KindiakMath

Category: O-Level Math

Return to Circle Geometry

Problem 1. Consider the circle below with equation $x^2 + y^2 = 1$ .

Given $x_0 > 0$ and $y_0 > 0$ , calculate the gradient of the tangent $T$ passing through the point $P(x_0, y_0)$ lying on the circle. Deduce that $OP \perp T$ .

(Click for Solution)

Solution. Since $(x_0, y_0)$ lies on the circle,

$x_0^2 + y_0^2 = 1.$

For any line with gradient $m$ passing through $(x_0, y_0)$ ,

$y = m(x-x_0) + y_0.$

When the line intersects the circle,

$x^2 + (m(x-x_0) + y_0)^2 = 1.$

Expanding the left-hand side,

$\begin{aligned} x^2 + m^2(x-x_0)^2 + 2m(x-x_0)y_0 + y_0^2 &= 1 \\ x^2 + m^2(x-x_0)^2 + 2m(x-x_0)y_0 + y_0^2- 1 &= 0 \\ x^2 + m^2(x-x_0)^2 + 2m(x-x_0)y_0 -x_0^2 &= 0, \end{aligned}$

where the last line follows from $x_0^2 + y_0^2 = 1$ .

Using the factorisation $x^2 - x_0^2 = (x-x_0)(x+x_0)$ , the left-hand side factorises to

$(x-x_0)(x+x_0 + m^2(x-x_0) + 2my_0) = 0.$

For $T$ to be a tangent with gradient $m_T$ , we require $x_0$ to be the only root. Hence, we must have

$x_0+x_0 + m_T^2(x_0-x_0) + 2my_0 = 0.$

Solving for $m$ ,

$\displaystyle 2x_0 + 2m_T \cdot y_0 = 0 \quad \Rightarrow \quad m_T = -\frac{x_0}{y_0}.$

On the other hand, $m_{OP} = y_0/x_0$ , so that

$\displaystyle m_T \cdot m_{OP} = -\frac{x_0}{y_0} \cdot \frac{y_0}{x_0} = -1.$

Therefore, $OP \perp T$ . We recover the tangent is perpendicular to radius property of circles.

Remark 1. The same argument holds for any general circle, with more careful book-keeping.

Problem 2. Given that $AB$ and $AC$ are tangents to the circle, show that $AB = AC$ and $AO$ bisects $\angle BOC$ .

This result is described by the phrase tangent from external point.

(Click for Solution)

Solution. By Problem 1, $AB \perp OB$ and $AC \perp OC$ . As radii, $OB = OC$ . As a common side, $AO = AO$ . Therefore, by the RHS Criterion,

$\Delta AOB \cong \Delta AOC.$

In particular, $AB = AC$ and $\angle AOB = \angle AOC$ . Hence, $AO$ bisects $\angle BOC$ .

Problem 3. Given that $ABE$ is tangent to the circle, show that $\angle ABC = \angle BDC$ . Deduce that $\angle EBD = \angle BCD$ . This result is known as the alternate segment theorem.

(Click for Solution)

Solution. Make the following constructions.

We aim to prove that $\alpha = \beta$ .

By Problem 1, $\angle AOB = 90^\circ$ . Hence,

$\angle OBC = 90^\circ - \alpha.$

Since $OB = OC$ as radii, $\Delta BOC$ is isosceles. Hence,

$\angle OCB = \angle OBC = 90^\circ - \alpha.$

Since the angle at the centre equals twice the angle at the circumference,

$\angle BOC = 2 \cdot \angle BDC = 2 \beta.$

Since angles in a triangle sum to $180^\circ$ ,

$\begin{aligned} \angle BOC + \angle OBC + \angle OCB &= 180^\circ. \end{aligned}$

Substituting the displays,

$\begin{aligned} 2\beta + (90^\circ - \alpha) + (90^\circ - \alpha) &= 180^\circ \\ 2 \beta + 180^\circ - 2\alpha &= 180^\circ \\ 2\beta - 2\alpha &= 0 \\ 2\beta &= 2\alpha \\ \beta &= \alpha. \end{aligned}$

Therefore, $\angle BDC = \alpha = \beta = \angle ABC$ , as required.

