# Experiments in Periodicity

When we wake up tomorrow, we know that the night has just past and day is creeping upon us. We know this because such event takes place everyday like a broken record. Plays over and over again. Observing such a phenomenon from a mathematical perspective, we say that we are looking at periodic functions.

As we can all remember from Trigonometry, the sine and cosine functions oscillate in a orderly fashion when we look at it graphically. By orderly fashion, I mean to say that we know the trend of the oscillation after observing it for a short period of time. We can also combine them, take the derivative, and integrate such functions to make a whole new function.

For this lab, we observe such periodic functions by calculating the area under the functions. In other words, we evaluate the definite integrals.

Previously, we discussed the absolute values on $Q$, the real numbers, Cauchy sequence, and basic process of computing the p-adic representation of a number. Now, we move ahead and discuss how to construct these p-adic numbers. We do this by using the absolute value in the usual sense as to how it is used with the reals.

$Definition$. The field of p-adic numbers $Q_p$ is defined to be the set of all equivalence classes of p-adic Cauchy sequence.

As mentioned earlier, we can represent p-adic numbers using the Cauchy sequence. With the given definition, now we know that for any p-adic Cauchy sequence, the sequence converges to a p-adic number. Natural question to arise would be to ask *how*. To address this *how* and to convince ourselves, we will consider a quick example before generalizing it.

Let’s take the 3-adic representation of $241$. We surly know that $241$ is a Cauchy sequence that converges to $241$ for {$241$, $241$, $241$, $…$} . The 3-adic representation of $241$ is $1 + 2*3^1 + 2*3^2 + 2*3^3 + 2*3^4$ or $(1, 2, 2, 2)$. This expansion represents the class of all Cauchy sequence equivalent to {$1$, $1+2*3^1$, ${1+2*3^1 + 2*3^3}$, $…$}. Looking closely, we notice the sequence as follows {$3$, $3^2$, $3^3$, $…$}. Taking the limit, we see that $lim_{n\to\infty}$ $3^n = 0$ for 3-adics. How? As we recall from absolute values of $Q$ for any prime $p$, we write $|p^0*k|=1$, $|p*k|=p^{-1}$, $|p^2*k|=p^{-2}$ and $|p^n*k|=p^{-n}$ in general for some constant $k$ as $n\to \infty$. It would make intuitive sense for $|3^n|=3^{-n}$ to approach zero as $n\to \infty$ because of the inverse. Let’s note the existence of the inverse relation we see here with the p-adic numbers. As $3^n$ grows without bounds, its 3-adic representation converges to zero.

Looking at the general case of p-adic Cauchy sequence, we see that any number in its p-adic representation will converge to zero and can be written as

{$a_n$} = {$a_0 * p^0, a_0 * p^0 + a_1*p^1, … , a_0 * p^0 + a_1*p^1 + a_2*p^2 + … + a_n*p^n$}

where $a_0, a_1, …, a_n$ represent the coefficients.  This p-adic expansion for the Cauchy sequence is abbreviated in the form of $a_0.a_1a_2a_{3,…,p}$. So, the 3-adic representation for $241$ can be abbreviated as $1.22\bar{2}_3$ where $0 \leq a_i \leq p-1$.

Now that we have a general knowledge of how the p-adic numbers work and how it’s constructed, we will consider another interesting representation of negative integers with p-adics. Let’s consider the p-adic expansion of $-1$ for any $p$. To start with, we begin by considering the 5-adic expansion for it.

Solution to $x + 1 = 0$ $mod$ $5$ $is$ $x = 4 = 4*5^0$

Solution to $x + 1 = 0$ $mod$ $5^2$ is $x = 24 = 4*5^0 + 4*5^1$

Solution to $x + 1 = 0$ $mod$ $5^3$ is $x = 124 = 4*5^0 + 4*5^1 + 4*5^2$

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Solution to $x + 1 = 0$ $mod$ $5^n$ is $x = 4*5^0 + 4*5^1 + 4*5^2 + … + 4*5^{(n-1)}$

Next, we consider the *7-adic* expansion of $-1$.

Solution to $x + 1 = 0$ $mod$ $7$ is $x = 6 = 6*7^0$

Solution to $x + 1 = 0$ $mod$ $7^2$ is $x = 48 = 6*7^0 + 6*7^1$

Solution to $x + 1 = 0$ $mod$ $7^3$ is $x = 342 = 6*7^0 + 6*7^1 + 6*7^2$

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Solution to $x + 1 = 0$ $mod$ $7^n$ is $x = 6*7^0 + 6*7^1 + 6*7^2 + … + 6*7^{(n-1)}$

$Conjecture$. The *p-adic* expansion of $-1$ for any $p$ is represented as $(p-1, p-1, p-1, …)$.

