Why the Square Root of 2 Is Irrational | Grade 8 Math | LevelKite

You already know that the equation $x^2 = 2$ has two solutions, $\sqrt{2}$ and $-\sqrt{2}$ , because $2$ is positive. The symbol $\sqrt{2}$ names the positive one: the positive number whose square is exactly $2$ .

Some square roots are whole numbers, such as $\sqrt{4} = 2$ . But $\sqrt{2}$ is not. Among nonnegative numbers (zero and the positive numbers), a larger number has a larger square. Since $1^2 = 1$ is less than $2$ and $2^2 = 4$ is more than $2$ , the number $\sqrt{2}$ lies between $1$ and $2$ .

So here is the question for this lesson: is there a fraction between $1$ and $2$ whose square is exactly $2$ ?

To answer that question, you need two terms.

A rational number is a number that you can write as a fraction $\frac{p}{q}$ , where $p$ and $q$ are integers (the numbers $\ldots, -2, -1, 0, 1, 2, \ldots$ ) and $q$ is not zero. This lesson calls such a fraction a fraction of integers. In decimal form, every rational number either terminates (its digits stop) or repeats (one block of digits repeats forever).

An irrational number is a number on the number line that is not rational. No fraction of integers equals it, and its decimal form neither terminates nor repeats.

The table shows some rational numbers. Notice that $\sqrt{4} = 2 = \frac{2}{1}$ is one of them.

Rational numbers. A bar marks the digits that repeat forever.

Fraction of integers	Decimal form	Terminates or repeats?
$\frac{2}{1}$	$2$	Terminates
$\frac{7}{5}$	$1.4$	Terminates
$\frac{-5}{2}$	$-2.5$	Terminates
$\frac{1}{3}$	$0.\overline{3} = 0.333\ldots$	Repeats: the $3$ repeats forever
$\frac{5}{11}$	$0.\overline{45} = 0.454545\ldots$	Repeats: the block $45$ repeats forever

Worked example

Test one fraction

Is the fraction $\frac{7}{5}$ equal to $\sqrt{2}$ ?

Decide what to check. The square root of 2 is the positive number whose square is 2. This fraction is positive, so it equals the square root of 2 exactly when its square equals 2.
$\left(\frac{7}{5}\right)^2 \overset{?}{=} 2$
Square the fraction. To multiply two fractions, multiply the numerators together and multiply the denominators together.
$\left(\frac{7}{5}\right)^2 = \frac{7}{5} \times \frac{7}{5} = \frac{7 \times 7}{5 \times 5} = \frac{49}{25}$
Write 2 as a fraction with the denominator 25 so that the two numbers are easy to compare. Multiplying the numerator and the denominator by the same nonzero number, 25, does not change the value.
$2 = \frac{2}{1} = \frac{2 \times 25}{1 \times 25} = \frac{50}{25}$
Compare the numerators, because the denominators match. 49 is not 50, so the square of the fraction is not 2. It falls short of 2 by one twenty-fifth.
$\frac{49}{25} \neq \frac{50}{25}$

Answer

No. $\frac{7}{5}$ is not equal to $\sqrt{2}$ , because its square is $\frac{49}{25}$ , which is $\frac{1}{25}$ less than $2$ . The fraction is only close to $\sqrt{2}$ .

Squaring worked as a test for $\frac{7}{5}$ , and it works for any fraction. Let $p$ and $q$ be positive integers, and suppose $\frac{p}{q} = \sqrt{2}$ .

Square both sides. Equal numbers have equal squares, squaring a fraction squares its numerator and its denominator, and $(\sqrt{2})^2 = 2$ . So $\frac{p^2}{q^2} = 2$ .
Multiply both sides by $q^2$ . This keeps the two sides equal, and $q^2$ is a positive integer because $q$ is not zero. You get $p^2 = 2q^2$ .

So $\frac{p}{q}$ can equal $\sqrt{2}$ only if $p^2 = 2q^2$ : the square of the numerator must be exactly double the square of the denominator. Call this the integer test. For $\frac{7}{5}$ , it asks whether $49$ equals $50$ .

You cannot test every fraction one at a time, because there are infinitely many. You need a reason that covers them all. Even and odd numbers give you one.

An even integer is $2$ times an integer, such as $12 = 2 \times 6$ . An odd integer is $1$ more than an even integer, such as $17 = 16 + 1$ . No integer is both, so an odd number is never equal to an even number.

Here is the fact you need: an odd integer always has an odd square. To see why, multiply two odd integers and write the first one as an even number plus $1$ . Multiply each part of that sum by the second odd integer:

$(\text{even} + 1) \times \text{odd} = (\text{even} \times \text{odd}) + \text{odd}$ .

The first part is even, because an even number times an integer is still $2$ times an integer. An even number plus an odd number is odd. So an odd integer times an odd integer is odd. For example, $3^2 = 9$ , $5^2 = 25$ and $7^2 = 49$ are all odd.

Worked example

A fraction with an odd numerator

The fraction $\frac{17}{12}$ is close to $\sqrt{2}$ . Can it be exactly equal to $\sqrt{2}$ ?