For the second result,

$\begin{aligned} \angle EBD &= 180^\circ - \angle ABC - \angle ABC\quad &(\text{adj}\ \angle\ \text{on st. line}) \\ &= 180^\circ - \angle ABC - \angle BDC \quad &(\angle ABC = \angle BDC) \\ &= \angle BCD.\quad &(\angle\ \text{sum of}\ \Delta) \end{aligned}$

—Joel Kindiak, 9 Jan 26, 1146H

January 9, 2026
Matchmaking Triangles
This is our motivating question for today:

What is the size of an angle in a triangle whose side lengths are equal to each other?

You might laugh at this question and reply: it is obviously

$180^\circ \div 3 = 60^\circ.$

And your final answer is right—but how did you know that all three angles are equal to each other?

To answer this question properly, we need to properly discuss congruent triangles, from which we can discuss isosceles triangles, of which equilateral triangles are a special case.

Definition 1. Consider the two triangles in the diagram below.

We say that $\Delta ABC$ is congruent to $\Delta A'B'C'$ , denoted $\Delta ABC \equiv \Delta A'B'C'$ , if their corresponding side lengths and angles equal each other:

$a = a', \quad b = b',\quad c = c',\quad \alpha = \alpha', \quad \beta = \beta',\quad \gamma = \gamma'.$

In ordered tuple notation:

$(a,b,c,\alpha,\beta,\gamma) = (a',b',c',\alpha',\beta',\gamma').$

The topic of congruent triangles is commonly presented as a collection of “cookbook recipes” to determine if two triangles are effectively the same.

But how do we know that these recipes actually work? It turns out that we have the necessary “meta-recipes” to construct these recipes, and that’s what we shall do.

Theorem 1 (RHS Criterion). Let $\Delta ABC, \Delta PQR$ be the following two right-angled triangles with $\angle BCA = \angle QRP = 90^\circ$ :

Then $\Delta ABC \equiv \Delta PQR$ if and only if

$c = r\quad \text{and} \quad (a= p \quad \text{or} \quad b=q).$

Proof. In the direction $(\Rightarrow)$ , we obtain all of the equalities by Definition 1.

For the direction $(\Leftarrow)$ , suppose

$\displaystyle \angle BCA = \angle QRP = 90^\circ,\quad c = r, \quad a=p.$

We will deal with the case $b = q$ later. Observe that exactly one of the following hold:

$\angle ABC = \angle PQR, \quad \angle ABC < \angle PQR, \quad \angle ABC > \angle QPR.$

We claim that the remaining two cases are impossible.

Suppose $\angle ABC < \angle PQR$ . Then we can overlay the triangles on their edges $BC = QR$ as follows:

Hence, $AP > 0$ so that $b < q$ . However, by Pythagoras’ theorem,

$b = \sqrt{c^2 - a^2} = \sqrt{r^2 - p^2} = q,$

a contradiction, ruling out the case $\angle ABC < \angle PQR$ .

The case $\angle ABC > \angle QPR$ can also be ruled out by relabelling $(A,B,C)$ with $(P,Q,R)$ and vice versa, and using the same reasons above (i.e. a symmetric argument).

Therefore, $\angle ABC = \angle PQR$ .

Finally, since angles in a triangle sum to $180^\circ$ ,

$\begin{aligned} \angle BAC &= 180^\circ - \angle ABC - \angle BCA \\ &= 180^\circ - \angle PQR - \angle QRP = \angle QPR. \end{aligned}$

Since all corresponding sides and angles equal each other, $\Delta ABC \equiv \Delta PQR$ .

For the case $b = q$ , we could use a symmetric argument, or take the following even shorter approach: by Pythagoras’ theorem,

$a^2 = c^2 - b^2 = r^2 - q^2 = p^2,$

so that $a > 0$ and $p > 0$ implies $a = p$ , therefore

$\displaystyle \angle BCA = \angle QRP = 90^\circ,\quad c = r, \quad a=p,$

which, as shown just now, implies $\Delta ABC \equiv \Delta PQR$ .

