# Number System-Real Numbers

Author:
Rahul Chandra Saha
M.Sc., Mathematics

### Revisiting Irrationals

Recall that irrationals are those numbers which fails to represent them as $\frac{p}{q}$, where $p$ and $q$ are integers and $q$ is non-zero.
There are some notions by which we can mark a number as rational or irrational by simply observing them. E.g. if$\sqrt{m}$ is rational if it is a perfect square,otherwise an irrational. To prove our claim we need to prove the next theorem first.

Theorem:  If $p$ is a prime number and $p$ divides $a^2$ then $p$ divides $a$, where $p$ is a positive integer.

Proof:
Let the prime factorisation of $a$ be $a=p_1 p_2 p_3….p_n$ where $p_1,p_2,…p_n$ are primes(not necessarily all are distinct).
Therefore, $a^2=(p_1 p_2 p_3….p_n)(p_1 p_2 p_3….p_n)$ $=p_1^2p_2^2….p_n^2$.
Now $p$ divides $a^2$. Thus by fundamental theorem of arithmetic, $p$ is one of the $p_i’s$. Say $p=p_k$.
Therefore $p$ divides $a$.(Q.E.D)

Think Yourself:
Q1. Can we replace the positive integer condition by integers only?

Let’s concentrate on our claim.

Theorem:
$\sqrt{2}$ is an irrational number.

Proof:
If possible, assume that $\sqrt{2}$ is rational. Then, $\sqrt{2}= \frac{r}{s}$ for some integers $r,s$. Upon dividing $r,s$ by $HCF(r,s)$ we will get two coprime numbers, say, $a,b$.
Then,
$\sqrt{2}= \frac{a}{b}$
$\implies 2b^2= a^2.$
$\implies 2$ divides $a^2.$
$\implies 2$ divides $a$(by previous theorem)
$\implies a=2c$ for some integer $c$.
Substituting $a$, we get $2b^2=4c^2 \implies b^2 = 2c^2$. This shows that $2$ divides $b^2$ and thus $b$. Which contradicts that $a,b$ are prime to each other as $2$ divides both $a$ and $b$. Thus our assumption is wrong. So $\sqrt{2}$ is an irrational number.(Q.E.D)

$\textbf{Note:}$ The statement can be proved replacing $2$ by $p$, where $p$ is not a perfect square with similar arguments.(Give a try!)

Example 1. Show that $5- \sqrt{2}$ is irrational.
Solution: If possible assume that $5- \sqrt{2}$ is rational. Then there is co-primes $a,b$ such that $5- \sqrt{2} = \frac{a}{b}.$ Then, $\sqrt{2}= \frac{5b-a}{b}.$ Since $a$ and $b$ are integers then so $5b$ and thus $5b-a$ are integers. Thus $\sqrt{2}$ is an rational number. Which is a contradiction. Thus $5- \sqrt{2}$ is irrational.

Example 2. Show that $2\sqrt{3}$ is irrational.
Try yourself.

### Revisiting Rationals and their decimal expansions

The rationals can be represented in two forms. One is the form $\frac{p}{q}$ and other is decimal representation of them. Let us see some examples,
$\frac{1}{2}= 0.5, \frac{2}{3}=0.66666…$
The first fraction terminates but the second one fails to terminate itself in decimal representation. What reasons causes this contrast? The answer is in the next theorems.

Theorem:
Let $x$ be a rational number whose decimal expansion terminates. Then $x$ can be expressed in the form, $\frac{p}{q}$, where $p,q$ are coprimes,and the prime factorisation of $q$ is of the form $2^n5^m$, where $n,m$ are non-negative integers.

Theorem:
Let $x=\frac{p}{q}$ be a rational number, such that the prime factorisation of $q$ is of the form $2^n5^m$, where $n,m$ are non-negative integers. Then $x$ has a decimal expansion which terminates.

E.g. Consider the fraction $\frac{57}{200}.$ Then $200=2^3 \times 5^2$.
Thus here $q= 2^3 \times 5^2$. Now, $\frac{57}{200}= 0.285$, which is terminating.

Theorem:
Let $x=\frac{p}{q}$ be a rational number, where $p,q$ are coprimes. If the prime factorisation of $q$ is not of the form $2^n5^m$, where $n,m$ are non-negative integers,then $x$ has a decimal expansion which is non-terminating repeating(recurring) and conversely.

E.g. Consider $\frac{7}{3}$. Then the denominator is not in the form $2^n5^m$. Also, $\frac{7}{3}=2.33333…$ is not terminating. Notice that 3 is repeating itself. This is an example of recurring decimal.

Think Yourself: If the numbers after decimal neither terminate nor recurring then what will be the number rational or irrational?