Category: O-Level Math

Divisibility Tricks
Let $n$ be an integer between $0$ and $9$ inclusive. It is obvious that:
- $n$ is even if and only if $n$ is one of the numbers $0, 2, 4, 6, 8$
- $n$ is a multiple of $3$ if and only if $n$ is one of the numbers $0, 3, 6, 9$ .
- $n$ is a multiple of $5$ if and only if $n$ is one of the numbers $0, 5$ .
Let $m$ be number between $1$ and $9$ inclusive. Define the two-digit number

$x(m,n) := 10m + n.$

For example, $x(6, 7) = 10 \times 6 + 7 = 67$ .

Problem 1. Show that $x(m, n)$ is even if and only if $n$ is even (i.e. $n$ is one of $0, 2, 4, 6, 8$ ).

(Click for Solution)

Solution. Write $x(m,n) = 10m + n$ so that

$n = x(m,n) - 10m.$

$(\Rightarrow)$ If $x(m,n)$ is even, then there exists an integer $k$ such that $x(m,n) = 2k$ . Hence,

$\begin{aligned} n &= x(m,n) - 10m \\ &= 2k - 10m \\ &= 2(k-5m). \end{aligned}$

Since $k - 5m$ is still an integer, $n$ must be even.

$(\Leftarrow)$ If $n$ is even, then there exists an integer $\ell$ such that $n = 2 \ell$ . Hence,

$\begin{aligned} x(m,n) &= 10m + n \\ &= 10m + 2\ell \\ &= 2(5m + \ell). \end{aligned}$

Since $5m + \ell$ is still an integer, $x(m,n)$ must be even.

Remark 1. The same proof can be modified to show that $x(m, n)$ is a multiple of $5$ if and only if $n$ is either $0$ or $5$ . Furthermore, $x(m, n)$ is a multiple of $10$ if and only if $n = 0$ . These results work for numbers regardless of the number of digits, for example:
- $123\, 456$ is even,
- $123\, 455$ is a multiple of $5$ ,
- $123\, 450$ is a multiple of $10$ .
We remark that the one-digit multiples of $3$ are

$\begin{aligned} 3 &= 3 \times 1,\\ 6 &= 3 \times 2,\\ 9 &= 3 \times 3. \end{aligned}$

Problem 2. Show that $x(m, n)$ is a multiple of $3$ if and only if $m+n$ is a multiple of $3$ .

(Click for Solution)

Solution. By definition,

$x(m,n) = 10m + n = 9m + (m+n),$

so that $m+n = x(m,n) - 9m$ .

$(\Rightarrow)$ If $x(m,n)$ is a multiple of $3$ , then there exists an integer $k$ such that $x(m,n) = 3k$ . Hence,

$\begin{aligned} m+n &= x(m,n) - 9m \\ &= 3k - 9m \\ &= 3(k-3m). \end{aligned}$

Since $k - 3m$ is still an integer, $m+n$ must be a multiple of $3$ .

$(\Leftarrow)$ If $m+n$ is a multiple of $3$ , then there exists an integer $\ell$ such that $m+n = 3 \ell$ . Hence,

$\begin{aligned} x(m,n) &= 9m + (m+n) \\ &= 9m + 3\ell \\ &= 3(3m + \ell). \end{aligned}$

Since $3m + \ell$ is still an integer, $x(m,n)$ must be a multiple of $3$ .

Example 1. $57$ is a multiple of $3$ , because

$57 \to 5+7 = 12 \to 1+2 = 3.$

This result works for numbers regardless of the number of digits: since

$12\, 357 \to 1+2+3+5+7 = 18 \to 1+8 = 9,$

$12\, 357$ is a multiple of $3$ .

Remark 2. The same proof can be modified to show that $x(m, n)$ is a multiple of $9$ if and only if $m+n$ is a multiple of $9$ . Furthermore, we can use more advanced techniques to reduce all calculations to the set of one-digit numbers $\{1,2,\dots, 9\}$ .

Now let $n_1, n_2, n_3$ be numbers in $0,1,2,\dots, 9$ , and $n_1 > 0$ . Define the three-digit number

$y(n_1,n_2,n_3) := 100n_1 + 10n_2 + n_3.$

For example

$y(4,2,0) = 100 \times 4 + 10 \times 2 + 0 = 420.$

We remark that the first five multiples of $4$ are

$\begin{aligned} 4 &= 4 \times 1,\\ 8 &= 4 \times 2,\\ 12 &= 4 \times 3, \\ 16 &= 4 \times 4, \\ 20 &= 4 \times 5. \end{aligned}$

Furthermore, $0 = 4 \times 0$ is a multiple of $4$ .

Problem 3. Show that:
- $x(n_2,n_3)$ is a multiple of $4$ if and only if there exists a positive integer $k$ such that $x(n_2 - 2k, n_3)$ is a multiple of $4$ ,
- $y(n_1, n_2, n_3)$ is a multiple of $4$ if and only if $x(n_2,n_3)$ is a multiple of $4$ .
(Click for Solution)

Solution. By definition, $x(n_2, n_3) = 10 n_2 + n_3$ and

$x(n_2 - 2k, n_3) = 10 (n_2-2k) + n_3 = 10n_2 + n_3 - 20k = x(n_2, n_3) - 4 \cdot 5k.$

Since $4 \cdot 5k$ is obviously a multiple of $4$ , $x(n_2, n_3)$ is a multiple of $4$ if and only if $x(n_2 - 2k, n_3)$ is.

By definition of $x(n_2, n_3)$ ,

$y(n_1,n_2,n_3) := 100n_1 + 10n_2 + n_3 = 4 \cdot 25n_1 + x(n_2,n_3).$

Therefore $x(n_2, n_3)$ is a multiple of $4$ if and only if

$y(n_1,n_2,n_3) - 4 \cdot 25n_1$

is a multiple of $4$ , which holds if and only if $y(n_1,n_2,n_3)$ is a multiple of $4$ .

Example 2. Since

$924 \to 24 \to 4,$

$924$ is a multiple of $4$ . In fact, this principle works for any number with more than $2$ digits: $123\, 924$ is also a multiple of $4$ since

$123\, 916 \to 16.$

We can reduce all calculations to the set $\{0, 4 , 8, 12, 16\}$ .

—Joel Kindiak, 16 Jan 26, 1432H
March 6, 2026
Shapes in Three Dimensions

The three-dimensional version of “area” and “perimeter” would be, rather unsurprisingly, “volume” and “surface area”.

We start with the simplest object: a prism.

Definition 1. A solid with height $h$ is called a prism if all of its cross-sections have the same shape. In the special case that the base of the prism is a circle, we call the solid a cylinder.

Just like how a rectangle is said to have area $(\text{base}) \times (\text{height})$ , a prism also is said to have volume

$(\text{base area}) \times (\text{height}).$

In that sense, a prism is similar to a “three-dimensional” rectangle, with many possible shapes for its base.

Corollary 1. The volume of a cylinder with height $h$ and base radius $r$ is $\pi r^2 h$ .

If there’s a “three-dimensional” rectangle, is there a “three-dimensional” triangle? Yes, and it’s called a pyramid.

Definition 2. A solid with height $h$ is called a pyramid if all of its cross-sections are similar to one another, and they “shrink” to a common point. In the special case that the base of the prism is a circle, we call the solid a cone.

This description is similar to that of a triangle: the bases are similar to one another and “shrink” to a common point:

We have seen that a triangle has area $\frac 12 \times (\text{base}) \times (\text{height})$ . What’s the formula for that of a pyramid?

Theorem 1. The volume of a prism is $\frac 13 \times (\text{base area}) \times (\text{height})$ .

Proof. Omitted as it requires integral calculus. Nevertheless, the factor $1/3$ is directly related to the $3$ dimensions in which we defined a pyramid.

Corollary 2. The volume of a cone with height $h$ and base radius $r$ is $\frac 13 \pi r^2 h$ .

Proof. By Theorem 1, the cone has a volume of

$\frac 13 \times (\text{base area}) \times (\text{height}) = \frac 13 \cdot \pi r^2 \cdot h = \frac 13 \pi r^2 h.$

Having discussed volumes, it’s worth looking into its partner, surface areas. As the name suggests, surface areas refer to the areas of the surfaces of a solid. But what happens when weird faces abound?

We can’t discuss them all, but there are some special cases.

Theorem 3. The curved surface area of a cylinder with height $h$ and radius $r$ is $2 \pi r h$ .

Proof. The curved surface area of a cylinder can be thought of as a “wrap” of a rectangle with base $2 \pi r$ and height $h$ :

Hence, the curved surface area is $2 \pi r \cdot h = 2 \pi r h$ .

Theorem 4. The curved surface area of a cone with height $h$ and radius $r$ is $\pi r \sqrt{r^2 + h^2}$ .

Proof. Let $\ell$ denote the slant height of the cone. The curved surface area of a cone can be thought of as a “wrap” of a sector with radius $\ell$ :

Now the length of the arc of the sector is the circumference of the original cone, namely, $2 \pi r$ . Hence, the total area of the sector is given by

$\displaystyle \frac{2 \pi r}{2 \pi \ell} \times \pi \ell^2 = \pi r \ell.$

Now by Pythagoras’ theorem,

$r^2 + h^2 = \ell^2 \quad \Rightarrow \quad \ell = \sqrt{r^2 + h^2}.$

Therefore, the curved surface area is $\pi r \sqrt{r^2 + h^2}$ , as required.

Finally we should discuss the geometry of the three-dimensional version of a circle: the sphere.