The RHS Criterion only describes congruence between right-angled triangles. Yet, more is true. Since right angles open our discussion on general angles, right-angled triangles also open our exploration of general triangles.

Theorem 2 (SAS Criterion). Let $\Delta ABC, \Delta PQR$ be the following two triangles:

Then $\Delta ABC \equiv \Delta PQR$ if and only if

$b = q, \quad c = r,\quad \angle CAB = \angle RPQ.$

Proof. The direction $(\Rightarrow)$ is trivial.

For the direction $(\Leftarrow)$ , the hypothesis tells us that

$AC = b = q = PR.$

Construct the perpendicular heights (i.e. the altitudes) below.

Using rotation, we can assume $AC \parallel PR$ . By construction,

$\Delta ABD, \quad \Delta BCD, \quad \Delta PQS, \quad \Delta QRS$

are all right-angled triangles. Our line of attack is as follows:
- Show that $BD = QS$ .
- Deduce that $\Delta ABD \equiv \Delta PQS$ using the RHS Criterion.
- Similarly, $\Delta BCD \equiv \Delta QRS$ .
- Conclude that $\Delta ABC \equiv \Delta PQR$ .
The first point is the most challenging.

Since $AC \perp BD$ and $AC \parallel PR$ and $PR \perp QS$ , we must have $BD \parallel QS$ . We already have $AB = PQ$ by hypothesis, so it remains to prove $BD = QS$ .

If $BD < QS$ , then aligning $B$ with $Q$ , we can draw the diagram below.

By Pythagoras’ theorem,

$c^2 = \sqrt{{AD}^2 + {BD}^2} < \sqrt{{PS}^2 + {QS}^2} = r^2.$

Since $c > 0$ and $r > 0$ , we have $c < r$ , a contradiction. Similarly, we can rule out the case $BD > QS$ . Therefore, $BD = QS$ .

By the RHS Criterion, $\Delta ABD \equiv \Delta PQS$ . In particular, $BD = QS$ .

We next establish $\Delta BCD \equiv \Delta QRS$ . By hypothesis, $BC = QR$ . In particular,

$\begin{aligned} CD &= BC - BD \\ &= QR - QS = RS. \end{aligned}$

By Pythagoras’ theorem,

$\begin{aligned} BC^2 &= {BD}^2 + {CD}^2 \\ &= {QS}^2 + {RS}^2 = QR^2. \end{aligned}$

Since $BC > 0$ and $QR > 0$ , we have $BC = QR$ . By the RHS Criterion again, $\Delta BCD \equiv \Delta QRS$ . In particular,

$\begin{aligned} \angle BCA &= \angle BCD \\ &= \angle QRS = \angle QRP. \end{aligned}$

Finally,

$\begin{aligned} \angle ABC &= \angle ABD + \angle CBD \\ &= \angle PQS + \angle RQS = \angle PQR. \end{aligned}$

Remark 1. This proof holds true in all three cases (i.e. $\angle ABC$ and $\angle PQR$ can be acute, right, or obtuse), so long as both of the other two angles are acute.

Once we have a congruence criterion test that involves both sides and angles, we can derive many other useful congruence criterion tests.

Theorem 3 (SSS Criterion). Let $\Delta ABC, \Delta PQR$ be the following two triangles:

Then $\Delta ABC \equiv \Delta PQR$ if and only if

$a=p,\quad b = q, \quad c = r.$

Proof. The direction $(\Rightarrow)$ is trivial.

For the direction $(\Leftarrow)$ , $AC = PR$ by hypothesis. Construct the altitudes once again.

If $a=p$ and $c= r$ , then by Pythagoras’ theorem, $BD < QS$ implies

$\begin{aligned} AC &= \sqrt{a^2 - d^2} + \sqrt{c^2 - d^2} \\ &> \sqrt{p^2 - s^2} + \sqrt{r^2 - s^2} = PR, \end{aligned}$

so that $AC > PR$ , a contradiction. Therefore, by a symmetric argument, we can conclude that $BD = QS$ .

