Irrational Numbers and Closure

What is an Irrational Number?

To understand irrational numbers, we first need to remember what a rational number is. A rational number can be written as a simple fraction $\frac{a}{b}$, where $a$ and $b$ are integers and $b \neq 0$. This includes integers, terminating decimals, and repeating decimals.

An irrational number is a real number that cannot be written as a simple fraction. Their decimal expansions never end and never repeat.

Examples:

• $\sqrt{5}$: Irrational. It is not a perfect square, so it cannot be simplified into a neat fraction.
• $0.\overline{3}$: Rational. The line means the 3 repeats forever, which is equivalent to the fraction $\frac{1}{3}$.
• $\pi$: Irrational. It is approximately $3.14159...$ with no repeating pattern.
• $-7$: Rational. It can be written as the fraction $\frac{-7}{1}$.

The Closure Property: Mixing Rational and Irrational Numbers

What happens when you add or multiply rational and irrational numbers together? There are two golden rules you need to know:

1. The sum of a rational number and an irrational number is always irrational. For example, if you add the rational number $4$ to the irrational number $\pi$, the result is $4 + \pi$. This new number is irrational.

2. The product of a nonzero rational number and an irrational number is always irrational. If you multiply $2$ (a rational number) and $\sqrt{5}$ (an irrational number), the result is $2\sqrt{5}$, which is irrational. Note that the rational number must be nonzero, because multiplying any number by $0$ results in $0$, which is rational.

Proving the Sum Rule

How do we know that adding a rational and an irrational always gives an irrational result? We can prove it using a method called proof by contradiction.

Let $r$ be a rational number and $x$ be an irrational number. We want to prove that their sum, $r + x$, is irrational.

1. Let's pretend the opposite is true: assume $r + x$ results in a rational number. Let's call this sum $s$. $r + x = s$
2. If we rearrange the equation to solve for our irrational number $x$, we get: $x = s - r$
3. We know that $s$ and $r$ are both rational. The difference between any two rational numbers is always another rational number.
4. Therefore, $s - r$ must be rational. But this means $x$ is rational!

This contradicts our starting fact that $x$ is irrational. Because our assumption led to an impossible conclusion, our original statement must be true: the sum of a rational and an irrational number is always irrational.