Definition 2. Define the sphere with centre and radius $r$ to be the set of points whose distance from the centre is $r$ .

Example 1. The Earth can be modelled as a sphere.

Source: Wikipedia

Theorem 5. The volume of a sphere with radius $r$ is $\frac 43 \pi r^3$ . The surface area of a sphere is $4 \pi r^2$ .

Proof. Omitted as it requires integral (and arguably, differential) calculus.

Having developed the many commonly-used formulas in high school geometry, we cannot avoid the dreaded T-word: trigonometry. Contrary to popular anxiety, trigonometry is simply a new language to describe the relationship between straightedges and curves (i.e. angles). More on that in the next post.

—Joel Kindiak, 7 Dec 25, 1931H

March 2, 2026
Basic Mensuration

We start by considering the area of a rectangle. If a rectangle has base $b$ (units) long and height $h$ , its area is $b \times h \equiv bh$ .

In fact, we used the rectangle intuition to list out the properties of real numbers, since we regarded $b, h$ as real numbers.

Theorem 1. Any triangle with base $b$ and height $h$ has area $\frac 12 bh$ .

Proof. We first suppose the triangle is right-angled:

By duplicating the triangle and rotating it, we obtain a rectangle with base $b$ and height $h$ :

Denoting the area of the triangle by $A$ ,

$2A = A + A = bh \quad \Rightarrow \quad A = \frac 12 bh.$

Now suppose the altitude lies inside the triangle:

Adding the areas, we obtain a total area of

$\frac 12 b_1 h + \frac 12 b_2 h = \frac 12 (b_1 + b_2)h = \frac 12 bh.$

Now we consider the case when the altitude lies outside of the triangle:

Denote the area of the original triangle by $A_1$ . Consider the new triangle with area $A_2$ as follows:

Since areas are additive, use the first case to obtain

$A_1 + \frac 12 ch = A_1 + A_2 = \frac 12 (b+c) h.$

Performing algebruh,

$A_1 = \frac 12 (b+c) h - \frac 12 c h = \frac 12 ((b+c)-c)h = \frac 12 bh.$

Theorem 2. Any trapezium with bases $b_1, b_2$ and height $h$ has area $\frac 12 (b_1 + b_2)h$ .

Proof. Consider the following trapezium without loss of generality:

Split the trapezium into two identical triangles:

Using Theorem 1, the area of the trapezium is given by

$\frac 12 b_1 h+\frac 12 b_2 h = \frac 12 (b_1 + b_2) h.$

Theorem 3. Any parallelogram with base $b$ and height $h$ has area $bh$ .

Proof. Consider the parallelogram below without loss of generality:

Since this parallelogram is a trapezium with bases $b_1 = b_2 = b$ , by Theorem 2, its area is given by

$\frac 12 (b_1 + b_2) h = \frac 12 (b+b)h = \frac 12 \cdot 2b \cdot h = bh.$

Most high school geometry problems boils down to applying these formulas one after another. However, they also include more “curvy” areas like circles. How do we compute such areas? We could memorise their formulas like $\pi r^2$ , but how do we get that formula in the first place?

Definition 1. Given a circle with circumference $C$ and radius $r$ , we define $\pi$ to be the constant

$\displaystyle \pi := \frac C{2r}.$

Equivalently, we recover the circumference formula $C = 2 \pi r$ . Numerically, $\pi \approx 3.142$ .

Theorem 4. The area of a circle with radius $r$ is $\pi r^2$ .

Proof Sketch. We will subdivide the circle into a collection of “curved triangles” or more technically, sectors:

Here, we use $12$ sub-divisions for convenience. We notice that the areas total to the circle area. By rearranging these sectors, we recover an approximate rectangle with approximate base $\pi r$ and approximate height $r$ :

Therefore, using the area of a rectangle, the circle has approximate area $\pi r \cdot r = \pi r^2$ . Increasing the number of subdivisions improves the approximation, and in the limit, we get an exact area of $\pi r^2$ .

Remark 1. A more complete construction can be formalised using integral calculus.

Finally, lets re-visit angles for completeness. Previously, we have usually used degrees— $360^\circ$ (i.e. $360$ degrees) to denote the angle of a “complete turn”. The reason for $360$ is convenience—the number $360$ has the factors $1,2,3,4,5,6$ that find many uses in astronomy and geology. In particular, $90^\circ$ denotes a $\frac 14$ -turn, $45^\circ$ denotes a $\frac 18$ -turn, so on and so forth.

Rather than use the convenience of $360$ , we might find it helpful to use the circumference of a circle. If a circle has radius $1$ , then its circumference is $2\pi$ :

Here’s our intuition: an angle of $2 \pi$ denotes a complete turn. In particular, $2 \pi = 360^\circ$ . In this case, we say that the angle is $2\pi$ radians, and automatically refer to “radians” when we do not write any unit.

Definition 2. An angle is said to be of $\theta$ radians if the corresponding curved segment (i.e. arc) in the unit circle has a length of $\theta$ .

Therefore, an angle of $2\pi$ radians corresponds to the whole circumference.

Lemma 1. Given any circle with radius $r$ , the angle corresponding to an arc with length $s$ is given by

$\displaystyle \theta = \frac sr.$

Equivalently, $s = r \theta$ . In particular, $C = r \cdot 2\pi = 2\pi r$ , agreeing with our prior intuitions on the circumference.

Proof Sketch. Consider the diagram below.

By regarding arcs as limits of straight lines, we can pass the intercept theorem for similar triangles to the limit and obtain

$\displaystyle \frac 1r = \frac{\theta}{s} \quad \Rightarrow \quad s =r \theta.$

Therefore, radians work for angles regardless of radius size.

Lemma 2. $2 \pi = 360^\circ$ . By algebra,

$\displaystyle 1 = \frac{1}{2\pi} \cdot 360^\circ, \quad 1^\circ = \frac{2\pi}{360}.$

Here, of course, $N^\circ := N \cdot 1^\circ$ .

Theorem 5. A sector with radius $r$ and interior angle $\theta$ (radians) has area $\frac 12 \theta r^2$ . In particular, with a full circle, we obtain the expected area $\pi r^2$ .

Proof. Since the sector sweeps $\theta/ 2\pi$ of the original circle, its area would correspond to

$\displaystyle \frac{ \theta }{ 2 \pi } \cdot \pi r^2 = \frac { \theta r^2 }2.$

For a full circle, set $\theta = 2\pi$ to obtain an area of $2\pi r^2/2 = \pi r^2$ .

Having explored areas, we need to think deeper. Literally. Next time, we will look at volumes and surface areas, the three-dimensional version of areas and perimeters.

Oh, let’s just conclude with perimeters:

Definition 3. The perimeter of a shape is the total length of its boundary.

Example 1. A rectangle with base $b, h$ has perimeter $2(b+h)$ :

Example 2. A circle with radius $r$ has perimeter $2 \pi r$ :

Indeed, the perimeter of a circle is simply its circumference.

Example 3. A half-circle (i.e. a semicircle) with radius $r$ has perimeter $\pi r + 2r$ :

—Joel Kindiak, 21 Nov 25, 1715H

February 26, 2026
About Higher Powers

This post isn’t about God; I write at length about God elsewhere.

We have explored expressions involving $x, x^2, x^3$ , and even binomials such as $(1+x)^n$ . Much of our exploration arises from interpreting powers as repeated multiplication in the following sense: given any positive number $a$ and positive integers $m, n$ ,

$a^{m+n} = a^m a^n, \quad (a^m)^n = a^{mn}.$

After all, denoting

$a^{n} = \underbrace{ a \times a \times \cdots \times a }_n,$

we have

$\begin{aligned} a^m a^n &= \underbrace{ a \times a \times \cdots \times a }_m \times \underbrace{ a \times a \times \cdots \times a }_n \\ &= \underbrace{ a \times a \times \cdots \times a \times a \times a \times \cdots \times a }_{m + n} \\ &= a^{m+n}.\end{aligned}$

Therefore,

$\begin{aligned} (a^m)^n &= \underbrace{ a^m \times a^m \times \cdots \times a^m }_n \\ &= a^{\footnotesize \overbrace{m + m + \cdots + m}^n} \\ &= a^{m \times n} \\ &= a^{mn}. \end{aligned}$

It turns out that we are allowed to regard $x, y$ as real numbers.

Theorem 1. Given any positive number $a > 0$ , for real numbers $x, y$ ,

$a^{x+y} = a^x a^y, \quad a^x > 0.$

We call $a^x$ the base– $a$ exponential of $x$ .

Proof. The complete construction of the exponential takes place here, using the tools in undergraduate real analysis.

Theorem 2 (Laws of Exponents). Using the exponential in Theorem 1, we have the following laws of exponents:

$\displaystyle a^0 = 1, \quad a^{1/n} = \sqrt[n]{a}, \quad a^{-x} = \frac 1{a^x}, \quad a^{x-y} = \frac{ a^x }{ a^y }.$

Here, $n$ refers to a positive integer, and $x,y$ refers to real numbers.