By the RHS Criterion, $\Delta ABD \equiv \Delta PQS$ , so that

$\angle CAB = \angle DAB = \angle SPQ = \angle RPQ.$

By hypothesis, $AB = PQ$ and $AC = PR$ . Hence, by the SAS Criterion,

$\Delta ABC \equiv \Delta PQR,$

as required.

Example 1. Consider the triangle $\Delta ABC$ below.

Prove that $\theta = 90^\circ$ if and only if $a^2 + b^2 = c^2$ . The direction $(\Leftarrow)$ is called the converse of Pythagoras’ theorem.

Solution. The direction $(\Rightarrow)$ is the already-proven vanilla Pythagoras’ theorem.

For the direction $(\Leftarrow)$ , suppose $a^2 + b^2 = c^2$ . Construct a right-angled triangle $\Delta PQR$ with base $b$ and height $a$ as follows:

By the vanilla Pythagoras’ theorem, $a^2 + b^2 = d^2$ . By hypothesis, $c^2 = d^2$ .

Since $c > 0$ and $d > 0$ , we have $c = d$ .

By the SSS Criterion, $\Delta ABC \equiv \Delta PQR$ . In particular,

$\theta = \angle BCA = QRP = 90^\circ.$

Theorem 4 (ASA Criterion). Let $\Delta ABC, \Delta PQR$ be the following two triangles:

Then $\Delta ABC \equiv \Delta PQR$ if and only if

$\angle BAC = \angle QPR, \quad b=q,\quad \angle BCA = \angle QRP.$

Proof. It suffices to prove $(\Leftarrow)$ . By hypothesis, since angles in a triangle sum to $180^\circ$ ,

$\begin{aligned} \angle ABC &= 180^\circ - \angle BAC - \angle BCA \\ &= 180^\circ - \angle QPR - \angle QRP = \angle PQR. \end{aligned}$

Similar to the proof of Theorem 2, we claim that $AB = PQ$ . Superimpose $\Delta ABC$ onto $\Delta PQR$ such that $A = P$ and $C = R$ .

Since the corresponding angles $\angle ACB = \angle PRQ$ and $AC = PR$ , we have $BC \parallel QR$ . Since $C = R$ , $BC$ passes through $R$ . By Playfair’s axiom, $B$ must lie on $QR$ . Using a symmetric argument, $B$ must lie on $PQ$ .

Therefore, $B$ lies on the intersection between the line $PQ$ and the line $QR$ . Since there exists one and only one intersection point between the two lines, namely $Q$ , we must have $B = Q$ , so that $BC = QR$ .

By hypothesis, $\angle BCA = \angle QRP$ and $AC = PR$ . Hence, by the SAS Criterion, $\Delta ABC \equiv \Delta PQR$ , as required.

Theorem 5 (AAS Criterion). Let $\Delta ABC, \Delta PQR$ be the following two triangles:

Then $\Delta ABC \equiv \Delta PQR$ if and only if

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

Proof. Since angles sum to $180^\circ$ ,

$\begin{aligned} \angle CAB &= 180^\circ - \angle ABC - \angle BCA \\ &= 180^\circ - \angle PQR - \angle QRP = \angle RPQ. \end{aligned}$

Therefore, by the ASA Criterion, $\Delta ABC \equiv \Delta PQR$ , as required.

Definition 2. A triangle is said to be isosceles if it has two sides with equal length.

Example 2. Consider the triangle $\Delta ABC$ below with base angles $\angle BAC$ and $\angle BCA$ .

Prove that $AB = BC$ if and only if $\angle BAC = \angle BCA$ . In this case, we say that the base angles of an isosceles triangle are equal.

Solution. Our proof boils down to the analysis of $\Delta ABC$ with its “rearranged” self $\Delta CBA$ . Clearly $\angle ABC = \angle CBA$ , since they denote the same angle.

We prove in two directions. For the direction $(\Rightarrow)$ , suppose $AB = BC$ . Almost trivially, $AB = CB$ and $BC = BA$ . By the SAS Criterion, $\Delta ABC \equiv \Delta CBA$ . In particular, $\angle BAC = \angle BCA$ .

