Return of the Square Root

Previously, we have seen that to solve a quadratic equation

$ax^2 + bx + c = 0,$

we can use the quadratic formula given by

$\displaystyle x = \frac{ -b \pm \sqrt{ \Delta } }{2a}, \quad \Delta := b^2 - 4ac.$

provided that $\Delta \geq 0$ , so that $\sqrt{\Delta}$ is a unique sensible real number.

Example 1. Given non-negative integers $a, b$ , show that $\sqrt{ab} = \sqrt a \sqrt b$ . Furthermore, if $b \neq 0$ , show that

$\displaystyle \sqrt{\frac ab} = \frac{\sqrt a}{\sqrt b}.$

Solution. We observe that

$\begin{aligned} (\sqrt a \sqrt b)^2 &= (\sqrt a \sqrt b) \cdot (\sqrt a \sqrt b) \\ &= (\sqrt a \cdot \sqrt a) \cdot (\sqrt b \cdot \sqrt b) \\ &= (\sqrt a)^2 \cdot (\sqrt b)^2 = a b.\end{aligned}$

Since $\sqrt a \sqrt b \geq 0$ ,

$\sqrt{ab} = |\sqrt a \sqrt b| = \sqrt a \sqrt b.$

For the second claim, we employ a similar strategy:

$\begin{aligned} \left( \frac{ \sqrt a }{ \sqrt b } \right)^2 = \frac{ \sqrt a }{ \sqrt b } \cdot \frac{ \sqrt a }{ \sqrt b } = \frac{(\sqrt a)^2}{(\sqrt b)^2} = \frac ab. \end{aligned}$

so that $\displaystyle \sqrt{\frac ab} = \frac{ \sqrt a }{ \sqrt b }$ .

Lemma 1. $\sqrt{\Delta}$ is an non-negative integer if and only if there exists a non-negative integer $n$ such that $\Delta = n^2$ , in which we write $\sqrt{\Delta} = n$ . In this case, we say that $\Delta$ is a perfect square.

Proof Sketch. Use prime factorisation and a proof by contradiction in the spirit of proving that the number $\sqrt 2$ is not a fraction (i.e. an irrational number).

Lemma 2. Suppose there exist positive integers $n, k$ such that $\Delta = k^2 \cdot n$ . Then $\sqrt{\Delta} = k \sqrt{n}$ .

Proof. By Example 1,

$\sqrt{\Delta} = \sqrt{k^2 \cdot n} = \sqrt{k^2} \cdot \sqrt n = k \sqrt n.$

Example 2. Evaluate the numbers $\sqrt 4, \sqrt 8, \sqrt{16}, \sqrt{32}$ in the form $a + b\sqrt 2$ , where $a,b$ are rational numbers.

Solution. We observe that $\sqrt 4 = \sqrt{2^2} = 2$ . By Lemma 2,

$\sqrt 8 = \sqrt{4 \cdot 2} = \sqrt 4 \sqrt 2 = 2\sqrt 2.$

Similarly, $\sqrt{16} = \sqrt{4^2} = 4$ , so that by Lemma 2 again,

$\sqrt{32} = \sqrt{16 \cdot 2} = \sqrt{16} \sqrt 2 = 4 \sqrt 2.$

Lemma 3. Suppose $\Delta$ is not a perfect square. If $a, b$ are rational numbers such that $a + b \sqrt{\Delta} = 0$ , then $a = b = 0$ . In particular, if $a+b \sqrt{\Delta} = c+d \sqrt{\Delta}$ , then $a=c$ and $b=d$ . We call this technique comparing of coefficients.

Proof Sketch. Follow the proof strategy in Lemma 1. Left as an exercise in proof by contradiction and baby number theory for interested readers.

Definition 1. Let $n$ be a positive natural number.

We say that $n$ square-free if none of the numbers $2^2, 3^2, 4^2, \dots$ are factors of $n$ .
A real number of the form $\sqrt n$ , where $n$ is square-free, is called a surd.
Given a square-free natural number $n$ , all numbers $a+b\sqrt n$ , where $a,b$ are rational, are real numbers. The collection of such numbers is called a quadratic field, commonly denoted $\mathbb Q(\sqrt n)$ .