Proof. Using Theorem 1,

$a^0 = a^{0+0} = a^0 \cdot a^0 = (a^0)^2 \quad \Rightarrow \quad (a^0)^2 - a^0 = 0.$

Factorising $a^0 > 0$ ,

$a^0 (a^0 - 1) = (a^0)^2 - a^0 = 0 \quad \Rightarrow \quad a^0 - 1 = 0 \quad \Rightarrow \quad \boxed{ a^0 = 1 }.$

Similarly, by applying Theorem 1 a total of $n$ times,

$(a^{1/n})^n = \underbrace{a^{1/n} \cdot a^{1/n} \cdot \cdots \cdot a^{1/n}}_n = a^{n \cdot (1/n)} = a\quad \Rightarrow \quad \boxed{ a^{1/n} = \sqrt[n]{a} }.$

Using Theorem 1 and the first result,

$\displaystyle a^x \cdot a^{-x} = a^{x+(-x)} = a^0 = 1.$

Finally, using Theorem 1 and the previous result,

$\displaystyle a^{x-y} = a^{x + (-y)} = a^x \cdot a^{-y} = a^x \cdot \frac 1{a^y} = \frac{ a^x }{ a^y } \quad \Rightarrow \quad \boxed{ a^{x-y} = \frac{ a^x }{ a^y } }.$

Just like how we devoted nontrivial time and energy to solve equations that look like $x^2 = 2$ (it is $x = \pm \sqrt 2$ ), we would also like to solve equations such as $2^x = 3$ . We observe that, rather straightforwardly, if $k$ is an integer, then

$2^x = 2^k \quad \Rightarrow \quad x = k.$

However, how do we solve equations such as $2^x = 3$ ? The idea is to use better and better decimal approximations of $x$ , and we give the “perfect” answer—the base- $2$ logarithm of $3$ . For a proper, technical construction, see the same post pertaining exponential functions.

Definition 1. Given real constants $a, b$ with $a > 0$ , the base- $a$ logarithm of $b$ is the unique number $x$ such that

$a^x = b.$

In this case, we denote $x := \log_a(b)$ .

Theorem 3 (Inverse Property). Given real constants $a, b$ with $a > 0$ ,

$a^{\log_a(b)} = b,\quad \log_a(a^b) = b.$

Proof. Define $x := \log_a(b)$ . By Definition 1,

$a^{\log_a(b)} = a^x = b.$

Similarly, define $y := \log_a(a^b)$ . By Definition 1,

$a^y = a^b \quad \Rightarrow \quad \log_a(a^b) = y = b.$

Thanks to the basic property described in Theorem 1, the laws of exponents, and the inverse property, we recover all of the laws of logarithms. But to do that, we need our special friend $e \approx 2.718$ .

Theorem 4. Given positive constants $a, x, y$ , we have the following laws of logarithms:

$\begin{aligned} \log_a(xy) &= \log_a(x) + \log_a(y),\\ \log_a\left( \frac xy \right) &= \log_a(x) - \log_a(y). \end{aligned}$

Furthermore $\log_a(1) = 0,\log_a(a) = 1$ , and for any $r > 0$ ,.

$\log_a(x^r) = r \log_a(x).$

Proof Sketch. We prove the first property, prove a special case for the last property, and relegate the others as an exercise in definition chasing. Using Theorem 1 and the inverse property,

$a^{\log_a(x) + \log_a(y)} = a^{\log_a(x)} \cdot a^{\log_a(y)} = xy.$

By Definition 1, $\log_a(xy) = \log_a(x) + \log_a(y)$ . For the last property, we prove the special case that $r$ is a positive integer:

$\begin{aligned} \log_a(x^r) &= \log_a(\underbrace{ x \cdot x \cdot \cdots \cdot x }_r )\\ &= \underbrace{ \log_a(x) + \log_a(x) + \cdots + \log_a(x) }_r = r \log_a(x), \end{aligned}$

The general case uses our technical definition of the exponential as outlined in the real-analysis posts.

Corollary 1. Given real constants $a, x, y$ and $a > 0$ ,

$(a^x)^y = a^{xy}.$

Proof. Using the last property in the laws of logarithms,

$\log_a((a^x)^y) = y \log_a(a^x) = y \cdot x \log_a(a) = y \cdot x \cdot 1 = xy.$

By the inverse property,

$(a^x)^y = a^{\log_a((a^x)^y)} = a^{xy},$

as required.

Example 1. By considering the real number $r := {\sqrt 2}^{\sqrt 2}$ , demonstrate the existence of irrational numbers $a,b$ such that $a^b$ is rational. You may assume without proof that $\sqrt 2$ is irrational.

Solution. There are two cases to consider: either $r$ is rational or not. If $r$ is rational, then setting $a = b = \sqrt 2$ yields

$a^b = {\sqrt 2}^{\sqrt 2} = r,$

which is rational. If $r$ is irrational, then setting $a = r = {\sqrt 2}^{\sqrt 2}$ and $b = \sqrt 2$ yields, by Corollary 1,

$a^b = r^{\sqrt 2} = {\left( {\sqrt 2}^{\sqrt 2} \right)}^{\sqrt 2} = {\sqrt 2}^{\sqrt 2 \cdot \sqrt 2} = {\sqrt 2}^2 = 2,$

which is rational.

No matter which scenario is ultimately true, we can always arrive at the conclusion that there exist irrational numbers $a, b$ such that $a^b$ is rational.

Theorem 5 (Change-of-Base). Given positive numbers $a, b, c$ ,

$\displaystyle \log_a(b) = \frac{\log_c(b)}{\log_c(a)}.$

In particular, denoting $\lg \equiv \log_{10}$ and $\ln \equiv \log_e$ , where $e \approx 2.718$ denotes the exponential unit,

$\displaystyle \log_a(b) = \frac{\lg(b)}{\lg(a)} = \frac{\ln(b)}{\ln(a)}.$

Proof. Using the inverse property,

$\begin{aligned} \log_a(b) \cdot \log_c(a) &= \log_c(a^{\log_a(b)}) = \log_c(b) \quad \Rightarrow \quad \boxed{ \log_a(b) = \frac{\log_c(b)}{\log_c(a)} }. \end{aligned}$

The base- $10$ logarithm finds much use in numerically estimating the logarithms. For instance, the pH of a solution that measures the acidity of that solution (crucial in ecological applications) is defined by the number $-\lg([\mathrm H^+])$ , where $[\mathrm H^+]$ denotes the concentration of protons in that solution. The base- $e$ logarithm finds much use in theoretical analysis, since the rate of change of the function $\ln(x)$ is $1/x$ .

Exponentials and logarithms form the cornerstone of any discussion involving rates of change, i.e. calculus, and infect every serious field of study in STEM: population growth in biology, cooling laws in thermodynamics, logistic growth in machine learning, so on and so forth.

Perhaps these mathematical higher powers point to an ultimate higher power?

For now, let’s return to earth and return to practical mathematics by discussing measurements of various tangible objects.

—Joel Kindiak, 28 Oct 25, 1523H

February 24, 2026
Quadratics Revisited
Previously, we have seen that given $a > 0$ , the graph of $y = ax^2$ looks like the diagram below.

For any point $P(x, y)$ on the graph, it has the same distance to the (focal) point $F(0, 1/(4a))$ and the (directrix) line $y = -1/(4a)$ . Hence, its shape is known as a parabola. In particular, it has a minimum point $(0, 0)$ .

If we shifted the graph rightwards by $h$ and upwards by $k$ , we recover a more general parabola with minimum point $(h, k)$ .

Of course, if $a < 0$ , then the whole graph gets “flipped” to the downside:

The diagram above shows the graph of $y = -x^2 + 4$ .

Example 1. Determine the $y$ -intercept of the quadratic graph $y = -x^2 + 4$ .

Solution. To determine the $y$ -intercept of the quadratic graph, we need to find the intersection of the graph $y = -x^2 + 4$ and the $y$ -axis, whose equation is $x = 0$ . Since the $x$ -values should match, we can substitute the latter equation into the former:

$y = -0^2 + 4 = 4.$

Therefore, our $y$ -intercept is $(0, 4)$ . No surprises there.

Example 2. Determine the two $x$ -intercepts of the quadratic graph $y = -x^2 + 4$ .

Solution. To determine the two $x$ -intercepts of the quadratic graph, we need to find the intersections of the graph $y = -x^2 + 4$ and the $x$ -axis, whose equation is $y = 0$ . Since the $y$ -values should match, we can substitute the latter equation into the former:

$0 = -x^2 + 4 \quad \Rightarrow \quad x^2 = 4.$

It is tempting at this stage to write $x = \sqrt 4 = 2$ so that our $x$ -intercept is $(2, 0)$ . However, this solution is only partially correct. Just like any solution in Singapore or US politics.

Notice in the graph that we have a second $x$ -intercept to the left of the $y$ -axis. This is because $x$ could have been a negative number as well. Notice that

$2^2 = 4\quad \text{and}\quad (-2)^2 = 4$

are both correct equations. (Furthermore, these are the only two correct equations). Therefore, if $x^2 = 4$ , we must conclude either possibility, namely, that $x = -2$ or $x = 2$ . To abbreviate, we use the $\pm$ notation: $x = \pm 2$ . This notation states that either $-2$ or $+2$ are plausible values that $x$ represents.

To answer our original question, since $x = \pm 2$ and $y = 0$ in both cases, there are two $x$ -intercepts: $(-2, 0)$ and $(2, 0)$ .

Example 3. Determine the $x$ -intercept(s) of the quadratic graph $y = x^2 - 4$ .

Solution. To determine the two $x$ -intercepts of the quadratic graph, we need to find the intersections of the graph $y = x^2 - 4$ and the $x$ -axis, whose equation is $y = 0$ . Since the $y$ -values should match, we can substitute the latter equation into the former:

$x^2 - 4 = 0 \quad \iff \quad 0 = -x^2 + 4.$

By Example 2, we have the solutions $x = \pm 2$ . Therefore, there are two $x$ -intercepts: $(-2, 0)$ and $(2, 0)$ .