The direction $(\Leftarrow)$ is proven similarly. Suppose $\angle BAC = \angle BCA$ . Then by relabelling, $\angle ACB = \angle CAB$ . Furthermore, $AC = CA$ , since they denote the same side. By the ASA Criterion, $\Delta ABC \equiv \Delta CBA$ . In particular, $AB = BA = BC$ , as required.

Definition 3. A triangle is said to be equilateral if all three sides have the same length.

Example 3. Determine the size of any interior angle in an equilateral triangle.

Proof. Denote the equilateral triangle by $\Delta ABC$ , and denote $\angle CAB = \theta$ .

Since $CA = CB$ , by Example 2,

$\angle CBA = \angle CAB = \theta.$

Since $AC = AB$ , by Example 2 again,

$\angle ACB = \angle ABC = \angle CBA = \theta.$

Since the angles in a triangle sum to $180^\circ$ ,

$\begin{aligned} \angle CAB + \angle CBA + \angle ACB &= 180^\circ \\ \theta + \theta + \theta &= 180^\circ \\ 3\theta &= 180^\circ \\ \theta &= 60^\circ. \end{aligned}$

Therefore,

$\angle CBA = \angle CAB = \angle ACB = 60^\circ.$

That is, each angle in an equilateral triangle has a size of $60^\circ$ .

If the geometry on a triangle is fascinating, the geometry on a circle is even more astounding! We will visit the geometry of circles in the next post.

—Joel Kindiak, 14 Oct 25, 1920H
January 7, 2026
The Curvy Gradient
Calculus, in the 21st century, continues to be the sorrow of most students required to learn it against their will. It doesn’t need to be this way, though.

Consider the graph of $y = x^2$ below.

Define the point $P_t$ on the graph by $(t, t^2)$ . Using algebra, we can evaluate the gradient of the tangent at $P_t$ to $2t$ .

Now consider the graph of $y =x^3$ .

Define the point $Q_t$ on the graph by $(t, t^3)$ . Using algebra, we can evaluate the gradient of the tangent at $Q_t$ to $3t^2$ .

We are going to generalise this observation.

Definition 1. Let $y = f(x)$ be the graph of a function. The derivative of $f(x)$ at $x = t$ , denoted $f'(t)$ , is defined to be the gradient of the tangent at $(t, f(t))$ .

Define the derivative of $f(x)$ by

$\displaystyle \frac{\mathrm d}{\mathrm dx}(f(x)) := f'(x).$

Since $y = f(x)$ , we can also write

$\displaystyle \frac{\mathrm dy}{\mathrm dx} \equiv \frac{\mathrm d}{\mathrm dx}(y) \equiv \frac{\mathrm d}{\mathrm dx}(f(x)).$

We describe this process as differentiating $f(x)$ with respect to $x$ .

Remark 1. Strictly speaking, the derivative is defined as a limit, and the gradient of the tangent is defined to be the derivative. However, we adopt the convention in Definition 1 for the sake of visual intuition.

Using more mathematical tools to establish a common pattern, we will define the derivative of $x^n$ , where $n$ is any rational number.

Theorem 1. The derivative of $x^n$ with respect to $x$ is given by

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^n) := nx^{n-1}.$

Proof. See this post for the more formal perspective, and see this exercise for the algebraic calculation. We see that

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^2) = 2x,\quad \frac{\mathrm d}{\mathrm dx}(x^3) = 3x^2,$

and these results agree with our investigation in earlier examples.

