The Real Numbers and the Cauchy Sequence

To further continue with our path into studying the p-acid numbers, we must be able to construct it to have a deeper understanding of it. In order to do so, we break from learning the properties and how the p-adic numbers work. Instead we will take time to observe what a Cauchy sequence is and it’s relation to the p-adics. Continuing on, we will first define the limits of the Cauchy sequence of the rational numbers.

$Definition$. For a given sequence of rational numbers ${a_n} = ({a_1, a_2,…})$, it is known as being real Cauchy sequence if for some $ε > 0$ , there exists a positive integer $N$ such that for all $i,j>N, |a_i-a_j|_{\infty} < ε$.

Let’s consider the following real number $1.4142135…$. We can represent it by the limit of the sequence as

$a_1 = 1 = 1*10^0$

$a_2 = 1.4 = 1*10^0 + 4*10^{-1}$

$a_3 = 1.41 = 1*10^0 + 4*10^{-1} + 1*10^{-2}$

$a_4 = 1.414 = 1*10^0 + 4*10^{-1} + 1*10^{-2} + 4*10^{-3}$

$…$

Can the reader recognize the sequence? The expansion of $1.4142135…$ is the converging Cauchy sequence representation of the constant $\sqrt{2}$. This sequence can also be written as $({1, 1.4, 1.41, 1.414, 1.4142, …})$. To be completely sure that $\sqrt{2}$ is a Cauchy sequence, next we set $ε = 0.05$. We see that $N = 2$ and verify this by computing $|1.41-1.414|_{\infty} = 0.004 < 0.05 = ε$ which is true.

If we are given two real Cauchy sequences ${a_n}$ and ${b_n}$, we know that the both of the sequences represent the same real number if ${|a_n – b_n|}$ converge to 0. This is fairly easy to demonstrate and see. Taking ${a_n} = ({1, 1, 1, 1, 1, …})$ and ${b_n} = ({0.9, 0.99, 0.999, 0.9999, 0.99999, …})$ as an example, we see that ${|a_n – b_n|} = ({0.1, 0.01, 0.001, 0.0001, 0.00001, …})$ converges to 0 as $n\rightarrow{\infty}$ . The Cauchy sequence here for ${a_n}$ can seem counter intuitive at first if the reader is countering such case for the first time. After all, how can $0.999\bar{9} = 1$ be true? To clearly understand $how$ this works, we observe the sum of the infinite series for $0.999\bar{9}$.

$$0.999\bar{9} = lim_{n\rightarrow\infty} 9 \frac{1}{10} + 9\frac{1}{100} + 9\frac{1}{1000} + … + 9\frac{1}{10^n}$$

Now we apply the geometric sum given by

$$1 + r + r^2 + … + r^{n-1} = \frac{1 – r^n}{1 – r}$$

to our sequence to get

$$  = \frac{9 *(1/10)}{1- \frac{1}{10}} = \frac{9*10}{10*9} = 1$$

Our $r$ value here is $1/10$ because that is our common ratio. As we can see, it is true that $0.999\bar{9}=1$ for the sequence of $0.999\bar{9}$. Note that this is true because we consider the *infinite* sequence of $0.999\bar{9}$. If we did have a cutoff, then we could say that $0.9999$ is approximately equal to $1$ with a slight error. But that’s clearly not the case here.

Now that we have observed the Cauchy sequences of rational numbers, we will next study the p-adic numbers and how we construct it with our thus far accumulated knowledge regarding absolute value and the Cauchy sequence.