Remark 1. Different graphs can yield the same roots.

Example 4. Determine the $x$ -intercept(s) of the quadratic graph $y = x^2 - 2x + c$ , where $c = -3, -1, 1, 2$ respectively.

Solution. We recall that $(x-1)^2 = x^2 - 2x + 1$ . Therefore,

$\begin{aligned} y = x^2 - 2x + c = (x-1)^2 - 1 + c. \end{aligned}$

To find the $x$ -intercepts of the graph, we set $y = 0$ :

$\begin{aligned} (x-1)^2 - 1 + c &= 0 \\ (x-1)^2 &= 1 - c. \end{aligned}$

Now we analyse this equation case-by-case.

If $c = -3$ , then $(x-1)^2 = 1-(-3) = 4$ . Taking square roots,

$x-1 = \pm \sqrt 4 = \pm 2 \quad \Rightarrow \quad x = 1 \pm 2.$

Hence, $x = -1$ or $x = 3$ , yielding the $x$ -intercepts $(-1, 0)$ and $(3, 0)$ .

If $c = -1$ , then $(x-1)^2 = 1-(-1) = 2$ . Taking square roots,

$x-1 = \pm \sqrt 2 \quad \Rightarrow \quad x = 1 \pm \sqrt 2.$

Hence, $x = 1 - \sqrt 2$ or $x = 1 + \sqrt 2$ , yielding the $x$ -intercepts $(1 - \sqrt 2, 0)$ and $(1 + \sqrt 2, 0)$ .

If $c = 1$ , then $(x-1)^2 = 1-1 = 0$ . Taking square roots,

$x-1 = 0 \quad \Rightarrow \quad x = 1.$

Hence, $x = 1$ , yielding only one $x$ -intercept $(1, 0)$ .

If $c = 3$ , then $(x-1)^2 = 1-3 = -2$ . Now, if $x$ is a real number, then $(x-1)$ is a real number too, which implies that $(x-1)^2 \geq 0 > -1$ . Therefore, $(x-1)^2 \neq -2$ . This can only mean that the equation $(x-1)^2 = -2$ has no real solutions, and therefore, the graph of $y = x^2 - 2x + 2$ no $x$ -intercepts.

The diagram below shows the graphs of all four cases.

Example 5. Given real numbers $h$ and $k$ , determine the number of $x$ -intercepts of the quadratic graph $y = (x-h)^2 + k$ , and determine them in terms of $h$ and $k$ if they exist.

Solution. Repeating the calculations in Example 4, we set $y = 0$ :

$(x-h)^2 + k = 0 \quad \Rightarrow \quad (x-h)^2 = -k.$

There are three cases: $k < 0$ , $k = 0$ , or $k > 0$ .

In the case $k < 0$ , $-k > 0$ , so that $\sqrt{-k} > 0$ . Hence,

$(x-h)^2 = -k \quad \Rightarrow \quad x = h \pm \sqrt{-k}.$

Therefore, the graph has two $x$ -intercepts $(h - \sqrt{-k}, 0)$ and $(h + \sqrt{-k}, 0)$ .

In the case $k = 0$ ,

$(x-h)^2 = -k \quad \Rightarrow \quad x = h \pm \sqrt{-k} = h \pm \sqrt{-0} = h.$

Therefore, the graph has only one $x$ -intercept $(h, 0)$ .

In the case $k > 0$ , $-k < 0$ , so that

$(x-h)^2 \geq 0 > -k.$

Therefore, the equation $(x-h)^2 = -k$ has no real solutions, and the graph has no $x$ -intercepts.

Remark 2. The line $x = h$ is called the line of symmetry of the graph, and the point $(h, k)$ is the turning point of the graph.

Definition 1. We say that $x_0$ is a real number solution of the quadratic equation $ax^2 + bx + c = 0$ if $ax_0^2 + bx_0 + c = 0$ . Equivalently,
- $x_0$ is a real number solution of the equation $ax^2 + bx + c = 0$ ,
- $x_0$ is a root of the function $ax^2 + bx + c$ ,
- $(x_0, 0)$ is an $x$ -intercept of the graph $y = ax^2 + bx + c$ .
Theorem 1. Let $ax^2 + bx + c = 0$ be a quadratic equation, where $a \neq 0$ . Define the discriminant $\Delta$ of the quadratic equation by

$\Delta := b^2 - 4ac.$

Then the equation $ax^2 + bx + c = 0$ has
- $2$ real number solutions if $\Delta > 0$ ,
- $1$ real number solution if $\Delta = 0$ , and
- $0$ real number solutions if $\Delta < 0$ .
In the first two cases, the roots $x_{\pm}$ are given by the quadratic formula:

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

Proof Sketch. We first leave it as an exercise to find constants $h, k$ in terms of $a,b, c$ such that

$\displaystyle y = ax^2 + bx + c \quad \iff \quad \frac ya = (x-h)^2 + k.$

This process is called completing the square, and done correctly, should yield the results

$\displaystyle h = - \frac b{2a},\quad k = -\frac{\Delta}{(2a)^2}.$

Then set $y = 0$ and apply the calculation in Example 5. In the case $a > 0$ and $\Delta \geq 0$ ,

$\displaystyle x = -h \pm \sqrt{-k} = -\frac{b}{2a} \pm \sqrt{\frac{\Delta}{(2a)^2}} = \frac{ -b \pm \sqrt{ \Delta } }{2a}.$

Once again, the square roots make their return. In Example 4, the equation $x^2 - 2x - 1 = 0$ has the roots given by $x = 1 \pm \sqrt 2$ . We will explore expressions of this form in the next post.

—Joel Kindiak, 24 Oct 25, 1256H
February 4, 2026
Solving Cubic Equations

Previously, we concluded the previous post with a simple question: how do we solve the cubic equation below?

$ax^3 + bx^2 + cx + d = 0$

Definition 1. Given real number constants $a,b,c,d$ with $a \neq 0$ , we call the graph $y = ax^3 + bx^2 + cx + d$ a cubic graph.

Using ideas in undergraduate mathematics, we can show that every cubic graph must have at least one real root.

Lemma 1. Given constants $a,b,c,d$ with $a \neq 0$ , suppose for any $x$ we have

$ax^3 + bx^2 + cx + d = 0.$

Then $a=b=c=d=0$ .

Proof. To illustrate the proof, observe that setting $x = 0$ yields $d = 0$ , so that

$ax^3 + bx^2 + cx = 0.$

For nonzero $x$ , we must have $ax^2 + bx + c = 0$ . Since this equation holds for any $x$ except $0$ , using the notion of limits in calculus, we are allowed to set $x = 0$ to deduce $c = 0$ . Repeat the procedure to deduce that $a = b = 0$ as well.

A more complete proof uses some linear algebra; namely the standard basis of polynomial space.

Now consider the cubic graph $y = ax^3 + bx^2 + cx + d$ . We observe that for any $r$ ,

$s := ar^3 + br^2 + cr + d$

is a uniquely defined real number. Since can write $s$ purely in terms of $r$ , we make the notation $s = f(r)$ , where

$f(x) = ax^3 + bx^2 + cx + d.$

We call $f$ a polynomial function, properly defined using discrete mathematics. In the case above, we say that $f$ is a cubic function. In fact, the analogous function

$f(x) = ax^2 + bx + c$

is called the quadratic function. The function $f(x) = ax+b$ is called a linear function, since $f(x)$ grows linearly in the following sense:

$\begin{aligned} f(x+k) &= a(x+k) + b \\ &= ax + b + ak \\ &= f(x) + ak. \end{aligned}$

Similarly the function $f(x) = a$ is called a constant function, since no matter what input number $x$ we plug in, the output is a constant number $a$ .

Lemma 2. For any real number $k$ and $n = 2, 3$ , $(x - k)$ is a factor of $(x^n - k^n)$ .

Proof. For the case $n = 2$ , we use the difference of squares property:

$x^2 - k^2 = (x-k)(x+k).$

Thus, $(x-k)$ is a factor of $(x^2 - k^2)$ . For the case $n = 3$ , we seek constants $p, q$ such that

$x^3 - k^3 = (x-k)(x^2 + px + q).$

On the right-hand side,

$\begin{aligned} (x-k)(x^2 + px + q) &= x(x^2 + px + q) - k(x^2 + px + q) \\ &= (x^3 + px^2 + qx) - (kx^2 + kpx + kq) \\ &= x^3 + (p-k)x^2 + (q-kp)x - kq \end{aligned}$

In order for the right-hand side to equal $(x^3 - k^3)$ , we must stipulate $p-k = 0 \iff p = k$ and

$q - kp = 0 \quad \iff \quad q = kp = k^2.$

Using these stipulations,

$\begin{aligned} (x-k)(x^2 + kx + k^2) &= x^3 + (k-k)x^2 + (k^2-k \cdot k)x - k \cdot k^2 \\ &= x^3 - k^3, \end{aligned}$

as required.

Theorem 1. For any cubic function, $(x-k)$ is a factor of $f(x) - f(k)$ .

Proof. Expanding $f(x) - f(k)$ yields

$\begin{aligned} f(x) - f(k) &= (ax^3 + bx^2 + cx + d) - (ak^3 + bk^2 + ck + d) \\ &= a(x^3 - k^3) + b(x^2 - k^2) + c(x-k). \end{aligned}$

By Lemma 2, since $(x-k)$ is a factor of both $(x^2 - k^2)$ and $(x^3 - k^3)$ , there exist polynomials $g(x), h(x)$ such that

$\begin{aligned} f(x) - f(k) &= a \cdot (x-k) \cdot g(x)+ b\cdot (x-k) \cdot h(x) + c(x-k) \\ &= (x-k) \cdot (a \cdot g(x)+ b \cdot h(x) + c). \end{aligned}$

Since $a \cdot g(x) + b \cdot h(x) + c$ is a polynomial, we have that $(x-k)$ is a factor of $(f(x) - f(k))$ .