Example 3. Evaluate the following derivatives:

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^4), \quad \frac{\mathrm d}{\mathrm dx}(x),\quad \frac{\mathrm d}{\mathrm dx}(1), \quad \frac{\mathrm d}{\mathrm dx} \left( \frac 1x \right),\quad \frac{\mathrm d}{\mathrm dx}(\sqrt x).$

Solution. The first expression is immediate using Theorem 1:

$\begin{aligned} \frac{ \mathrm d}{\mathrm dx}(x^4) &= 4x^{4-1} \\ &= 4x^3. \end{aligned}$

The second expression requires the two laws of exponents $x = x^1$ and $x^0 = 1$ :

$\begin{aligned} \frac{ \mathrm d}{\mathrm dx}(x) &= \frac{ \mathrm d}{\mathrm dx}(x^1) \\ &= 1x^{1-1} \\ &= 1x^0 \\ &= 1 \cdot 1 = 1. \end{aligned}$

The third expression requires $1 = x^0$ :

$\displaystyle \begin{aligned} \frac{ \mathrm d}{\mathrm dx}(1) &= \frac{ \mathrm d}{\mathrm dx}(x^0) \\ &= 0x^{0-1} \\ &= 0x^{-1} = 0. \end{aligned}$

The fourth expression requires $1/x = x^{-1}$ and more generally, $x^{-n} = 1/x^n$ :

$\begin{aligned} \frac{ \mathrm d}{\mathrm dx}\left( \frac 1x \right) &= \frac{ \mathrm d}{\mathrm dx}(x^{-1}) \\ &= (-1)x^{-1-1} \\ &= -x^{-2} = -\frac 1{x^2}. \end{aligned}$

The fifth expression requires $\sqrt x = x^{1/2}$ and $x^{-n} = 1/x^n$ :

$\begin{aligned} \frac{ \mathrm d}{\mathrm dx} (\sqrt x) &= \frac{ \mathrm d}{\mathrm dx}(x^{\frac 12}) \\ &= {\textstyle \frac 12} \cdot x^{\frac 12-1} \\ &= {\textstyle \frac 12} \cdot x^{-\frac 12} \\ &=\frac 1{2} \cdot \frac 1{x^{\frac 12}} \\ &= \frac 12 \cdot \frac 1{\sqrt x} = \frac 1{2\sqrt x}. \end{aligned}$

Is it possible to evaluate derivatives of combinations of these functions? Yes!

Theorem 2. Given functions $f(x), g(x)$ and constants $a, b$ ,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(a \cdot f(x) + b \cdot g(x)) = a \cdot \frac{\mathrm d}{\mathrm dx}(f(x)) + b \cdot \frac{\mathrm d}{\mathrm dx}(g(x)).$

Proof. See this post from a calculus perspective and this post from a linear-algebra perspective. This result is known as the linearity of the derivative.

Example 4. Given functions $f(x), g(x), h(x)$ , show that

$\displaystyle \frac{\mathrm d}{\mathrm dx}( f(x) + g(x) + h(x) ) = \frac{\mathrm d}{\mathrm dx}(f(x)) + \frac{\mathrm d}{\mathrm dx}(g(x)) + \frac{\mathrm d}{\mathrm dx}(h(x)).$

Solution. Using the linearity of the derivative,

$\begin{aligned} &\frac{\mathrm d}{\mathrm dx}( f(x) + g(x) + h(x) ) \\ &= \frac{\mathrm d}{\mathrm dx}( (f(x) + g(x)) + h(x) ) \\ &= \frac{\mathrm d}{\mathrm dx}(f(x) + g(x)) + \frac{\mathrm d}{\mathrm dx}(h(x)) \\ &= \left(\frac{\mathrm d}{\mathrm dx}(f(x)) + \frac{\mathrm d}{\mathrm dx}(g(x))\right) + \frac{\mathrm d}{\mathrm dx}(h(x)) \\ &= \frac{\mathrm d}{\mathrm dx}(f(x)) + \frac{\mathrm d}{\mathrm dx}(g(x)) + \frac{\mathrm d}{\mathrm dx}(h(x)).\end{aligned}$

Example 5. Evaluate the following derivatives:

$\displaystyle \frac{\mathrm d}{\mathrm dx}( x^2 + 2x + 1 ),\quad \frac{\mathrm d}{\mathrm dx}( (x+1)^2 ),\quad \frac{\mathrm d}{\mathrm dx}( (x+1)^3 ).$

Solution. For the first expression, we use linearity as per Example 4:

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}( x^2 + 2x + 1 ) &= \frac{\mathrm d}{\mathrm dx}(x^2) + 2 \cdot \frac{\mathrm d}{\mathrm dx}(x) + \frac{\mathrm d}{\mathrm dx}(1). \end{aligned}$

Using the results in Theorem 1 and Example 3, we have

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}( x^2 + 2x + 1 ) &= \frac{\mathrm d}{\mathrm dx}(x^2) + 2 \cdot \frac{\mathrm d}{\mathrm dx}(x) + \frac{\mathrm d}{\mathrm dx}(1) \\ &= 2x + 2 \cdot 1 + 0 \\ &= 2x + 2. \end{aligned}$

For the second expression, we expand $(x+1)^2$ , then use the previous answer:

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}( (x+1)^2 ) = \frac{\mathrm d}{\mathrm dx}( x^2 + 2x + 1 ) = 2x + 2 \end{aligned}$

For the third expression, we first expand $(x+1)^3$ :

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

We then use linearity and established results as per Theorem 1 and Example 3:

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}( (x+1)^3 ) &= \frac{\mathrm d}{\mathrm dx}( x^3 + 3x^2 + 3x + 1 ) \\ &= \frac{\mathrm d}{\mathrm dx}(x^3) + 3 \cdot \frac{\mathrm d}{\mathrm dx}(x^2) + 3 \cdot \frac{\mathrm d}{\mathrm dx}(x) + \frac{\mathrm d}{\mathrm dx}(1) \\ &= 3x^2 + 3 \cdot 2x + 3 \cdot 1 + 0 \\ &= 3x^2 + 6x + 3. \end{aligned}$

To shorten notation, we write $f'(x) = \displaystyle \frac{\mathrm d}{\mathrm dx}(f(x))$ .

Example 6. Given functions $f(x), g(x)$ , show that

$\displaystyle \frac{\mathrm d}{\mathrm dx}( f(x) - g(x) ) = \displaystyle f'(x) - g'(x).$

Solution. Since $f(x) - g(x) = f(x) + (-1) \cdot g(x)$ , we use linearity to derive

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}( f(x) - g(x) ) &= \frac{\mathrm d}{\mathrm dx}( f(x) + (- 1) \cdot g(x) ) \\ &= \frac{\mathrm d}{\mathrm dx}( f(x) ) + (- 1) \cdot \frac{\mathrm d}{\mathrm dx} ( g(x) ) \\ &= f'(x) + (-1) \cdot g'(x) \\ &= f'(x) - g'(x).\end{aligned}$

Example 7. Let $a, b$ be positive constants. In economics, the revenue $R$ earned from selling $x$ units of a good is given by

$R \equiv R(x) = x(a - bx).$

Define the two quantities related to the revenue:
- The average revenue is defined by $\mathrm{AR}(x) := R(x)/x$ .
- The marginal revenue is defined by $\mathrm{MR}(x) := R'(x)$ .
Show that the $x$ -intercept of $y = \mathrm{MR}(x)$ is half of the $x$ -intercept of $y = \mathrm{AR}(x)$ .

Solution. By algebra, $\mathrm{AR}(x) = a-bx$ and

$\mathrm{R}(x) = x(a-bx) = ax - bx^2.$

Using linearity,

$\begin{aligned} \mathrm{MR}(x) &= \frac{\mathrm d}{\mathrm dx}(R(x)) \\ &= \frac{\mathrm d}{\mathrm dx}(ax - bx^2) \\ &= a \cdot \frac{\mathrm d}{\mathrm dx}(x) - b \cdot \frac{\mathrm d}{\mathrm dx}(x^2) \\ &= a \cdot 1 - b \cdot 2x \\ &= a - 2bx. \end{aligned}$

Let $x_1$ denote the $x$ -intercept of $y = \mathrm{AR}(x)$ and $x_2$ denote the $x$ -intercept of $y = \mathrm{MR}(x)$ .

