One of the biggest misconceptions people have about infinity is that they think that a number like pi has infinitely more digits in it than any number with a finite number of digits, so the set of irrational numbers must have way more numbers than the set of rational numbers where all the numbers terminate. But this is not true, given a number like pi, we can form an infinite set of rational numbers that terminate, from the first digits of our number pi, like this…
Even though it is true that the irrational numbers go on to infinity, it is also true that the rational numbers will chase them all they way. And from this fact, we can find a beautiful symmetry between the numbers that terminate and the numbers that don’t terminate.
Given any complete set of finite binary numbers between 0 and 1, of length n, it will be true that the quantity of numbers that terminate with a zero at the n’th position, will be equal to the quantity of numbers that terminate with a 1.
We can call the numbers that terminate with a 1, “maximally great” because they have as many digits as possible in a system with n digits. All the other numbers are not maximally great because they have less than n digits in them.
For example, consider this set…
0.0 0.1 0.01 0.11
We will have 2 maximally great numbers and 2 not maximally great numbers. We can go to the next digit...
0.001 0.101 0.011 0.111
Then we will have 4 maximally great numbers and the 4 not maximally great numbers above. We can go to the next digit...
Then we will have 8 maximally great numbers and the 8 not maximally great numbers above.
You can repeat this forever and the number of maximally great numbers will always be the same as the not maximally great numbers.
The infinite set of not maximally great numbers will be the set of rational numbers that terminate while the infinite set of maximally great numbers will be the union of the set of rational numbers that repeat forever and the set of irrational numbers. So the set of irrational numbers can’t have a greater cardinality than the set of rational numbers.
Let i1 & i2 be irrational numbers so that 0< i1 < i2 <1 then there will always exist a rational number r so that i1 < r < i2.
Therefore, it will always be possible to find a rational number between any two irrational numbers, so the cardinality of the irrational numbers can’t be greater than the rational numbers.
If you form a list of decimal numbers (between 0 and 1) that can only be represented with an infinite number of digits (ex; 1/3, pi-3, e-2, root(2)-1 etc.) then for each of those numbers you can construct an infinite number of decimal numbers that have a finite number of digits.
For example; given the number 0.333... you can construct the infinite set {0.3, 0.33, 0.333, ...}, given the number pi - 3, you can construct the infinite set {0.1, 0.14, 0.141, ...}
Since we can extract an infinite quantity of numbers that terminate from every number that does not terminate, then it would be impossible for the set of numbers with an infinite number of digits to have a greater cardinality than the set of numbers that can be represented with a finite number of digits.
Since all the numbers on the left are finite in length and all the numbers on the right are infinite in length, we can always find a unique rational number by taking the first n digits from the irrational number until we have a number that is not on the left.
If you were to then diagonalize the list of irrational numbers on the right, and change every digit to something else, you would get a number not on that list but you could then find a rational number not on the left side to pair with the new number that you created.
Therefore the cardinality of the irrational numbers can not be greater than the rational numbers.
As for the power set being of a greater cardinality than the original set that it is made of. This is true for any finite set, but not infinite sets.
Cantor had a proof by contradiction to show that a set can’t be put into a 1-1 with it’s power set, but we can make a one to one pairing of the natural numbers with it’s power set like this.
We can then construct a special set S that is made up of all the natural numbers that are not members of the set that they are paired with.
Then the question will be 'what natural number z will be paired with S' ?
Since none of our natural numbers are members of the sets that they are paired with, we will have to say that z is not a member of the set it is paired with either. This gives us a problem, if z is not a member of S then it should be a member of S. But, when we examine the set S we find that it is the set of natural numbers, it is the universal set, so z will be a member of the universal set. (all numbers are)
It is possible to list the real numbers (between 0 and 1) like this, where D(x,y) is a digit and D(x,y) = {1,2,3,4,5,6,7,8,9 or 0} and no row has all the same values for all the D’s.
Then as n -> infinity, you will have a list of all the real numbers.
Cantor created a number by taking the diagonal of this list…
D = 0. D(1,1) D(2,2) D(3,3) … D(n,n)
He then created a new number by mutating every digit in the diagonal and claimed that this number is not in the list of real numbers, when in fact it was just not in the first n rows of the list, and it was in some row beyond row n. So he did not create a number not in the list of real numbers.