Remark 1. The result of Theorem 1 still holds for polynomials whose highest power is larger than $3$ . Details here.

Lemma 3. Given any linear, quadratic, or cubic function $f$ and real number $k$ , there exists a unique polynomial $g$ and a unique real number $r$ such that

$f(x) = (x-k) \cdot g(x) + r.$

We call $r$ the remainder of $f$ after dividing by $(x-k)$ .

Proof. See Problem 3 in this post.

Theorem 2. Given any linear, quadratic, or cubic function $f$ , the remainder of $f$ after dividing by $(x-k)$ is $f(k)$ . Furthermore, $k$ is a root of $f$ if and only if $(x-k)$ is a factor of $f(x)$ .

Proof. By Theorem 1, $(x-k)$ is a factor of $f(x) - f(k)$ . Hence, there exists a polynomial $g$ such that

$f(x) - f(k) = (x-k) \cdot g(x)\quad \iff \quad f(x) = (x-k) \cdot g(x) + f(k).$

By Lemma 3, since $g$ and $f(k)$ are unique, $f(k)$ is the remainder of $f$ after dividing by $(x-k)$ is $f(k)$ . Furthermore, $(x-k)$ is a factor of $f(x)$ if and only if $f(k) = 0$ , which holds if and only if $k$ is a root of $f$ .

Remark 2. The first result is called the remainder theorem, while the second result is called the factor theorem. Furthermore, this result holds for polynomials whose highest power of $x$ (i.e. degree) is larger than $3$ .

Example 1. Show that the equation $x^3 - 6x^2 + 11x - 6 = 0$ has one solution given by $x = 1$ . Hence, solve the equation completely.

Solution. Define the function $f(x) = x^3 - 6x^2 + 11x - 6$ . We first observe that

$\begin{aligned} f(1) &= 1^3 - 6 \cdot 1^2 + 11 \cdot 1 - 6 \\ &= 1 - 6 + 11 - 6 \\ &= 0. \end{aligned}$

Therefore, $1$ is a root of $f$ . By the factor theorem, $(x-1)$ is a factor of $f(x)$ . Therefore, there exist real numbers $p, q$ such that

$x^3 - 6x^2 + 11x - 6 = (x-1)(x^2 + px + q).$

Expanding the right-hand side,

$\begin{aligned} x^3 - 6x^2 + 11x - 6 &= (x-1)(x^2 + px + q) \\ &= (x^3 + px^2 + qx) - (x^2 + px + q) \\ &= x^3 + (p-1)x^2 + (q-p)x - q. \end{aligned}$

Comparing the coefficients of $x^2$ and $1$ respectively,

$p-1 = -6 \quad \Rightarrow \quad p = -5,$

and $q = 6$ . Therefore,

$x^3 - 6x^2 + 11x - 6 = (x-1)(x^2 - 5x + 6).$

To solve the equation, we set the left-hand side equal to $0$ :

$(x-1)(x^2 - 5x + 6) = 0.$

Therefore, $x-1 = 0$ or $x^2 - 5x + 6 = 0$ . In the former, $x = 1$ . In the latter, we use the quadratic equation:

$\displaystyle x = \frac{ -(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 6} }{ 2 \cdot 1} = \frac{5 \pm 1}{2}.$

Therefore, $x = 2$ or $x = 3$ . Therefore, the equation

$x^3 - 6x^2 + 11x - 6 = 0$

has three solutions, namely: $x = 1, 2, 3$ .

In particular, we can, somewhat reasonably, solve all cubic equations.

Theorem 3. Given real constants $a,b,c,d$ and $a \neq 0$ , there exists real constants $x_0, p, q$ such that equation

$ax^3 + bx^2 + cx + d = (x - x_0)(ax^2 + px + q).$

Proof. Define $f(x) = ax^3 + bx^2 + cx + d$ . Using the intermediate value theorem in calculus and real analysis, we can show that $x_0$ must be a root of $f$ . By the factor theorem, $(x-x_0)$ is a factor of $f$ . Hence, the factorisation holds.

Remark 3. For a more systematic approach to solve cubic equations, i.e. some kind of cubic formula, we need to use complex numbers.

Corollary 1. Every cubic equation has either one real solution, two real solutions, or three real solutions.

Example 2. Determine the (possibly complex) 3rd roots of unity—the solutions to the equation $z^3 = 1$ .

Solution. Rather obviously, $z = 1$ is a solution of the equation $z^3 - 1 = 0$ . By the factor theorem, $(z-1)$ is a factor of $(z^3 - 1)$ . Hence, there exist real constants $p, q$ such that

$\begin{aligned} z^3 - 1 &= (z-1)(z^2 + pz + q) \\ &= z^3 + (p-1)z^2 + (q-1)z - q. \end{aligned}$

Comparing coefficients, $p = q = 1$ , so that

$z^3 - 1 = (z-1)(z^2 + z + 1).$

To solve the equation $z^2 + z + 1 = 0$ , we use the quadratic formula:

$\displaystyle z = \frac{ -1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1} }{ 2 \cdot 1 } = \frac{ - 1 \pm \sqrt {-3} }{ 2 } = \textstyle \frac 12 (-1 \pm i\sqrt 3).$

Therefore, $z = 1, \frac 12 (-1 + i\sqrt 3), \frac 12 (-1 - i\sqrt 3)$ .

Remark 4. Defining the primitive cube root of $1$ by $\zeta_3 := \frac 12 (-1 + i\sqrt 3)$ , we can show that

$\zeta_3^2 = \frac 12 (-1 - i\sqrt 3),\quad \zeta_3^3 = 1.$

This idea is leveraged in the mathematical study called Galois theory, and more broadly, abstract algebra. We call the set $\{ 1, \zeta_3 , \zeta_3^2 \}$ a cyclic group under multiplication, and thus Abelian, since

$\{ 1, \zeta_3 , \zeta_3^2 \} = \{ \zeta_3^n : n \in \mathbb Z \}.$

We can establish this connection using some basic divisibility ideas in introductory number theory.

We could experiment with polynomials of higher degrees, but it turns out that we can’t do better than a degree-four polynomial, by an advanced result known as the Abel–Ruffini theorem.

What is for sure is that for any degree- $n$ polynomial, we must have $n$ complex roots—this is called the fundamental theorem of algebra. By the intermediate value theorem again, if $n$ is odd, we are guaranteed at least one real root.

All’s to say this: making sense of polynomials with higher powers is really, really hard. We won’t be able to do it much justice in our current discussion, but that’s okay. Let’s start small.

We leave it as an exercise to verify the following algebraic expansions are correct:

$\begin{aligned} (1+x)^1 &= 1+x, \\ (1+x)^2 &= 1 + 2x + x^2, \\ (1+x)^3 &= 1 + 3x + 3x^2 + x^3, \\ (1+x)^4 &= 1 + 4x + 6x^2 + 4x^3 + x^4. \end{aligned}$

How do we expand the binomial $(1+x)^5$ ? We explore the answer to this question when discussing the binomial theorem.

—Joel Kindiak, 27 Oct 25, 1214H

January 30, 2026
Visualising Vectors
So far, we have discussed numbers that:
- either have no direction, or
- have one dimension of direction.
We use positive numbers to represent either quantities with no direction, or quantities with some reasonable notion of “increase”. The larger the positive number, the larger the quantity.

Definition 1. For any positive number $x$ , define the magnitude of $x$ by $|x| := x$ .

For the second case, we use negative numbers, namely numbers of the form $-x$ for some positive number $x$ , to represent quantities with some reasonable notion of “decrease”.
- The quantity $x$ describes the magnitude of the decrease.
Since $x = -(-x)$ , we have the following more refined definition of a magnitude.

Definition 2. For any real number $x$ , define the magnitude of $x$ by

$|x| := \begin{cases} x, & x \geq 0, \\ -x, & x < 0. \end{cases}$

We also call $|x|$ the absolute value of $x$ .
- Note that we define $|0| = 0$ ; indeed, the number $0$ should denote some quantity with non-existent size.
Many a time, however, since we live in a three-dimensional world, it helps to have quantities that describe three-dimensional change. While what follows easily extends to three dimensions, we will keep discussions simple by working with just two dimensions.

Definition 3. Define a two-dimensional vector by the object $\begin{bmatrix} x \\ y \end{bmatrix}$ , visualised using a two-dimensional arrow in the $x$ – $y$ plane.

Using Pythagoras’ theorem, define the magnitude or the norm of the vector $\begin{bmatrix} x \\ y \end{bmatrix}$ by

$\left\| \begin{bmatrix} x \\ y \end{bmatrix} \right\| := \sqrt{x^2 + y^2}.$

Example 1. Using the diagram above,

$\|\mathbf u\| = \left\| \begin{bmatrix} 2 \\ 1 \end{bmatrix} \right\| = \sqrt{2^2 + 1^2} = \sqrt{5}.$

We leave it as an exercise to verify that

$\|\mathbf v\| = \sqrt{10},\quad \|\mathbf w\| = \sqrt{8},\quad \| \mathbf z\| = \sqrt{13}.$

Example 2. Let $x$ be a real number. Show that $\left\| \begin{bmatrix} x \\ 0 \end{bmatrix} \right\| = |x|$ .

