Permalink Submitted by Raymond.cavalla... on December 16, 2019

I guess what you say must be the consensus in math community, but still the math does not feel right for me.

Let’s start with comparing the countable fractions with the uncountable Booleans.

In the countable fractions the axis are both natural numbers, so they have both the same “amount” or “proportion”, that means that by taking a diagonal, you are getting the same “rate”, that seems counter intuitive dealing with infinite but it means that each item on the rows will interact with all the itens in the cols and vice versa.

On the Boolean side if you had a finite amount n in the cols (numbers) you would have a 2^n rows, so taking a diagonal would leave a lot of solutions off the diagonal, if you think in other way, as all Booleans would represent an integer number so you can see that taking all the Booleans is infinite but countable.

So I conclude that the diagonal analysis shows that something is uncountable or there is a difference in the rate of the axis.

In the case of the halting table Tn(i) , I came to understand that the n “amount” or “rate” has more itens than the natural number infinity i, but that is not all, the “halt program” could not be a universal Turing machine unless it could take at least n*i inputs.

## math

I guess what you say must be the consensus in math community, but still the math does not feel right for me.

Let’s start with comparing the countable fractions with the uncountable Booleans.

In the countable fractions the axis are both natural numbers, so they have both the same “amount” or “proportion”, that means that by taking a diagonal, you are getting the same “rate”, that seems counter intuitive dealing with infinite but it means that each item on the rows will interact with all the itens in the cols and vice versa.

On the Boolean side if you had a finite amount n in the cols (numbers) you would have a 2^n rows, so taking a diagonal would leave a lot of solutions off the diagonal, if you think in other way, as all Booleans would represent an integer number so you can see that taking all the Booleans is infinite but countable.

So I conclude that the diagonal analysis shows that something is uncountable or there is a difference in the rate of the axis.

In the case of the halting table Tn(i) , I came to understand that the n “amount” or “rate” has more itens than the natural number infinity i, but that is not all, the “halt program” could not be a universal Turing machine unless it could take at least n*i inputs.

but I have no bases for that is just a thought.