Cantor could have just padded all the columns with trailing zero’s so then he could have extended the diagonal across 10^n columns, but then the list would have 10^(10^n) rows. This strategy would be self defeating.
What he actually did was rather strange, he assumed that n = 10^n when n is infinite, and only then was he able to diagonalize the matrix with n columns and 10^n rows. But, if your proof depends on the identity n = 10^n and your conclusion is that identity is wrong then he defeats his own argument.
Either way Cantor’s proof fails. There are no multiple infinities.
P1 All real numbers have an infinite number of digits. (some have an infinite number of trailing zero’s)
P2 All natural numbers have a finite number of digits.
From each of the numbers in the set of real numbers we can form a unique natural number by taking the first string that is not on our list of natural numbers. We can add this unique natural number to our list of natural numbers, and continue this forever with all the real numbers. Therefore we can find a unique natural number to pair with every real number. And the cardinality of the real numbers can’t be greater than that of the natural numbers.
Example:
A) 1 -> 0.14159265… B) 14 -> 0.14159266… C) 141 -> 0.14159267… D) 1410 -> 0.141
When you continue this forever, you will have all the real numbers between 0 and 1 that do not terminate. A proof of this conjecture would be nice, but that is what God created grad students for.
Cantor’s goal was to find a unique natural number for every real number and vice versa, it was not to find a 1 to 1 correspondence between the naturals and reals (though that would have established his point, it was a sufficient condition but not a necessary condition). This is known as the Schröder–Bernstein theorem. For two sets to be equipotent, you need only show that there exists injective functions f : A → B and g : B → A then |A| = |B|
It is easy to find a unique real number given a natural number, e.g. 12 → 12.0 or 12 → 12.14159… or generally n → n.xyz… where n is a natural number and xyz… is any string of digits. Or, we can find a real number in (0,1) for every natural number. e.g. 1230 → 0.0321 or 985356295141 → 0.141592653589
But, finding a unique natural number given a real number is not as easy, yet it can be done.
Cantor assumed that his goal was to find a one to one, (this was his blind spot) when his goal was actually to just find a unique natural number to correspond to each and every real number in any set of real numbers, and this is doable.
Cantor assumed that it was necessary to list all the natural numbers, in order and on one list. This assumption is erroneous. It is like a rich man going to a small country auction, if he bids all his money ($1 million) on the first item, then he will be able to buy only that item, but if he bids in smaller increments, he will be able to buy all the items up for auction.
Cantor could have paired his purported list of ‘ALL’ real numbers in (0,1) to a small subset of the natural numbers like this.
Then when he finds, by diagonalization, that he has not listed all the real numbers we could simply say “so what, we have not listed all the natural numbers either”. So, Cantors proof is thus inconclusive.
We can then ask Cantor to take the diagonals of the elements in his list, change the digits and make a second infinite list of real numbers that were not on his first list. And we can then pair those real numbers to some other subset of the natural numbers like this.
Again, we can ask Cantor to take the diagonals of the elements in his second list, change the digits and make a new infinite list of real numbers, that were not on this second list (being careful to not duplicate any numbers from his first list). And we can then pair those real numbers to yet another subset of the natural numbers like this.
And, naturally, we can keep this game going forever. The interesting thing about this is, that when the prefix of our natural numbers approaches infinity, then it will become harder to find a real number, by changing the digits of the diagonal numbers, that is not on one of the infinite quantity of lists of real numbers. Also notice that most of the natural numbers like 333, 777, 123789 etc do not appear anywhere in our pairings, but that is okay, since we have enough natural numbers to pair to our sets of irrational numbers without them. All this tells us is that it is possible that the set natural numbers is bigger than that of the set of real numbers in (0,1) or that they are equal in size.
Now that we have cast doubt on Cantors diagonal proof, we can show how it is possible to find a function that does pair up the natural numbers to the irrational numbers. (the rational numbers can be paired, so we will ignore them for now)
We will generate a set of random irrational numbers between 0 and 1, our list might look like this.
Then we can imagine a deck of cards where the first card has the number 1 on it, the second card has a 2, third a 3 and so on, a stack with all the natural numbers in order.