In particular, if $a,b,c$ are integers, then $\Delta = b^2 - 4ac$ is an integer as well, and the quadratic equation has roots

$\displaystyle x = -\frac b{2a} \pm \frac 1{2a}\sqrt{\Delta}.$

By Lemma 2, we still obtain two real and distinct roots. By Lemma 3, we will not be able to represent them purely using fractions.

Example 3. Evaluate $\sqrt{2/3}$ in the form $a + b\sqrt 6$ , where $a,b$ are rational numbers.

Solution. By Example 1,

$\displaystyle \sqrt{ \frac 23 } = \frac{\sqrt 2}{\sqrt 3}.$

By observing that $\displaystyle 1 = \frac{\sqrt 3}{\sqrt 3}$ ,

$\displaystyle \sqrt{ \frac 23 } = \frac{\sqrt 2}{\sqrt 3} \cdot \frac{\sqrt 3}{\sqrt 3} = \frac{\sqrt 2 \sqrt 3}{(\sqrt 3)^2} = \frac{\sqrt{2 \cdot 3}}{3} = \frac 13 \sqrt 6.$

Therefore, $\sqrt{2/3} = \frac 13 \sqrt 6$ .

Example 4. Given rational numbers $a,b$ and a non-negative integer $n$ , show that

$(a+b\sqrt n)(a - b\sqrt n)$

is a rational number. Hence, evaluate $\displaystyle \frac 1{3+2\sqrt 2}$ in the form $a + b\sqrt 2$ , where $a,b$ are rational numbers.

Solution. Using the difference of squares formula,

$(a+b\sqrt n)(a - b\sqrt n) = a^2 - (b\sqrt n)^2 = a^2 - b^2 n.$

Since $a,b,n$ are rational numbers, so is $a^2 - b^2 n$ .

Similar to Example 3, we observe that $\displaystyle 1 = \frac{3 - 2\sqrt 2}{3 - 2\sqrt 2}$ , so that

$\begin{aligned} \frac 1{3+2\sqrt 2} &= \frac 1{3+2\sqrt 2} \cdot \frac{ 3-2\sqrt 2 }{3-2\sqrt 2} \\ &= \frac {1 \cdot (3-2\sqrt 2)}{ (3-2\sqrt 2)(3-2\sqrt 2) } \\ &= \frac{ 3 - 2\sqrt 2 }{ 3^2 - (2\sqrt 2)^2 } \\ &= \frac{ 3 - 2\sqrt 2 }{ 9 - 8 } \\ &= 3-2\sqrt 2. \end{aligned}$

Remark 1. The technique in Examples 3 and 4 is called rationalising the denominator.

Example 5. Calculate the length of a square with area $(3 + 2 \sqrt 2)$ sqaured-units in the form $a + b\sqrt 2$ , where $a,b$ are rational numbers.

Solution. Let $(a + b \sqrt 2)$ denote the length of the square. Since the square has area $(3 + 2 \sqrt 2)$ ,

$\begin{aligned}3 + 2 \sqrt 2 &= (a + b \sqrt 2)^2\\ &= a^2 + 2a(b\sqrt 2) + (b\sqrt 2)^2 \\ &= a^2 + (2ab)\sqrt 2 + b^2 \cdot 2 \\ &= (a^2 + 2b^2) + (2ab) \sqrt 2. \end{aligned}$

By comparing of coefficients as per Lemma 3,

$a^2 + 2b^2 = 3\quad \text{and}\quad 2ab = 2.$

Thus, we reduce the problem to solving a pair of equations, and we shall do so by substitution. Making $b$ the subject in the second equation,

$\displaystyle 2ab = 2 \quad \Rightarrow \quad b = \frac{2}{2a} = \frac 1a.$

Substituting this value of $b$ into the first equation,

$\begin{aligned} a^2 + 2\left( \frac 1a \right)^2 &= 3 \\ a^2 + 2 \cdot \frac 1{a^2} &= 3 \\ a^2 \cdot a^2 +2 &= 3 a^2 \\ (a^2)^2 + 2 &= 3a^2 \\ (a^2)^2 - 3a^2 + 2 &= 0. \end{aligned}$

Since $\Delta = (-3)^2 - 4 \cdot 1 \cdot 2 = 1 = 1^2$ is a perfect square, we could solve by either factorisation or the quadratic formula. We shall use the latter since we are lazy:

$\displaystyle a^2 = \frac{-(-3) \pm \sqrt{\Delta}}{2 \cdot 1} = \frac{3 \pm 1}{2}.$

Hence, $a^2 = 1$ or $a^2 = 2$ .

