "countable" to be the case, well, you can't have your intuition go your way.

please understand you escargot[/QUOTE]

I have a much simpler explanation.
What is a natural number? It's a finite sequence of digits. What is the totality of natural numbers? An infinite sequence of digits, ordered in a certain way.
What is a irrational number? An infinite sequence of digits. Every irrational number corresponds to a certain permutation of the set of natural numbers. If you take all these permutations together, it is obvious there are more irrationals than naturals. You effectively run into an infinity of infinities

Originally Posted by Plotato

i knew you would still be confused about this, oh, how the germans have stumped the french once again

mathematically, countability is defined of whether you can assign 1, 2, etc. to whatever you want to map to on a one-to-one basis, no element left unjerked. if the list is infinitely long, then, as unintuitive as it sounds, then a countably infinite list is full... matched with infinite elements, on a one-to-one basis.

Cantor's argument comes with the above presupposition that you can build such a list where you can map every integer (1 -> 0.1232, etc) with a real number once (therefore making it countable). the order of the list doesn't matter. now why wouldn't such a list already have the whatever number produced by diagonalization? no, i can, through diagonalization, come up with a new number that won't have an sole integer assigned to. i know it sounds unintuitive, but i will make bold the affirmative. the case is: i have produced a new number of which there is not an assignment of an integer, therefore proving my idea that such a list is countable wrong. guess its not countable. this idea also implies the varying sizes to infinity, as we can't match one-to-one every element from one set of numbers to another without leaving something behind...

the thing with math is that you have to explicitly define, prove and state everything in a sheer logical manner. the definition of "countable" is mentioned earlier, and if i, for some reason, can't make that one-to-one complete map from the integers to whatever, the definition is now uncountable by contradiction. intuition doesn't matter. somethings in math are just unintuitively the case. yes, every number that would exist exists between 0 and 1... can you "count" them all, given the definition of "countable"? if you're still thinking that you could always generate a new integer for a new real number produced, well that would just break the definition of "countable" that we have, which is a pretty good one, based off of mathematical reasoning. if we hold the definition of "countable" to be the case, well, you can't have your intuition go your way.

please understand you escargot

I have a much simpler explanation.
What is a natural number? It's a finite sequence of digits. What is the totality of natural numbers? An infinite sequence of digits, ordered in a certain way.
What is a irrational number? An infinite sequence of digits. Every irrational number corresponds to a certain permutation of the set of natural numbers. If you take all these permutations together, it is obvious there are more irrationals than naturals. You effectively run into an infinity of infinities

Fun fact: you can extend this indefinitely. There are also "numbers" that have an [I]uncountably infinite [/I]number of digits after the decimal point, in which case this set of numbers has a cardinality that is equal to (N!)!, aka equal to the cardinality of the set of all the subsets of irrational numbers.

Fun fact: you can extend this indefinitely. There are also "numbers" that have an uncountably infinite number of digits after the decimal point, in which case this set of numbers has a cardinality that is equal to (N!)!, aka equal to the cardinality of the set of all the subsets of irrational numbers.

@[URL="https://www.sc2mafia.com/forum/member.php?u=29377"]Marshmallow Marshall[/URL] are you still confused about this, Frenchman? I can offer a more intuitive explanation:

The set of integers are numbers with a distance equal to 1 between them. If you draw a tick at every single natural number along a line, you will see a distance of 1 between each. However, this cannot be done with irrational numbers because every irrational number consists of a countable infinity of digits after the decimal point. This means that in order to reach the immediate neighbour of a given irrational number, you have to add or subtract an amount of the form 0.00000.....01 = 1 / infinity = 0 from each. You cannot do this.

Basically the irrational numbers are an infinity of infinities. In fact, if you were to say that the rational/integral numbers have a cardinality of, say, N, then the irrational numbers have a cardinality that is "conceptually" equal to 2 ^ N: if you were to take the set of all possible subsets (including the whole set) of integral numbers, this set of sets would have the same cardinality as the set of irrationals: basically, there are as many subsets of the integral numbers as there are rational numbers.

This number is bewilderingly large because this infinity is the [B]BINARY POWER [/B]​of another infinity.

@Marshmallow Marshall are you still confused about this, Frenchman? I can offer a more intuitive explanation:

The set of integers are numbers with a distance equal to 1 between them. If you draw a tick at every single natural number along a line, you will see a distance of 1 between each. However, this cannot be done with irrational numbers because every irrational number consists of a countable infinity of digits after the decimal point. This means that in order to reach the immediate neighbour of a given irrational number, you have to add or subtract an amount of the form 0.00000.....01 = 1 / infinity = 0 from each. You cannot do this.

Basically the irrational numbers are an infinity of infinities. In fact, if you were to say that the rational/integral numbers have a cardinality of, say, N, then the irrational numbers have a cardinality that is "conceptually" equal to 2 ^ N: if you were to take the set of all possible subsets (including the whole set) of integral numbers, this set of sets would have the same cardinality as the set of irrationals: basically, there are as many subsets of the integral numbers as there are rational numbers.

This number is bewilderingly large because this infinity is the BINARY POWER ​of another infinity.