Infinite and finite representation of the same real number

46 Views Asked by At

I'm reading The Foundations of Mathematics and trying to understand the proof of the finite and infinite representation of the same real number.


For example, if a number x has two different decimal expansions, then, without loss in generality, we can take $x = a_0 · a_1 . . . a_{n–1}a_n . . . = a_0 · a_1 . . . a_{n-1}b_n . . .$ where $a_n < b_n$.

Multiply through by $10^n$ to get $a_0a_1 . . . a_{n–1}a_n · a_{n+1} . . . = a_0a_1 . . . a_{n–1}b_n · a_{n+1}. . .$ where $a_n < b_n$.

Subtracting the whole number $a_0a_1 . . . a_{n–1}a_n$ gives $0 · a_{n+1} . . . = k · b_{n+1}$ . . . where $k = b_{n+1} – a_{n+1} > 0$ is a positive integer.

But the first decimal is $0 · a_{n+1} . . . < 0 · 999 . . . ≤ 1$ and the second exceeds the positive integer k. So they can be equal only if $k = 1$ and both decimals represent the same limiting value 1. In this case, $a_{n+1} = a_{n+2} = · · · = 9, b_{n+1} = b_{n+2} = · · · = 0$ and $b_n = a_n + 1$.


Specifically, how do they produce $0 · a_{n+1} . . . = k · b_{n+1}$ . . . where $k = b_{n+1} – a_{n+1} > 0$ is a positive integer? Shouldn't it be $0 · a_{n+1} . . . = k · a_{n+1} . . .$ where $k = b_{n} – a_{n}$ ?

Also which decimals are meant by "the first" and "the second"?

Link to the full explanation