# 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.

Anyone who has not touched upon the p-adic numbers might find the topic rather confusing. Truth be told, it is confusing. The arithmetic part of the process can be a bit challenging due to its abstraction. Assuming that the reader is not familiar with the p-adic numbers, we will consider an example to clarify the matter.

Let’s consider the equation $x+64=0$. Any one individual might tell us that the solution is $x=-64$ when we ask them to solve the equation and of course, they would be right. But we wish to know the 5-adic representation of $-64$. We follow the following procedure:

Solution to $x + 64 = 0$ $mod$ $5$ is $x = 1 = 1 * 5^0$

Solution to $x + 64 = 0$ $mod$ $5^2$ is $x = 11 = 1 * 5^0 + 2 * 5^1$

Solution to $x + 64 = 0$ $mod$ $5^3$ is $x = 61 = 1 * 50^0 + 2 * 5^1 + 2 * 5^2$

Solution to $x + 64 = 0$ $mod$ $5^4$ is $x = 561 = 1 * 50^0 + 2 * 5^1 + 2 * 5^2 + 4 * 5^3$

Solution to $x + 64 = 0$ $mod$ $5^5$ is $x = 3061 = 1 * 50^0 + 2 * 5^1 + 2 * 5^2 + 4 * 5^3 + 4 * 5^4$

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Solution to $x + 64 = 0$ $mod$ $5^n$ is $x = 1 *5^0 + 2 * 5^1 . . . 4 * 5^{n-1}$

For  $x + 64 = 0$ $mod$ $5$, we can instantly see that the solution is $x = 1$ since $1 + 64 = 65$ is evenly divisible by $5$. Notice that $x = 1$ can be written as $x = 1 * 5^0$. Now for $x + 64 = 0$ $mod$ $5^2$, we know that when we add $11$ to $64$ which gives us $75$, it will be divisible by $25$. Once again, notice that we can write $x=11$ as $11 = 1 * 5^0 + 2 * 5^1$. The same procedure applies from $x + 64 = 0$ $mod$ $5^3$ all the way up to $x + 64 = 0$ $mod$ $5^n$. In general, we take our $x$ value for $x + 64 = 0$ $mod$ $5^n$ and write the linear combination for it in the form of $x = 1 *5^0 + 2 * 5^1 . . . 4 * 5^{n-1}$.

Following the same procedure, we represent $-64$ in its 5-adic representation as $(1,2,2,4,4,…)$. Although quite interesting, the arithmetic for p-adic representation can be a tedious process. Now that we have seen an example, next we will consider p-adic representation of rational numbers and absolute value on $Q$.

From the start, most students are introduced to rational and irrational numbers. We are taught how to construct these numbers by setting axioms and properties. We also learn about prime numbers and the properties associated with it. But at the end before most students graduate, they may not even know that there exists a whole set of different type of numbers known as the p-adic numbers.

The p-acid numbers builds upon the arithmetic of the rational numbers. As for the arithmetic of p-acid numbers, where $p$ is a prime, the process is different than what we have been used to seeing. The difference comes from the alternative definition of absolute values on $Q$.

Find the $7$-acid expansion of $-1$.

# Square root of 2 is irrational?

My name is Farighon. I have a deep interest in the field of Combinatorics and Number Theory although I do like to take time to study Complex Analysis when I do have some free time. Since the inception of my interest in mathematics, I have always been interested and fascinated by proofs for theorems, lemmas, and propositions. But I would be lying if I said that I understood each proof that I have come across. However, there is one proof that has made perfect sense. It is none other than the proof for the square root of 2.

The proof for the square root of 2 being irrational has been one of the primary interests of ancient mathematicians starting with the Babylonian’s. Then the ancient Indians. Although later on it was proven by a simple yet elegant proof that square root of 2 is irrational, the hunt for determining the square root of 2 to as many decimal places as possible is an ongoing task for mathematicians teamed up with Computer Scientists. After all, who can expect a mathematician to be ever satisfied when perfection is what shapes and disciplines them?

The article for the proof of $\sqrt(2)$ can be found at the following article for the curious reader: http://en.wikipedia.org/wiki/Square_root_of_2.