The first number on our random list is 0.5123453… so we can pair it with the fifth card in our deck. The second number on our list is 0.5125674… since we have used the 5 card, we must now use card 15. (the first two digits in reverse) The third number on our list is 0.5124195… since we have used the 5 and 15 card, we must now use card 215. (the first three digits in reverse) The fourth number on our list is 0.5127676… since we have used the 5, 15 and 215 card, we must now use card 7215. (the first four digits in reverse) The fifth number on our list is 0.5124295… since we have used the 5, 15 and 215 card, we must now use card 4215. (the first four digits in reverse) And so on.
If you ponder on it for a while, you will see that there will never be an irrational number that we can’t find a natural number to pair with it.
We could sort the natural numbers and keep the irrational that it is paired with, together with it, and then look at the nth digit of the nth real number that it is associated with. For example,
then change the diagonal digits to something else and we would get an irrational number that is not in our list of irrational numbers.
Since the number that we create is different from every number in our list of irrational numbers, it must be different at some finite point. (it is impossible for two irrational numbers to have the first infinite number of digits in common, then to be followed by some digits that are different.)
Since this diagonal defines a finite string of digits, that has not been seen on the right side of our set, then we know that same finite string does not appear on the left side either, since the left side is simply a reflection of the digits that have appeared on the right side. So we know that there exists a natural number equivalent to that string (or a substring of that string) that will be available to pair with the irrational diagonal number that introduced that finite string. Therefore, the cardinality of the two sets is the same.
That is about it, if you have any questions or comments, I would be glad to hear from you.
It has been said that all infinite sets are endless, but some are more endless than others. Hmmm?
A proof is not about language, it is about understanding. If you have a perfect proof that you don’t understand, what good is it. It would simply allow you to believe something that you don’t understand, this is called faith based mathematics.
Consider this “proof” that concludes the cardinality of the irrationals are less than or equal to a set of rationals. Is it formal? Is it valid? Is it understandable?
Maximally great numbers
ReplyDeleteOne of the biggest misconceptions people have about infinity is that they think that a number like pi has infinitely more digits in it than any number with a finite number of digits, so the set of irrational numbers must have way more numbers than the set of rational numbers where all the numbers terminate. But this is not true, given a number like pi, we can form an infinite set of rational numbers that terminate, from the first digits of our number pi, like this…
{3.14159…} -> {3, 3.1, 3.14, 3.141, 3.1415, 3.14159, …}
Even though it is true that the irrational numbers go on to infinity, it is also true that the rational numbers will chase them all they way. And from this fact, we can find a beautiful symmetry between the numbers that terminate and the numbers that don’t terminate.
Given any complete set of finite binary numbers between 0 and 1, of length n, it will be true that the quantity of numbers that terminate with a zero at the n’th position, will be equal to the quantity of numbers that terminate with a 1.
We can call the numbers that terminate with a 1, “maximally great” because they have as many digits as possible in a system with n digits. All the other numbers are not maximally great because they have less than n digits in them.
For example, consider this set…
0.0
0.1
0.01
0.11
We will have 2 maximally great numbers and 2 not maximally great numbers. We can go to the next digit...
0.001
0.101
0.011
0.111
Then we will have 4 maximally great numbers and the 4 not maximally great numbers above. We can go to the next digit...
0.0001
0.1001
0.0101
0.1101
0.0011
0.1011
0.0111
0.1111
Then we will have 8 maximally great numbers and the 8 not maximally great numbers above.
You can repeat this forever and the number of maximally great numbers will always be the same as the not maximally great numbers.
The infinite set of not maximally great numbers will be the set of rational numbers that terminate while the infinite set of maximally great numbers will be the union of the set of rational numbers that repeat forever and the set of irrational numbers. So the set of irrational numbers can’t have a greater cardinality than the set of rational numbers.
Let i1 & i2 be irrational numbers so that 0< i1 < i2 <1 then there will always exist a rational number r so that i1 < r < i2.
ReplyDeleteTherefore, it will always be possible to find a rational number between any two irrational numbers, so the cardinality of the irrational numbers can’t be greater than the rational numbers.
You can find a version of this argument here…
https://youtu.be/0XBPR066jTY?t=44m37s
Infinite Sets from Infinitely Long Numbers
ReplyDeleteIf you form a list of decimal numbers (between 0 and 1) that can only be represented with an infinite number of digits (ex; 1/3, pi-3, e-2, root(2)-1 etc.) then for each of those numbers you can construct an infinite number of decimal numbers that have a finite number of digits.