Use the integer test. The numerator 17 and the denominator 12 are positive integers, so the fraction can equal the square root of 2 only if the square of 17 is exactly double the square of 12.
$17^2 \overset{?}{=} 2 \times 12^2$
Look at the left side. 17 is odd, and an odd integer always has an odd square, so the left side is odd.
$17^2 = 17 \times 17 \quad \text{is odd}$
Look at the right side. It is 2 times an integer, so it is even.
$2 \times 12^2 \quad \text{is even}$
Compare. An odd number is never equal to an even number, so the two sides cannot be equal. Working out both sides confirms it: one is 289 and the other is 288.
$17^2 = 289 \quad \text{and} \quad 2 \times 12^2 = 2 \times 144 = 288$

Answer

No. $\frac{17}{12}$ is not equal to $\sqrt{2}$ , because $17^2$ is odd and $2 \times 12^2$ is even. The same reasoning rules out every fraction of positive integers with an odd numerator.

Worked example

An even numerator and an odd denominator

Can the fraction $\frac{10}{7}$ be exactly equal to $\sqrt{2}$ ?

Use the integer test again. The numerator 10 and the denominator 7 are positive integers, so the fraction can equal the square root of 2 only if the square of 10 is exactly double the square of 7.
$10^2 \overset{?}{=} 2 \times 7^2$
Look at the left side. 10 is even, so write it as 2 times 5. Multiplication can be done in any order, so put the two 2s together. The square is then 4 times an integer, and a number like that is called a multiple of 4.
$10^2 = (2 \times 5) \times (2 \times 5) = (2 \times 2) \times (5 \times 5) = 4 \times 25 = 100$
Look at the right side. 7 is odd, so its square, 49, is odd. Twice an odd number is never a multiple of 4: if it were, half of it would be even, but half of it is the odd number. Dividing 98 by 4 confirms this, because it leaves a remainder of 2.
$2 \times 7^2 = 2 \times 49 = 98 = 4 \times 24 + 2$
Compare. A multiple of 4 is never equal to a number that is not a multiple of 4, so the two sides cannot be equal.
$100 \neq 98$

Answer

No. $\frac{10}{7}$ is not equal to $\sqrt{2}$ , because $10^2 = 100$ is a multiple of $4$ and $2 \times 7^2 = 98$ is not. The same reasoning rules out every fraction of positive integers with an even numerator and an odd denominator.

Every fraction p/q of positive integers is one of these three kinds

Kind of fraction	What you find	Example
$p$ is odd	$p^2$ is odd, but $2q^2$ is even. They cannot be equal.	$\frac{17}{12}$ : $17^2 = 289$ , $2 \times 12^2 = 288$
$p$ is even and $q$ is odd	$p = 2k$ for an integer $k$ , so $p^2 = 4k^2$ is a multiple of $4$ . $2q^2$ is twice an odd number, so it is not.	$\frac{10}{7}$ : $10^2 = 100$ , $2 \times 7^2 = 98$
$p$ and $q$ are both even	Divide both by $2$ ; the value stays the same. Repeat until one is odd, then use a row above.	$\frac{14}{10} = \frac{7}{5}$ , and $7$ is odd: first row

Now put the pieces together. Take any fraction of positive integers. If the numerator and the denominator are both even, divide both by $2$ . Dividing both by the same nonzero number does not change the fraction's value. Repeat until at least one of them is odd. That must happen, because positive integers cannot keep getting smaller forever.

The fraction is now in the first or second row of the table, so it fails the integer test and is not equal to $\sqrt{2}$ . The fraction you started with has the same value, so it is not equal to $\sqrt{2}$ either.

So no fraction of positive integers equals $\sqrt{2}$ . Since $\sqrt{2}$ is positive, you could write any fraction of integers equal to it with a positive numerator and a positive denominator (if both were negative, multiply both by $-1$ , which keeps the value). That means no fraction of integers at all equals $\sqrt{2}$ .

Key concept

The square root of 2 is irrational

$\sqrt{2}$ is the positive number whose square is exactly $2$ , and it is irrational: you cannot write it as $\frac{p}{q}$ for any integers $p$ and $q$ with $q \neq 0$ . A fraction of positive integers could equal $\sqrt{2}$ only if $p^2 = 2q^2$ , and even-and-odd reasoning shows that this never happens.

Because $\sqrt{2}$ is irrational, its decimal form neither terminates nor repeats. Any terminating decimal or fraction of integers that you write for it is an approximation (close, but not exact), so mark it with $\approx$ . For the exact value, write $\sqrt{2}$ .

\sqrt{2} \neq \frac{p}{q} \quad \text{for all integers } p \text{ and } q \text{ with } q \neq 0

Mistake to avoid: trusting the calculator screen. A calculator shows $\sqrt{2}$ as $1.414213562$ . It is tempting to think the decimal stops there, which would make $\sqrt{2}$ equal to $\frac{1414213562}{1000000000}$ , a rational number. This fails because the calculator rounds the number to fit its screen. Check by squaring: $1.414213562^2 = 1.999999998944727844$ , which is close to $2$ but is not $2$ . More digits will never fix this. Every terminating decimal is a fraction of integers, such as $1.414 = \frac{1414}{1000}$ , and no fraction of integers equals $\sqrt{2}$ . Write $\sqrt{2} \approx 1.414213562$ , not $\sqrt{2} = 1.414213562$ .

Takeaway. No fraction of integers has a square of exactly $2$ , so $\sqrt{2}$ is irrational, and its decimal form neither terminates nor repeats. A value such as $1.414$ is only an approximation, so write $\sqrt{2} \approx 1.414$ , and write $\sqrt{2}$ itself when you need the exact value. This is a fact about $\sqrt{2}$ , not about every square root: $\sqrt{4} = 2$ is rational.

Next, you will estimate square roots and cube roots, and you will find and correct errors in working with roots.