Solution. By Definition 3,

$\left\| \begin{bmatrix} x \\ 0 \end{bmatrix} \right\| = \sqrt{x^2 + 0^2} = \sqrt{x^2}.$

Now we consider two cases:
- If $x \geq 0$ , then $\sqrt{x^2} = x$ .
- If $x < 0$ , then $-x > 0$ and $\sqrt{x^2} = \sqrt{(-x)^2} = -x$ .
By Definition 2, $\sqrt{x^2} = |x|$ . Therefore,

$\left\| \begin{bmatrix} x \\ 0 \end{bmatrix} \right\| = \sqrt{x^2} = |x|.$

Remark 1. Example 2 illustrates vectors as extensions of the numbers that we are familiar with (not without its limitations). Hence, we can describe $|x|$ as the norm or the magnitude of $x$ .

This characterisation of vectors turns out to be incredibly useful in making sense of two-dimensional quantities. However, we need to define meaningful calculations to actually use them.

Consider the vectors $\mathbf u = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$ and $\mathbf v = \begin{bmatrix} -1 \\ 3 \end{bmatrix}$ below.

What do we mean by $\mathbf u + \mathbf v$ ? Intuitively, it means that starting from the point $(0, 0)$ , we first travel according to $\mathbf u$ , then continue our journey according to $\mathbf v$ . This process is equivalent to sliding the ‘tail’ of $\mathbf v$ to the ‘tip’ of $\mathbf u$ , also known as tip-to-tail addition.

An equivalent interpretation is to create a parallelogram using $\mathbf u$ and $\mathbf v$ as sides, and $\mathbf u + \mathbf v$ , by definition is the ‘final point’ regardless how we travel. In either case, we notice that

$\begin{bmatrix}2 \\ 1 \end{bmatrix} + \begin{bmatrix} -1 \\ 3 \end{bmatrix} = \mathbf u + \mathbf v = \begin{bmatrix}1 \\ 4\end{bmatrix} = \begin{bmatrix} 2 + (-1) \\ 1 + 3 \end{bmatrix}.$

Therefore, we are justified in making the following definition for vector addition. We include scalar multiplication using similar intuitions.

Definition 4. Define vector addition and scalar multiplication as follows:

$\begin{bmatrix} x_1 \\ y_1\end{bmatrix} + \begin{bmatrix} x_2 \\ y_2\end{bmatrix} := \begin{bmatrix} x_1+x_2 \\ y_1+y_2\end{bmatrix},\quad \alpha \begin{bmatrix} x \\ y \end{bmatrix} := \begin{bmatrix} \alpha x \\ \alpha y \end{bmatrix}.$

In particular, given the two-dimensional vectors $\mathbf u, \mathbf v$ , define:
- $-\mathbf v := (-1)\mathbf v$ , and
- $\mathbf u -\mathbf v := \mathbf u + (-\mathbf v)$ .
Example 3. Let $\mathbf v$ be any two-dimensional vector. Define the vector $\mathbf 0 := \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ . Evaluate separately the quantities $\mathbf v + \mathbf 0$ , $0 \mathbf v$ , and $\mathbf v - \mathbf v$ .

Solution. Suppose $\mathbf v = \begin{bmatrix} x \\ y \end{bmatrix}$ . By the definition of vector addition,

$\begin{aligned}\mathbf v + \mathbf 0 &= \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} x + 0 \\ y + 0 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} = \mathbf v.\end{aligned}$

By the definition of scalar multiplication,

$\begin{aligned} 0\mathbf v &= 0\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0x \\ 0y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \mathbf 0.\end{aligned}$

Finally by the definition of vector subtraction and scalar multiplication,

$\begin{aligned}\mathbf v - \mathbf v &= \mathbf v + (-1)\mathbf v \\ &= \begin{bmatrix} x \\ y \end{bmatrix} + (-1) \begin{bmatrix} x \\ y \end{bmatrix} \\ &= \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} (-1)x \\ (-1)y \end{bmatrix} \\ &= \begin{bmatrix} x + (-1)x \\ y + (-1)y \end{bmatrix} \\ &= \begin{bmatrix} x -x \\ y -y \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \mathbf 0.\end{aligned}$

For much more detail and insight, check out my fuller suite of posts on linear algebra here. Linear algebra, at its core, is the very first bridge between geometry and algebra that any student encounters.

Theorem 1. Consider the line $\ell : y = mx + c$ . Then there exist two-dimensional vectors $\mathbf r_0, \mathbf d$ such that

$\begin{bmatrix} x \\ y \end{bmatrix} = \mathbf r_0 + \lambda \mathbf d, \quad \lambda \in \mathbb R.$

In this case, we call the vector $\mathbf d$ the direction vector of $\ell$ .

Proof Sketch. Define $\mathbf r_0 = \begin{bmatrix} 0 \\ c \end{bmatrix}$ and $\mathbf d = \begin{bmatrix} 1 \\ d \end{bmatrix}$ , and verify that the equation holds.

Theorem 2. Define the lines $\ell_1 : y = m_1 x$ and $\ell_2 : y = m_2 x$ , where $m_1 \neq 0$ . Then $\ell_1 \perp \ell_2$ if and only if $m_1 \cdot m_2 = -1$ .

Proof Sketch. Consider the diagram below.

Using Pythagoras’ theorem,

$OA = \sqrt{1 + m_1^2},\quad OB = \sqrt{1 + m_2^2},\quad AB = \sqrt{(m_1 - m_2)^2}.$

Therefore, by Pythagoras’ theorem and its converse, $OA \perp OB$ if and only if $OA^2 + OB^2 = AB^2$ :

$\begin{aligned} (1 + m_1^2) + (1 + m_2^2) &= (m_1 - m_2)^2.\end{aligned}$

We leave it as an exercise in algebra to simplify this equation to

$m_1 \cdot m_2 = -1.$

Theorem 3. Define the lines $\ell_1 : y = m_1 x$ and $\ell_2 : y = m_2 x + c$ , where $m_1\neq 0$ . Then $\ell_1 \parallel \ell_2$ if and only if $m_1 = m_2$ .

Remark 2. Observe the deliberate omission of a diagram in Theorem 3. The power of vectors (i.e. linear algebra) is to describe geometry without a need for visual representation (though the latter will be useful for us in the process of proving the result).

Proof Sketch. Define the line $\ell_3 : y = -(1/m_1 ) x$ . By Theorem 2, since $m_1 \cdot (-1/m_1 ) = -1$ , $\ell_1 \perp \ell_3$ .

Since the interior angles of a pair of lines sum to $180^\circ$ if and only if the lines are parallel,

$\begin{aligned} \ell_1 \parallel \ell_2 \quad &\iff \quad \ell_2 \perp \ell_3 \\ &\iff \quad m_2 \cdot \left(-\frac 1{m_1 } \right) = -1 \\ & \iff \quad m_1 = m_2 . \end{aligned}$

Corollary 1. Consider the lines

$\begin{aligned} \ell_1 : y &= m_1 x + c_1, \\ \ell_2 : y &= m_2 x + c_2, \end{aligned}$

where $m_1\neq 0$ . Then $\ell_1 \parallel \ell_2$ if and only if $m_1 = m_2$ .

Proof. Define $\ell_3 : y = m_1 x$ . By Theorem 3, $\ell_1 \parallel \ell_3$ . Then by Theorem 3 again,

$\begin{aligned} \ell_1 \parallel \ell_2\quad &\iff \quad \ell_2 \parallel \ell_3 \\ &\iff \quad m_1 = m_2. \end{aligned}$

Theorem 4. Let $\mathbf r_1,\mathbf r_2, \mathbf d_1,\mathbf d_2$ be two-dimensional vectors and $\mathbf d_1 \neq \mathbf 0, \mathbf d_2 \neq \mathbf 0$ . Using Theorem 1, consider the lines $\ell_1,\ell_2$ defined by

$\begin{aligned} \ell_1 : \mathbf r &= \mathbf r_1 + \lambda \mathbf d_1,\quad \lambda \in \mathbb R, \\ \ell_2 : \mathbf r &= \mathbf r_2 + \mu \mathbf d_2,\quad \mu \in \mathbb R, \end{aligned}$

where we abbreviate $\mathbf r = \begin{bmatrix} x \\ y \end{bmatrix}$ . Then $\ell_1 \parallel \ell_2$ if and only if there exists some real number $k$ such that $\mathbf d_2 = k\mathbf d_1$ .

Proof Sketch. Use Theorems 1 and 3.

There are many more implications of thinking in terms of vectors, but we conclude with the famous intercept theorem.

Lemma 1. Given two points $A, B$ , denote the vector starting at $A$ and ending at $B$ by $\overrightarrow{AB}$ .

Then $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ .

Proof. Using vector addition,

$\begin{aligned} \overrightarrow{AB} &= \overrightarrow{AO} + \overrightarrow{OB} \\ &= (-1)\, \overrightarrow{OA} + \overrightarrow{OB} \\ &= \overrightarrow{OB} + (-1)\, \overrightarrow{OA} \\ &= \overrightarrow{OB} - \overrightarrow{OA}. \end{aligned}$

Theorem 5 (Intercept Theorem). Given three distinct points $O,A,B$ and positive numbers $k_1, k_2$ , define the points $C, D$ by $\overrightarrow{OC} = k_1 \overrightarrow{OA}$ and $\overrightarrow{OD} = k_2 \overrightarrow{OB}$ below.

(Here, we assume $k_1 > 1, k_2 > 1$ for simplicity.)

