# is it possible to proof that this number is not rational

It is an idea I had when reading the proof that $(0,1)$ is uncountable. There the numbers in $(0,1)$ are written into a list in decimal expansion and then the diagonal is modified and the resulting number is a number not on the list.

Now instead consider $S= \mathbb Q \cap (0,1)$. Like in the proof about $(0,1)$ write the numbers into a list in decimal expansion. Add $1$ to every digit on the diagonal and compute the remainder modulo $10$. I am trying to proof that this new diagonal number is not rational but without using the knowledge that the rationals are countable. Here is the proof:

There are two cases: A rational has either a finite expansion or is periodic. Let the digits in the expansions be called $a_{mn}$. Let the new diagonal number after the modification be $d_n$.

In the first case: If $a_{nn}$ is finite then $d_n$ is finite and $d_n = 0$ for $n>N$ for an $N$. Then, it is possible to find $n+1$ different periodic rational with no $0$ in the expansion. But this is a contradiction. Therefore $a_{n n}$ can not be finite.

In the second case: if $a_{nn}$ is a periodic rational with period $c_1 c_2 \dots c_N$. Then because of a same argument like the finite case this period can not be constant ($N=1$). It is forced to contain all digits in $\{0, ..., 9\}$. But what now? Is it possible to finish this proof without using that the rationals are countable?

• My first intuition: no, it is not possible to show that the diagonal number has not a periodic representation, as nothing is said on how the rationals were ordered in the first place. – Carlos Eugenio Thompson Pinzón Nov 1 '13 at 11:39
• "If $a_{n,n}$ is finte" ? $a_{n,n}$ is a number, how could it be "finite"? I think you mean that the sequence $(a_{n,n})$ is eventually zero, but that doesn't imply that $(d_n)$ is eventually zero, in fact it implies the contrary. – Jack M Nov 1 '13 at 12:14

You start by explicitly assuming that $S=\mathbb Q\cap (0,1)$ is countable as you write all these numbers into a list. If your list does not contain all elements of $S$ it is well possible that the antidiagonal number is in fact one of the rationals not in the list. For example if $S$ consists only of those rational numbers having at least one $2$ in their decimal expansion, it might happen that the diagonal number is simply $0.222\ldots =\frac29$ and your antidiagonal number becomes $0.333\ldots=\frac13$, which is rational but not an element of $S$. On the other hand, if $S$ really contains all eventually periodic decimal expansions then it is clear that the antidiagonal is not eventually periodic as it differs from each single element of $S$.
By the way, you should have a closer look at how you define your antidiagonal number: You might accidentally end up with $0.000\ldots=0$ or $0.999\ldots = 1$
• But is it not possible to write uncountably many numbers into a list? Are all lists in mathematics always assumed to be countable? Imagine a list with index $i$ in $\mathbb R$. Is it wrong? – blue Nov 1 '13 at 11:57
• When you list numbers, you are only using indexes $i$ in $\mathbb{N}$. – Shaun Ault Nov 1 '13 at 12:04
When you want to prove something without the assumption that the rationals in $\ ]0,1[\$are countable you should not begin with the assumption that you have an explicit counting $m\to r_m=(a_{mn})_{n\geq1}$ of them.
Nevertheless you can use your knowledge that the decimal expansion of any rational is eventually periodic, even in the case where there are two representations for the same number. It is enough to exhibit a sequence of digits which is not eventually periodic. E.g., the decimal fraction $$0.101001000100001000001000000100000001\ldots$$ encodes an irrational number.