For example; given the number 0.333... you can construct the infinite set {0.3, 0.33, 0.333, ...}, given the number pi - 3, you can construct the infinite set {0.1, 0.14, 0.141, ...}
Since we can extract an infinite quantity of numbers that terminate from every number that does not terminate, then it would be impossible for the set of numbers with an infinite number of digits to have a greater cardinality than the set of numbers that can be represented with a finite number of digits.
Cardinality of the set of irrational numbers
ReplyDeleteGiven the set of irrational numbers then for every number in that set we can find a rational counterpart in a subset of the rational numbers.
Therefore it is impossible for the set of irrational numbers to have a cardinality greater than the set of rational numbers.
How to do it?
We have a function or algorithm that takes an irrational number and returns a rational number that is not on our list of rational numbers.
We begin with our first irrational number i = 0.314159265359...
We will then search our list of rational numbers for the rational number 0.3
Since the list is empty we will not find it, so we can pair the two numbers.
0.3 -> 0.314159265359…
Then we get another irrational number i = 0.3141421356237…
We will again search our list for the rational number 0.3
Since the number is already on our list of rational numbers, we will get the next digit in our irrational number
We will then search our list for the rational number 0.31
Since the list does not contain the number 0.31, then we can pair the two numbers.
So our lists will now look like this
0.3 -> 0.314159265359…
0.31 -> 0.3141421356237…
We will add one more irrational number i = 0.314271828182846…
We will again search our list for the rational number 0.3
Since the number is already on our list of rational numbers, we will get the next digit in our irrational number
We will then search our list for the rational number 0.31
Since the number is already on our list of rational numbers, we will get the next digit in our irrational number
We will then search our list for the rational number 0.314
Since the list does not contain the number 0.314, then we can pair the two numbers.
So our lists will now look like this
0.3 -> 0.314159265359…
0.31 -> 0.3141421356237…
0.314 -> 0.314271828182846…
Since all the numbers on the left are finite in length and all the numbers on the right are infinite in length, we can always find a unique rational number by taking the first n digits from the irrational number until we have a number that is not on the left.
If you were to then diagonalize the list of irrational numbers on the right, and change every digit to something else, you would get a number not on that list but you could then find a rational number not on the left side to pair with the new number that you created.
Therefore the cardinality of the irrational numbers can not be greater than the rational numbers.
As for the power set being of a greater cardinality than the original set that it is made of. This is true for any finite set, but not infinite sets.
ReplyDeleteCantor had a proof by contradiction to show that a set can’t be put into a 1-1 with it’s power set, but we can make a one to one pairing of the natural numbers with it’s power set like this.
1 <-> {}
2 <-> {1}
3 <-> {2}
4 <-> {2,1}
5 <-> {3}
6 <-> {3,1}
7 <-> {3,2}
8 <-> {3,2,1}
9 <-> {4}
etc…
We can then construct a special set S that is made up of all the natural numbers that are not members of the set that they are paired with.
Then the question will be 'what natural number z will be paired with S' ?
Since none of our natural numbers are members of the sets that they are paired with, we will have to say that z is not a member of the set it is paired with either. This gives us a problem, if z is not a member of S then it should be a member of S.
But, when we examine the set S we find that it is the set of natural numbers, it is the universal set, so z will be a member of the universal set. (all numbers are)
There is no contradiction.
It is possible to list the real numbers (between 0 and 1) like this, where D(x,y) is a digit and D(x,y) = {1,2,3,4,5,6,7,8,9 or 0} and no row has all the same values for all the D’s.
ReplyDeleteRow (1) = 0. D(1,1) D(2,1) D(3,1) … D(n,1)
Row (2) = 0. D(1,2) D(2,2) D(3,2) … D(n,2)
Row (3) = 0. D(1,3) D(2,3) D(3,3) … D(n,3)
...
Row (n) = 0. D(1,n) D(2,n) D(3,n) … D(n,n)
Row (n+1) = 0. D(1,n+1) D(2,n+1) D(3,n+1) … D(n,n+1)
...
Row (10^n) = 0. D(1,10^n) D(2,10^n) D(3,10^n) … D(n,10^n)
Then as n -> infinity, you will have a list of all the real numbers.
Cantor created a number by taking the diagonal of this list…
D = 0. D(1,1) D(2,2) D(3,3) … D(n,n)
He then created a new number by mutating every digit in the diagonal and claimed that this number is not in the list of real numbers, when in fact it was just not in the first n rows of the list, and it was in some row beyond row n. So he did not create a number not in the list of real numbers.
