A proof about rational numbers

Question

The rational numbers are exactly those numbers whose decimal representations either terminate (i.e. all digits are 0 from some point onwards) or repeat.
How can I prove this statement?
Any help would be appreciated.Thank you!

Answer 1

Maybe going through an example will help you with this. We can rewrite real numbers which have a repeating, occurrence of numbers to the right of the decimal as a fraction. For example, the number $0.125125125125...=0.\overline{125}$. Define this number as $x=0.125125125125...$. Thus, $1000x=125.125125125...$ . Furthermore, $1000x-x=999x=125.\overline{125}-0.\overline{125}$. Looking at the far right expression, we know $125.\overline{125}-0.\overline{125}=125+0.\overline{125}-0.\overline{125}=125$. Thus, $999x=125$. Thus, $x=\frac{125}{999}$ which is a fractional representation of a real number. I should also add that just simplifying a terminating sequence of numbers to the right of the decimal should be apparent. For example, $0.36=36/100$. When numbers are to the left of the decimal, we find a common denominator. Hope this helps.

This might be more technical than what you want but here it goes. Take a repeating sequence of numbers we will say $0.\overline{x_1x_2...x_n}$ for some $n\in \mathbb{N}$ where all the xs are numbers in the set $\lbrace 0 , 1, 2, 3, 4, 5, 6, 7, 8, 9\rbrace$. Define this number as $y=0.\overline{x_1x_2...x_n}$. Thus, $10^ny=x_1x_2...x_n.\overline{x_1x_2...x_n}$. Whence, $(10^n-1)y=x_1x_2...x_n$. Thus, $y=\frac{x_1x_2...x_n}{10^n-1}$. Since $x_1x_2...x_n\in \mathbb{Z}$ and $10^n-1\in \mathbb{Z}^{\neq 0}$, then $y$ must be a rational number. I would then say without loss of generality you could do the same thing where this number is negative.

This proposition is used for the case where there are numbers to the left of the decimal. It should be easy to see that in the case that one of the numbers was negative that the same would occur.

Proposition: The sum of two rational numbers is rational.

Proof: Let $\frac{a}{b}$ be a rational number where $a$ is an integer and $b$ is an integer not equal to zero. Also, let $\frac{c}{d}$ be a rational number where $c$ is an integer and $d$ is an integer not equal to zero. Thus, $\frac{a}{b}+\frac{c}{d}=\frac{ad}{bd}+\frac{bc}{bd}=\frac{ad+cd}{bd}$. Thus, the numerator and denominator for the far right expression are both integers where the denominator does not equal zero. Thus, the theorem has been proven.