2.2 Decimal Expansions

When we think about numbers we usually think of a list of digits e.g. \(2024\) or \(1001\).

How does this picture of the integers compare with our description in terms of the axioms?

We would like to prove that every positive integer has a decimal expansion. We will give a proof of this later. Let us begin by asking what the “decimal expansion” of an integer means.

The axioms introduce the numbers \(0\) and \(1\), and we define the other possible digits \(2,3,\dots,9\) by

\(2:=1+1\)	\(3:=2+1\)	\(4:=3+1\)	\(5:=4+1\)	\(6:=5+1\)	\(7:=6+1\)	\(8:=7+1\)	\(9:=8+1\)

When we write \(2024\) this is a list of digits \(2,0,2,4\) which in reverse order are the units, tens, hundreds, thousands (and so on if we have a larger number). So really we are writing the number as a sum \(2\cdot 10^3 + 0\cdot 10^2+2\cdot 10^1 + 4\cdot 10^0\) (where of course \(10:=9+1\)):

Thousands	Hundreds	Tens	Units
\(2\)	\(0\)	\(2\)	\(4\)
\(2\cdot 10^3\)	\(0\cdot 10^2\)	\(2\cdot 10^1\)	\(4\cdot10^0\)

For an integer \(n\) its decimal expansion is a list of numbers \(d_k,d_{k-1},\dots,d_0\) with the properties that + each digit \(d_i\) is an integer such that \(0\leq d_i\leq 9\) + the first digit \(d_k\) is non-zero + \(n\) is equal to the following sum: \[n=\sum_{i=0}^k d_i\cdot 10^i\] The notation \(\sum_{i=0}^k d_i\cdot 10^i\) means the sum \(d_0\cdot 10^0 + d_1\cdot 10^1+\dots+d_{k-1}\cdot 10^{k-1}+d_k\cdot 10^k\). Of course we have also introduced the notation of powers to indicate repeated multiplication.

Note the following standard convention on powers: \[a^{b^c}\,\text{ means }\,a^{\left(b^c\right)}\] The reason why we interpret \(a^{b^c}\) in this way instead of as \((a^b)^c\) is that \((a^b)^c\) can be rewritten without brackets as \(a^{bc}\).

Implicit in the definition of the decimal expansion for a positive integer is the fact that the digits \(1,2,\dots 9\) are all positive. How do we know this? We’ll show that \(0<1\), from which we could also show that \(0<2,3\) etc.