Irrational numbers and the real line

If you draw a square of side $1$ and measure its diagonal, you get a length whose square is exactly $2$. So the diagonal is $\sqrt{2}$. The ancient Greeks fully expected this to be a fraction. It is not. The discovery that $\sqrt{2}$ cannot be written as $\dfrac{p}{q}$ was so shocking that legend says the discoverer was thrown overboard at sea.

Concept

A real number that cannot be written as $\dfrac{p}{q}$ for integers $p, q$ is called irrational. Famous irrationals include

$\sqrt{2}, \quad \sqrt{3}, \quad \sqrt{5}, \quad \pi, \quad e.$

Why $\sqrt{2}$ is irrational. Suppose, for the sake of argument, $\sqrt{2} = \frac{p}{q}$ in lowest terms. Squaring: $2 = \frac{p^2}{q^2}$, so $p^2 = 2 q^2$. The right side is even, so $p^2$ is even, so $p$ is even. Write $p = 2k$. Then $4k^2 = 2q^2$ gives $q^2 = 2k^2$, so $q^2$ is even, so $q$ is even. But then both $p$ and $q$ are even , contradicting "lowest terms". The original assumption must be wrong. So $\sqrt{2}$ is not rational.

Decimal test. A real number is irrational iff its decimal expansion neither terminates nor recurs. Examples (computed):

$\sqrt{2} = 1.4142135\dots, \quad \pi = 3.1415926\dots$

In each case the digits go on forever with no repeating block.

The real number line. Together, rationals and irrationals fill every single point of the number line, leaving no gaps. This complete number system is called the real numbers $\mathbb{R}$. Every length, every speed, every measurement you can imagine corresponds to a real number.

A useful image: imagine the rationals as a dense dust sprinkled on the line , between any two of them there are infinitely many more. But this dust still does not cover everything. The irrationals fill in the rest.

Operations and irrationals.

• Rational $+$ irrational $=$ irrational. (Adding a fraction to $\sqrt{2}$ cannot turn it into a fraction.)
• Rational $\times$ irrational $=$ irrational, as long as the rational is non-zero.
• Irrational $+$ irrational may be rational. Example: $\sqrt{2} + (1 - \sqrt{2}) = 1$.
• Irrational $\times$ irrational may be rational. Example: $\sqrt{2} \times \sqrt{2} = 2$.

Plotting $\sqrt{2}$ on the line. Construct a right triangle with legs $1$ and $1$. The hypotenuse has length $\sqrt{2}$ by Pythagoras. Lay the hypotenuse along the number line from $0$ , you have a precise geometric construction of $\sqrt{2}$.

Worked examples

Example 1. Is $\sqrt{9}$ rational or irrational?

• $\sqrt{9} = 3$, which is rational (in fact, an integer).
• Only square roots of non-squares (like $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}, \sqrt{10}, \dots$) are irrational.

Example 2. Show $5 - \sqrt{2}$ is irrational.

• Suppose it were rational. Call it $r = \frac{p}{q}$.
• Then $\sqrt{2} = 5 - r$. But $5 - r$ is rational (difference of two rationals), so $\sqrt{2}$ would be rational , contradiction.

Example 3. Place $\sqrt{2}$ and $\sqrt{3}$ between two consecutive integers.

• $1^2 = 1 < 2 < 4 = 2^2$, so $1 < \sqrt{2} < 2$.
• $1^2 = 1 < 3 < 4 = 2^2$, so $1 < \sqrt{3} < 2$.
• Both lie between $1$ and $2$, with $\sqrt{2} < \sqrt{3}$.

Example 4. Decide whether $0.10110111011110\dots$ (one more $1$ each time) is rational.

• The pattern never repeats with a fixed block (the $1$'s keep growing).
• So the decimal is non-terminating non-recurring, hence irrational.

Try it yourself

1. Identify each as rational or irrational: $\sqrt{16}, \sqrt{17}, \pi, 0.1\overline{6}, 2 + \sqrt{2}$.
2. Find two irrational numbers between $1$ and $2$.
3. Place $\sqrt{5}$ between two consecutive integers.
4. Show $3\sqrt{2}$ is irrational.
5. Find a rational number between $\sqrt{2}$ and $\sqrt{3}$.
6. If $a$ and $b$ are both irrational, is $a - b$ always irrational? Give an example to justify.
7. Why is $\sqrt{2} + \sqrt{3}$ irrational? (You may take this as believable , the proof is harder.)
8. Construct a length equal to $\sqrt{5}$ using two right triangles. Sketch your construction.

Activity / Insight

The irrational walk. Start at $0$ on a number line drawn on paper. Use a ruler to mark $1, 2, 3, \dots$ Now using a compass and Pythagoras, transfer the lengths $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \dots$ to the same line. (Try the "spiral of Theodorus" , successive right triangles each adding $1$ to the previous leg.) The plot fills with irrationals between the integers, showing that the rationals alone really were not enough.