Cantor could have just padded all the columns with trailing zero’s so then he could have extended the diagonal across 10^n columns, but then the list would have 10^(10^n) rows. This strategy would be self defeating.
What he actually did was rather strange, he assumed that n = 10^n when n is infinite, and only then was he able to diagonalize the matrix with n columns and 10^n rows. But, if your proof depends on the identity n = 10^n and your conclusion is that identity is wrong then he defeats his own argument.
Either way Cantor’s proof fails. There are no multiple infinities.
P1 All real numbers have an infinite number of digits. (some have an infinite number of trailing zero’s)
ReplyDeleteP2 All natural numbers have a finite number of digits.
From each of the numbers in the set of real numbers we can form a unique natural number by taking the first string that is not on our list of natural numbers. We can add this unique natural number to our list of natural numbers, and continue this forever with all the real numbers. Therefore we can find a unique natural number to pair with every real number. And the cardinality of the real numbers can’t be greater than that of the natural numbers.
Example:
A) 1 -> 0.14159265…
B) 14 -> 0.14159266…
C) 141 -> 0.14159267…
D) 1410 -> 0.141
I think that all the real numbers (that don’t terminate) can be listed in many orders, here is one such order...
ReplyDeleteTo list all the real numbers in order we need to use Cantor’s snake procedure. (I like it better in binary, but it is easier to understand in base 10)
First write down 1
0.1
Then 2 and 3
0.13
0.2
Then 4,5 and 6
0.136
0.25
0.4
Then 7,8,9 and 1 (the 10 is split in bottom and top row)
0.1361
0.259
0.48
0.7
Then 0, 11 and 12
0.13612
0.2591
0.481
0.71
0.0
Then 13,14, 15
0.136125
0.25911
0.4814
0.711
0.03
0.1
Then 16,17,18 and 1 (from 19)
0.1361251
0.259118
0.48141
0.7117
0.031
0.16
0.1
Then 9,20, 21, 22 and 2 ( from 23)
0.13612512
0.2591182
0.481412
0.71171
0.0312
0.160
0.12
0.9
Then 3, 24, 25, 26 and 27
0.136125127
0.25911822
0.4814126
0.711712
0.03125
0.1602
0.124
0.92
0.3
When you continue this forever, you will have all the real numbers between 0 and 1 that do not terminate. A proof of this conjecture would be nice, but that is what God created grad students for.
0.1361251272340762899019124
0.259118223735547099110113
0.48141261349661888511751
0.7117123624537989018123
0.031250348560787110113
0.16023514527879401641
0.1249347469786017122
0.922404516769110113
0.38346368785391531
0.2394506659006121
0.335267784110113
084596549281421
0.4156783005120
0.586439110113
0.55782171311
0.6329004129
0.781110112
0.19071201
0.0003128
0.110112
0.61191
0.2117
0.112
0.81
0.6
Cantor’s goal was to find a unique natural number for every real number and vice versa, it was not to find a 1 to 1 correspondence between the naturals and reals (though that would have established his point, it was a sufficient condition but not a necessary condition). This is known as the Schröder–Bernstein theorem. For two sets to be equipotent, you need only show that there exists injective functions f : A → B and g : B → A then |A| = |B|
ReplyDeleteIt is easy to find a unique real number given a natural number, e.g. 12 → 12.0 or 12 → 12.14159… or generally n → n.xyz… where n is a natural number and xyz… is any string of digits. Or, we can find a real number in (0,1) for every natural number. e.g. 1230 → 0.0321 or 985356295141 → 0.141592653589
But, finding a unique natural number given a real number is not as easy, yet it can be done.
Cantor assumed that his goal was to find a one to one, (this was his blind spot) when his goal was actually to just find a unique natural number to correspond to each and every real number in any set of real numbers, and this is doable.
Cantor assumed that it was necessary to list all the natural numbers, in order and on one list. This assumption is erroneous. It is like a rich man going to a small country auction, if he bids all his money ($1 million) on the first item, then he will be able to buy only that item, but if he bids in smaller increments, he will be able to buy all the items up for auction.
Cantor could have paired his purported list of ‘ALL’ real numbers in (0,1) to a small subset of the natural numbers like this.