In the latter case, $a^2 = 2 \Rightarrow a = \pm \sqrt 2$ , which is a contradiction since $a$ is a rational number and $\sqrt 2$ is not a rational number, therefore we reject this case, and conclude that $a^2 = 1$ . Hence, $a = \pm 1$ . Substituting into the expression for $b$ ,

$\displaystyle b = \frac 1a = \frac 1{\pm 1} = \pm 1.$

Therefore, $a = \pm 1 \Rightarrow b = \pm 1$ , which abbreviates the following two cases:

if $a = 1$ , then $b = 1$ , and
if $a = -1$ , then $b = -1$ .

Hence, the length of the square is either $1+\sqrt 2$ or $-1-\sqrt 2$ . Since lengths are non-negative and $-1 - \sqrt 2 < 0 + 0 = 0$ , we reject the latter case, and conclude that the length of the square is $(1 + \sqrt 2)$ units. In more mathematical jargon,

$\sqrt{3 + 2\sqrt 2} = 1 + \sqrt 2.$

Example 6. Calculate the length of the base of a right-isosceles triangle with longest side length (i.e. hypotenuse) $1$ .

Solution. Denote the desired side length by $x \geq 0$ . By Pythagoras’ theorem,

$\begin{aligned} x^2 + x^2 &= 1^2 \\ 2x^2 &= 1 \\ x^2 &= \frac 12. \end{aligned}$

Taking square roots,

$\displaystyle x = \sqrt{\frac 12} = \frac{\sqrt 1}{\sqrt 2} = \frac 1{\sqrt 2}.$

By rationalising the denominator,

$\displaystyle x = \frac 1{\sqrt 2} = \frac 1{\sqrt 2} \cdot \frac{\sqrt 2}{\sqrt 2} = \frac{\sqrt 2}{2} = \frac 12 \sqrt 2.$

Therefore, the triangle has base length $\frac 12 \sqrt 2$ .

Remark 2. As far as possible, we express answers involving surds in the form $a + b\sqrt n$ , where $a,b$ are simplified rational numbers and $n$ is square-free. In linear algebra, numbers of this form are called linear combinations of the basis $\{ 1, \sqrt n \}$ .

While understanding surds of the form $\sqrt n$ , where $n$ is non-negative, has its uses in simplifying otherwise complicated numerical expressions, its power becomes beefed up significantly if $n$ is negative, for example, if we consider numbers of the form $\sqrt{-1}$ .

You would scream at me for committing this mathematical crime. We have emphasised repeatedly that there is no real number $x$ such that $x^2 = -1$ , so how in good faith and conscience can we write $\sqrt{-1}$ and not cringe like the 6-7 kid? You are absolutely right—but we are assuming that the only numbers that we can talk about are real numbers.

Definition 1. Call the number $i := \sqrt{-1}$ the imaginary unit, defined by the “false” equation $i^2 = -1$ .

For more interested readers: legitimately defined using techniques in linear algebra here).
A number of the form $a + bi \equiv a + b\sqrt{-1}$ , where $a, b$ are real numbers, is called complex.

Since $0^2 = 0 \neq -1$ , we have $i \neq 0$ .

While we need a rather distinct visual idea for what $\sqrt{-1}$ means, if we allow the calculation $i^2 = -1$ , we recover many of the properties discussed in the previous lemmas and examples.

Theorem 1. The complex numbers satisfy the following properties:

If $k > 0$ , then $\sqrt{-k} = i \sqrt{k}$ .
If $a,b$ are real numbers such that $a+bi = 0$ , then $a=b=0$ .
Given real numbers $a,b$ , $\displaystyle \frac 1{a+bi}$ is still a complex number.
Given real numbers $a,b$ , $\sqrt{a+bi}$ is still a complex number.