Then $AB \parallel CD$ if and only if $k_1 = k_2$ . In this case,

$\displaystyle \frac{OA}{OC} = \frac{OB}{OD} = \frac{AB}{CD}.$

Proof Sketch. Denote $\overrightarrow{OA} = \mathbf a$ and $\overrightarrow{OB} = \mathbf b$ . By Lemma 1,

$\overrightarrow{AB} = \mathbf b - \mathbf a,\quad \overrightarrow{CD} = k_2 \mathbf b - k_1 \mathbf a.$

In the direction $(\Leftarrow)$ , suppose $k_1 = k_2 =: k$ . Then

$\overrightarrow{CD} = k \mathbf b - k \mathbf a = k(\mathbf b - \mathbf a) = k \overrightarrow{AB}.$

By Theorem 4, $AB \parallel CD$ .

In the direction $(\Rightarrow)$ , Theorem 4 yields some real number $k$ such that

$\begin{aligned} \overrightarrow{CD} &= k \overrightarrow{AB} \\ k_2 \mathbf b - k_1 \mathbf a &= k\mathbf b - k\mathbf a \\ (k - k_1) \mathbf a &= (k - k_2)\mathbf b \end{aligned}$

Now $\mathbf a, \mathbf b$ are non-zero. If $k_2 \neq k$ , then

$\begin{aligned} \mathbf b &= \frac{ k - k_1 }{ k - k_2 }\, \mathbf a \end{aligned}$

implies that $OA \parallel OB$ , a contradiction. Therefore, we must have $k_2 = k$ . Similarly, $k_1 = k$ . Therefore, $k_1 = k_2 = k$ , in which case,

$\displaystyle OC = k \cdot OA \quad \Rightarrow \quad \frac{OA}{OC} = \frac 1{k}.$

In a similar manner with the other sides,

$\displaystyle \frac{OA}{OC} = \frac{OB}{OD} = \frac 1{k} = \frac{AC}{CD}.$

Corollary 2 (Midpoint Theorem). If we have

$\displaystyle \frac{OA}{OC} = \frac{OB}{OD} = \frac 12,$

then $AB \parallel CD$ and $AB/CD = 1/2$ .

Proof. By hypothesis, set $k_1 = k_2 = 2$ in Theorem 5 to obtain $AB \parallel CD$ , and consequently, $AB/CD = 1/2$ .

In this case, we call $\Delta OAB$ and $\Delta OCD$ similar triangles, which we will revisit later on.

Using this idea of describing shapes using coordinates, we turn to parabolas, and namely, analyse graphs of the form $y = ax^2 + bx + c$ .

—Joel Kindiak, 22 Oct 25, 2217H
January 28, 2026
Similar Triangles

Having discussed a great deal about congruent triangles, we now turn our attention to a relaxation of congruence—similarities. Similarities are, no pun intended, similar ideas to congruence, but far more useful in helping us use small-scale items to represent large-scale items, such as maps and scale models.

Lemma 1. If $\Delta ABC \equiv \Delta PQR$ , then

$\displaystyle \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = 1.$

Definition 1 (SSS Similarity Definition). We say that the triangles $\Delta ABC, \Delta PQR$ are similar, denoted $\Delta ABC \sim \Delta PQR$ , if there exists some positive constant $k$ such that

$\displaystyle \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = k.$

Therefore, congruence becomes a special case of similarity via the special case $k = 1$ .

Since congruent triangles have angles that match, do similar triangles have angles match? Not only do they match, but that they have to match.

Lemma 2. We have $\Delta ABC \sim \Delta PQR$ if and only if their angles match:

$\angle ABC = \angle PQR,\quad \angle BCA = \angle QRP,\quad \angle BAC = \angle QPR.$

Proof Sketch. Suppose without loss of generality that $AB < PQ$ . For setup, translate and rotate $\Delta PQR$ where necessary so that $A = P$ and $B$ lies on $PQ$ .

In the direction $(\Rightarrow)$ , suppose $\Delta ABC \sim \Delta PQR$ . Then

$\displaystyle \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR}.$

A technical proof by contradiction shows that $C$ must lie on $PR$ . By the intercept theorem, $BC \parallel QR$ . Using corresponding angles,

$\angle ABC = \angle PQR,\quad \angle BCA = \angle QRP.$

Since the angles in a triangle sum to $180^\circ$ , $\angle BAC = \angle QPR$ .

In the direction $(\Leftarrow)$ , since $\angle BAC = \angle QPR$ , $C$ must lie on $PR$ . Since

$\angle ABC = \angle PQR,\quad \angle BCA = \angle QRP,$

using corresponding angles, $BC \parallel QR$ . By the intercept theorem again,

$\displaystyle \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR},$

yielding $\Delta ABC \sim \Delta PQR$ , as required.

Theorem 1 (AA Similarity Criterion). We have $\Delta ABC \sim \Delta PQR$ if and only if at least two of the following equalities hold:

$\angle ABC = \angle PQR,\quad \angle BCA = \angle QRP,\quad \angle BAC = \angle QPR.$

Proof. It suffices to prove the direction $(\Leftarrow)$ . Since angles in a triangle sum to $180^\circ$ , we get all three equalities if we know at least two of them hold. By Lemma 2, the result holds.

Theorem 2 (SAS Similarity Criterion). We have $\Delta ABC \sim \Delta PQR$ if and only if

$\displaystyle \angle ABC = \angle PQR,\quad \frac{AB}{PQ} = \frac{BC}{QR}.$

Proof Sketch. It suffices to prove the direction $(\Leftarrow)$ . Since $\angle ABC = \angle PQR$ , if $A$ lies on $PQ$ , then $C$ must lie on $QR$ .

By the intercept theorem,

$\displaystyle \frac{AB}{PQ} = \frac{BC}{QR} \quad \Rightarrow \quad AC \parallel PQ.$

Using corresponding angles, $\angle BAC = \angle QPR$ . By the AA Similarity Criterion, $\Delta ABC \sim \Delta PQR$ .

Similar triangles are used all the time in establishing interesting result in plane geometry.

Example 1 (Tangent-Secant Theorem). In the diagram below, $\angle ABC = \angle BDC$ (this is called the alternate segment theorem).

Show that $AB^2 = AC \cdot AD$ , known as the tangent-secant theorem.

Solution. Since $\angle ABC = \angle BDC = \angle BDA$ and $\angle BAC = \angle DBA$ , by the AA Similarity Criterion, $\Delta ABC \sim \Delta BDA$ . In particular,

$\displaystyle \frac{AB}{BD} = \frac{AC}{BA} = \frac{AC}{AB} \quad \Rightarrow \quad AB^2 = AC \cdot AD.$

Example 2. In the diagram below, $A,B,C,D$ are distinct points that lie on a circle.

Given that $K$ is the intersection of $AC$ and $BD$ , show that

$AK \cdot CK = BK \cdot DK.$

Solution. For the first claim, since angles in the same segment are equal (i.e. the butterfly theorem), $\angle ABK = \angle DCK$ . Since vertically opposite angles are equal, $\angle AKB = \angle DKC$ . By the AA Similarity Criterion, $\Delta ABK \sim \Delta DCK$ . In particular,

$\displaystyle \frac{ AK }{ DK } = \frac{ BK }{ CK } \quad \Rightarrow \quad AK \cdot CK = BK \cdot DK.$

Okay, let’s return to earth one more time.

Definition 2. Denote the scale $1 : N$ to mean that $1$ unit of length in some representation (e.g. a map, scale model, etc) represents $N$ units of length in real life.

Example 3. Suppose $1$ cm on a map represents $5$ km in real life. Determine the scale in terms of $1 : N$ .

Solution. Since $1\, \text m = 100\, \text{cm}$ ,

$\begin{aligned} 5\, \text{km} &= (5 \times 10^3) \times 1\, \text{m} \\ &= (5 \times 10^3) \times 100\, \text{cm} \\ &= 500\, 000\, \text{cm}. \end{aligned}$

Hence, the required scale is

$\begin{aligned}1\, \text{cm} : 5\, \text{km} &= 1\, \text{cm} : 500\, 000\, \text{cm} = 1 : 500\, 000.\end{aligned}$

Lemma 2. Suppose we have a scale of $1 : N$ . Then the area scale is given by $1 : N^2$ . That is, $1$ square unit of area represents $N^2$ square units of area in real life.

Proof Sketch. We can use the Riemann integration approach of approximating objects using a combination of squares. Since a square with length and height $1$ represents a real-life square with length and height $N$ , the real-life square has area $N \times N = N^2$ .

Lemma 3. Suppose we have a scale of $1 : N$ . Then the volume scale is given by $1 : N^3$ . That is, $1$ cubic unit of area represents $N^3$ cubic units of area in real life.

Proof Sketch. Follow the idea in Lemma 2.

Likewise, using triangles to approximate shapes, two objects are similar if they share the same shape, and thus share some kind of similarity ratio in Definition 1. Therefore, the results following Definition 2 hold, allowing us to use scale models to represent real-life objects. From these ideas we obtain products like world globe maps and merchandise of varying sizes.

To further leverage these similarity properties, it helps for us to have simple formulas for well-known shapes. While we cannot prove them all in O-Level mathematics, we can state them and relegate their proofs to integral calculus. We explore these formulas in the next post.

—Joel Kindiak, 4 Nov 25, 1515H

January 26, 2026
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 via the equation

$\displaystyle \frac{a-c}{d-b} = \sqrt{\Delta}.$

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 (or basic algebraic expansion),

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

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 would like 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.
Proof Sketch. Adapt the solutions in the previous lemmas and examples.

