Cantor’s argument also allows us to “explicitly” construct irrational numbers. Let’s take a list which includes all the rational numbers between \(0\) and \(1\). First we add \(1\), and then we take \(a/b\) first over the possible denominators \(b > 1\) and then taking all possible numerators \(a\) from \(1\) to \(b-1\). Let’s not even worry about duplicates! So we could take \(r_1, r_2, r_3, \ldots \) to be:
\[1/2,1/3,2/3,1/4,2/4,3/4,1/5,2/5,3/5,4/5,1/6,2/6,3/6,4/6,5/6,1/7, \ldots \]
Taking the opposite convention to class, let’s consider the decimal expansion of these numbers which, if it ends in \(999999\ldots\) we round up. Now we take a real number \(r\) defined as follows:
- If the \(n\)th digit of \(r_n\) is equal to to \(1\), we make \(r\) have the digit \(2\),
- If the \(n\)th digit of \(r_n\) is *not* equal to to \(1\), we make \(r\) have the digit \(1\),
So here is the list:
\[\begin{aligned}
r_1 & = 0.5000000000 \ldots \\
r_2 & = 0.3333333333 \ldots \\
r_3 & = 0.6666666666 \ldots \\
r_4 & = 0.2500000000 \ldots \\
r_5 & = 0.5000000000 \ldots \\
r_6 & = 0.7500000000 \ldots \\
r_7 & = 0.2000000000 \ldots \\
r_8 & = 0.4000000000 \ldots \\
r_9 & = 0.6000000000 \ldots \\
r_{10} & = 0.8000000000 \ldots \\
\end{aligned}
\]
and so say using the mathematica code
\[ \begin{split}
\texttt{Q := Flatten[Table[a/b, {b, 2, 20}, {a, 1, b – 1}]]} \\
\texttt{Table[If[Mod[Floor[Q[[n]] 10^n], 10] == 1, 1, 2], {n, 1, Length[Q]}]}
\end{split}
\]
We find that
\[r = 0.222222222222222211212222222212222222222222222212222221222222222222222\ldots \]
is an irrational number, because if it were rational in \(0,1\) then it would be equal to \(r_n\) for some \(n\),
but the \(n\)th digit of \(r\) is not equal to the \(n\)th digit of \(r_n\) by construction.
Another fact about decimal expansions of rational numbers is that they are (eventually) repeating. This gives (yet another) way to construct them using decimal expansions.