4 → 0.5123453…
44 → 0.5125674…
444 → 0.5124195…
4444 → 0.5123676…
44444 → 0.5127295…
…
Then when he finds, by diagonalization, that he has not listed all the real numbers we could simply say “so what, we have not listed all the natural numbers either”. So, Cantors proof is thus inconclusive.
We can then ask Cantor to take the diagonals of the elements in his list, change the digits and make a second infinite list of real numbers that were not on his first list. And we can then pair those real numbers to some other subset of the natural numbers like this.
14 → r1
144 → r2
1444 → r3
14444 → r4
144444 → r5
…
Again, we can ask Cantor to take the diagonals of the elements in his second list, change the digits and make a new infinite list of real numbers, that were not on this second list (being careful to not duplicate any numbers from his first list). And we can then pair those real numbers to yet another subset of the natural numbers like this.
24 → rr1
244 → rr2
2444 → rr3
24444 → rr4
244444 → rr5
…
And, naturally, we can keep this game going forever. The interesting thing about this is, that when the prefix of our natural numbers approaches infinity, then it will become harder to find a real number, by changing the digits of the diagonal numbers, that is not on one of the infinite quantity of lists of real numbers. Also notice that most of the natural numbers like 333, 777, 123789 etc do not appear anywhere in our pairings, but that is okay, since we have enough natural numbers to pair to our sets of irrational numbers without them. All this tells us is that it is possible that the set natural numbers is bigger than that of the set of real numbers in (0,1) or that they are equal in size.
Now that we have cast doubt on Cantors diagonal proof, we can show how it is possible to find a function that does pair up the natural numbers to the irrational numbers. (the rational numbers can be paired, so we will ignore them for now)
ReplyDeleteWe will generate a set of random irrational numbers between 0 and 1, our list might look like this.
0.5123453…
0.5125674…
0.5124195…
0.5127676…
0.5124295…
0.05127295…
…
Then we can imagine a deck of cards where the first card has the number 1 on it, the second card has a 2, third a 3 and so on, a stack with all the natural numbers in order.
The first number on our random list is 0.5123453… so we can pair it with the fifth card in our deck.
The second number on our list is 0.5125674… since we have used the 5 card, we must now use card 15. (the first two digits in reverse)
The third number on our list is 0.5124195… since we have used the 5 and 15 card, we must now use card 215. (the first three digits in reverse)
The fourth number on our list is 0.5127676… since we have used the 5, 15 and 215 card, we must now use card 7215. (the first four digits in reverse)
The fifth number on our list is 0.5124295… since we have used the 5, 15 and 215 card, we must now use card 4215. (the first four digits in reverse)
And so on.
So we have our pairing.
5 → 0.5123453…
15 → 0.5125674…
215 → 0.5124195…
7215 → 0.5127676…
4215 → 0.5124295…
50 → 0.05127295…
…
If you ponder on it for a while, you will see that there will never be an irrational number that we can’t find a natural number to pair with it.
We could sort the natural numbers and keep the irrational that it is paired with, together with it, and then look at the nth digit of the nth real number that it is associated with. For example,
1 → 0.14159265359798…
2 → 0.23606797749979…
3 → 0.3166247903554…
4 → 0.414213562373095…
5 → 0.5123453…
…
then change the diagonal digits to something else and we would get an irrational number that is not in our list of irrational numbers.
Since the number that we create is different from every number in our list of irrational numbers, it must be different at some finite point. (it is impossible for two irrational numbers to have the first infinite number of digits in common, then to be followed by some digits that are different.)
Since this diagonal defines a finite string of digits, that has not been seen on the right side of our set, then we know that same finite string does not appear on the left side either, since the left side is simply a reflection of the digits that have appeared on the right side. So we know that there exists a natural number equivalent to that string (or a substring of that string) that will be available to pair with the irrational diagonal number that introduced that finite string. Therefore, the cardinality of the two sets is the same.
That is about it, if you have any questions or comments, I would be glad to hear from you.
It has been said that all infinite sets are endless, but some are more endless than others. Hmmm?
A proof is not about language, it is about understanding. If you have a perfect proof that you don’t understand, what good is it. It would simply allow you to believe something that you don’t understand, this is called faith based mathematics.
ReplyDeleteConsider this “proof” that concludes the cardinality of the irrationals are less than or equal to a set of rationals. Is it formal? Is it valid? Is it understandable?
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