Can someone put this decimal expansion equation into layman's terms please?

Question

I am studying this page on decimal expansion. I understand everything up until the strange equation following this sentence.

The strange equation is directly above this sentence. I do not think it cares that I am typing about it or that you are looking at it and reading about it. Now, I found a couple of possible meanings for the three horizontal lines.

According to this page this is a "congruence relation". This same page defines this symbol and other symbols as meaning "is equal to by definition to" which I think means basically a synonym. Maybe both of these definitions mean the same thing I do not know because this type of math is obviously not my forte. This is why I pointed my web browser to math.stackexchange.com.

Also does the | symbol mean "or" as in a programming language or is this a divisor, the division symbol?

My theory, based on the help from a colleague work, is that this means r equals p divided by q. Is this correct?

Then the equation is followed by other equations that somehow prove the equation presumably. These equations are as follows:

How do these equations relate to the r equals p divided by q equation? Can someone provide a simple example using these equations or show how they work and what they are used for?

Thank-you for reading this. I apologize for my severe ignorance in these matters.

Answer 1

$r \equiv p/q$ simply means "$r$ is equivalent to $\dfrac{p}{q}$". That is, they're saying let $r$ be some rational number $\dfrac{p}{q}$. They then go on to say that if $r $ has a finite decimal expansion (the digits don't go on endlessly after the decimal point), then

$$\begin{array} \ r &=& \frac{a_1}{10}+\frac{a_2}{10^2}+ \dots + \frac{a_n}{10^n} \\ &=& \frac{a_1 10^{n-1} + a_2 10^{n-2} + \dots + a_n}{10^n} \\ &=& \frac{a_1 10^{n-1} + a_2 10^{n-2} + \dots + a_n}{2^n \cdot 5^n} \\ &=& \frac{p}{2^\alpha 5^\beta} \end{array}$$

with $p \not\equiv 0 \text{ mod } 5$ and $p \not\equiv 0 \text{ mod } 2$ (meaning $p$ is not divisible by $5$ or $2$).

As an arbitrary example, consider $r=\frac{83148}{100000}=0.83148$. We then have $$\begin{array} \ r &=& 0.8 + 0.03+0.001+0.0004+0.00008\\ &=& \frac{8}{10}+\frac{3}{10^2}+\frac{1}{10^3}+\frac{4}{10^4}+\frac{8}{10^5} \\ &=& \frac{8 \cdot 10^{4} + 3 \cdot 10^{3} + 1 \cdot 10^{2} + 4 \cdot 10^{1} + 8 }{10^5} \\ &=& \frac{8 \cdot 10^{4} + 3 \cdot 10^{3} + 1 \cdot 10^{2} + 4 \cdot 10^{1} + 8 }{2^5 \cdot 5^5} \\ &=& \frac{20787}{2^3 5^5} \end{array}$$

The end goal was to show that all numbers with finite decimal expansions can be written in the above form, that is, the form $\frac{p}{2^\alpha 5^\beta}$.

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Although another instance of $\equiv$ which would make perhaps more sense arrises from modular arithmetic. Consider: $$8 \equiv 3 \ (\text {mod } 5)$$

This means that $5|(3-8)$, or that the remainder of $\dfrac {3}{5}$ is the same as $\dfrac {8}{5}$. So, perhaps now you see how they're "equivalent".