Definition 2. Define the conjugate of a complex number by $(a+bi)^* := a-bi$ .

Example 7. Let $z$ be a complex number. Show that $z z^*$ is a real number. Deduce that

$zz^* = 0 \quad \Rightarrow \quad z = 0 + 0i.$

Solution. Write $z = a+bi$ , where $a,b$ are real numbers. Using Example 4,

$\begin{aligned} zz^* &= (a+bi)(a-bi) \\ &= a^2 - b^2 \cdot (-1) \\ &= a^2 + b^2, \end{aligned}$

which is a real number since $a,b$ are real numbers. In particular, if $zz^* = 0$ , then $a^2 + b^2 = 0$ . Since $a,b$ are real numbers,

$0 \leq a^2 = a^2 + 0 \leq a^2 + b^2 = 0.$

Therefore, $a^2 = 0$ implies that $a = 0$ . Similarly, $b = 0$ , so that

$z = a+bi = 0 + 0i.$

In particular, we can make the following claim.

Theorem 2. The solutions to the equation $ax^2 + bx + c = 0$ are given by

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

In particular, the solutions are
- real and distinct if $\Delta > 0$ ,
- real and repeated if $\Delta = 0$ ,
- complex conjugates if $\Delta < 0$ .
Any further discussion on complex numbers is relegated to A-Level Mathematics, and so we will only explore this idea in detail when we get there.

For now, we can ask a humble yet daunting question: given real numbers $a,b,c,d$ where $a \neq 0$ , how can we solve the cubic equation below?

$ax^3 + bx^2 + cx + d = 0$

We will explore this idea next time using polynomials.

—Joel Kindiak, 25 Oct 25, 1443H
January 23, 2026
The Rules of Shapes

If the sum of angles in a $3$ -sided polygon (i.e. a $3$ -gon / tri-gon / triangle) is $180^\circ$ , what is the sum of angles in an $n$ -sided polygon (i.e. an $n$ -gon)?

Theorem 1. The sum of interior angles in an $n$ -gon is $(n-2) \times 180^\circ$ .

Proof. In the case $n = 3$ , we have an angle sum of $180^\circ = (3-2) \times 180^\circ$ , as required.

In the case $n = 4$ , the key observation is to split the $4$ -gon into a triangle and a $3$ -gon.

In Case 1, there is no reflex interior angle. In Case 2, there is at least one reflex interior angle. In either case, since the interior angle sum of a $3$ -gon is $180^\circ$ , the $4$ -gon must have an interior angle sum of

$\begin{aligned} 180^\circ + (3-2) \times 180^\circ &= (1 + (3-2)) \times 180^\circ \\ &= (4-2) \times 180^\circ = 360^\circ. \end{aligned}$

Roughly speaking, we can generalise this case using mathematical induction.

Suppose that a $k$ -gon has interior angle sum of $(k-2) \times 180^\circ$ . Then we can split the $(k+1)$ -gon into a triangle and a $k$ -gon. Since the triangle has interior angle sum of $180^\circ$ and the $k$ -gon has interior angle sum of $(k-2) \times 180^\circ$ , the $(k+1)$ -gon has interior angle sum of

$\begin{aligned} 180^\circ + (k-2) \times 180^\circ &= (1 + (k-2)) \times 180^\circ \\ &= ((k+1) - 2) \times 180^\circ. \end{aligned}$

Therefore, if the result holds for $k = 3$ , it holds for $k = 4$ , and subsequently, $k = 5$ , and so on and so forth. Therefore, any $n$ -gon must have an interior angle sum of $(n-2) \times 180^\circ$ .

Definition 1. An $n$ -sided polygon is convex if it does not have any reflex interior angle.

Theorem 2. The sum of exterior angles for any convex $n$ -gon is $360^\circ$ .

Proof. Recall that the exterior angle is the angle that lies on the same straight line as the interior angle that it is adjacent to (the diagram shows the case of a $5$ -gon, and there’s no requirement that the polygon be regular).

Let $\alpha_k, \beta_k$ denote the $k$ -th interior and exterior angle respectively. Since adjacent non-overlapping angles on a straight line sum to $180^\circ$ ,

$\alpha_k + \beta_k = 180^\circ.$

Adding all $n$ pairs of angles together,

$\begin{aligned} (\alpha_1 + \beta_1) + \cdots + (\alpha_n +\beta_n) &= \underbrace{ 180^\circ + \cdots + 180^\circ }_{n} \\ (\alpha_1 + \cdots + \alpha_n) + (\beta_1 + \cdots + \beta_n) &= n \times 180^\circ \\ \text{(sum of interior angles)} + \text{(sum of exterior angles)} &= n \times 180^\circ \end{aligned}$

By Theorem 1,

$\text{(sum of interior angles)} = (n-2) \times 180^\circ.$

Therefore, substituting into the original equation,

$\begin{aligned}\text{(sum of exterior angles)} &= n \times 180^\circ - \text{(sum of interior angles)} \\ &= n \times 180^\circ - (n-2) \times 180^\circ \\ &= (n - (n-2)) \times 180^\circ \\ &= 2 \times 180^\circ = 360^\circ. \end{aligned}$

Definition 2. A polygon is regular if all of its interior angles are equal.

Corollary 1. The interior angle of a regular $n$ -gon is $(1-2/n) \times 180^\circ$ .

What about circles? Circles have an interesting cousin that we will care about a lot. Recall that a circle is a set of points that are located a fixed distance $r$ , called the radius, away some fixed centre point.

Definition 3. Let $\ell$ be a point and $F$ be a point not on $\ell$ . A parabola with focus $F$ and directrix $\ell$ is the set of points whose distance to $F$ and $\ell$ are equal.

Theorem 3. Fix $d > 0$ . The equation of a parabola with focus $F(0, d)$ and directrix $y = -d$ is given by $x^2 = 4dy$ . In particular, it contains the point $(0, 0)$ .

Proof. Let $P(x,y)$ be any point on the parabola. Using Pythagoras’ theorem,

$FP = \sqrt{ (x - 0)^2 + (y - d)^2 }.$

By definition of the distance to a line, its distance from $\ell$ is given by $\sqrt{(y+d)^2}$ . By the definition of a parabola,

$\begin{aligned} \sqrt{ (x - 0)^2 + (y - d)^2 } &= \sqrt{(y+d)^2} \\ (x-0)^2 + (y-d)^2 &= (y+d)^2 \\ x^2 + y^2 - 2dy + d^2 &= y^2 + 2dy + d^2 \\ x^2 &= 4dy. \end{aligned}$

Remark 1. Defining $a := 1/(4d)$ , we get the equation $y = ax^2$ . Since it must pass through the points $(0, 0)$ and $(1,a) = (1, 1/(4d))$ , $d$ decreases as $a$ increases, so that the resulting parabola becomes steeper.

Theorem 4. Let $a > 0$ , $b, c$ be constants. Then the quadratic graph with equation $y = ax^2 + bx + c$ will always be a parabola with some focus $F(h, d)$ and horizontal directrix $y = k - d$ , where $h,k$ are constants in terms of $a,b,c$ .

Proof. We claim that there exists real constants $h, k$ in terms of $a, b, c$ such that

$y = a(x-h)^2 + k\quad \iff \quad y - k = a(x-h)^2.$

In this case, we can define $d := 1/(4a)$ and obtain a focus of $F(h, d)$ and a directrix with equation $y-k = -d \iff y = k - d$ .

To make our case, expand the right-hand side:

$\begin{aligned} a(x-h)^2 + k &= a(x^2 - 2xh + h^2) + k \\ &= ax^2 - (2ah)x + (ah^2 + k) \\ &= ax^2 + bx + c. \end{aligned}$

Therefore, ensure $b = -(2ah)$ and $c = ah^2 + k$ by setting

$\displaystyle h := -\frac{b}{2a}, \quad k := c - ah^2.$

If we expand the right-hand side of $k$ , we get

$\begin{aligned} k &= c - a \left( - \frac b{2a} \right)^2 \\ &= c - a \cdot \frac{b^2}{4a^2} \\ &= c - \frac{b^2}{4a} = -\frac{b^2 -4ac}{4a}, \end{aligned}$

where the quantity $\Delta := b^2 - 4ac$ is called the discriminant of the quadratic function $ax^2 + bx + c$ . Finally, we have $(h, k)$ as the turning point of the parabola.

Remark 1. The derivation of $h, k$ is known as completing the square on the right-hand side.

Corollary 2. The graph of $y = \sqrt x$ lies on a parabola with focus $(1/4, 0)$ and directrix $x = -1/4$ .

Proof. By considering the equation $y = x^2$ , we set $(a,b,c) = (1,0,0)$ in Theorem 4 to obtain $(h,k, d) = (0,0, 1/4)$ , so that we recover a focus of $(0, 1/4)$ and directrix of $y = -1/4$ . Now replace $(x,y)$ with $(y,x)$ to obtain the desired result.

The cousins of the circle and the parabola would be the ellipse and hyperbola respectively, and together, these shapes constitute the four famous conic sections. Deriving their equations follows in spirit with the parabola, albeit requiring a little more tedious bookkeeping. Perhaps these graphs are better left as an exercise.

Before we proceed to the next large sub-topic in secondary school mathematics, namely algebra, it might help us to explore its bridge, linear algebra, with a little more detail. The main object of interest would be the vector, and vectors turn out to help us conceptualise many geometric ideas using coordinates and algebraic calculations.

—Joel Kindiak, 18 Oct 25, 2134H

January 21, 2026