We first solve $\mathrm{AR}(x_1) = 0$ :

$\begin{aligned} \mathrm{AR}(x_1) &= 0 \\ a - bx_1 &= 0 \\ x_1 &= \frac ab. \end{aligned}$

We next solve $\mathrm{MR}(x_2) = 0$ :

$\begin{aligned} \mathrm{MR}(x_2) &= 0 \\ a - 2bx_2 &= 0 \\ x_2 &= \frac a{2b}. \end{aligned}$

Hence, $2x_2 = 1/b = x_1$ , so that $x_2 = x_1/2$ .

What other combinations can we differentiate? Given functions $f(x), g(x)$ , we can differentiate the following:

$\displaystyle f(x) \cdot g(x),\quad \frac{ f(x) }{ g(x) },\quad f(g(x)).$

Tragically, they don’t follow the neat rules that we think they do.

Example 8. Which of the following equations are true?

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}( f(x) \cdot g(x) ) &= f'(x) \cdot g'(x), \\ \frac{\mathrm d}{\mathrm dx} \left( \frac{ f(x) }{ g(x) } \right) &= \frac{ f'(x) }{ g'(x) }, \\ \frac{\mathrm d}{\mathrm dx} \left( f(g(x)) \right) &= f'(g(x)). \end{aligned}$

Justify your answer.

Solution. Sadly, none of them are true.

We will use the counter-example $f(x) = x^3$ and $g(x) = x^2$ . Using Theorem 1, $f'(x) = 3x^2$ and $g'(x) = 2x$ .

For the first equation,

$\displaystyle \frac{\mathrm d}{\mathrm dx}( f(x) \cdot g(x) ) = \frac{\mathrm d}{\mathrm dx}(x^3 \cdot x^2) = \frac{\mathrm d}{\mathrm dx}(x^5) = 5x^4,$

however,

$\displaystyle f'(x) \cdot g'(x) = 3x^2 \cdot 2x = 6x^3.$

Therefore,

$\displaystyle \frac{\mathrm d}{\mathrm dx}( f(x) \cdot g(x) ) = 6x^3 \neq 5x^4 = f'(x) \cdot g'(x).$

For the second result, we leave it as an exercise to check that

$\displaystyle \frac{\mathrm d}{\mathrm dx} \left( \frac{ f(x) }{ g(x) } \right) = 1 \neq \frac{3x}{2} = \frac{ f'(x) }{ g'(x) }.$

For the third result, we leave it as an exercise to check that

$\displaystyle \frac{\mathrm d}{\mathrm dx} ( f( g(x) ) ) = 6x^5 \neq 3x^4 = f'(g(x)).$

To deal with these latter three results, we will need to look at the three musketeers of differentiation techniques: the chain rule, the product rule, and the quotient rule. For more complete proofs, see this post on the chain rule and this post for the product and quotient rules.

We will explore this differentiation trio next time.

—Joel Kindiak, 7 Jan 26, 1450H
January 7, 2026
Algebraic Calculus

Consider the graph of $y = x^2$ below.

Define the point $P_t$ on the graph by $(t, t^2)$ .

Use this graph to answer Problems 1–4.

Problem 1. Evaluate the gradient of $P_1 P_2$ .

(Click for Solution)

Solution. Since $P_1(1, 1)$ and $P_2(2, 4)$ , the gradient is given by

$\displaystyle m_{P_1P_2} = \frac{4-1}{2-1} = \frac 31 = 3.$

Problem 2. Given $h$ , evaluate the gradient of $P_1 P_{1+h}$ .

Verify your answer in Problem 1.

(Click for Solution)

Solution. Since $P_{1+h}(1+h, (1+h)^2)$ , the gradient is given by

$\begin{aligned} m_{P_1P_{1+h}} &= \frac{(1+h)^2-1}{(1+h)-1} \\ &= \frac{(1+2h+h^2)-1}{(1+h)-1} \\ &= \frac{2h+h^2}{h} \\ &= \frac{(2+h)\cdot h}{h} \\ &= 2+h. \end{aligned}$

This answer agrees with Problem 1, since setting $h = 1$ gives

$\begin{aligned} m_{P_1P_{1+h}} &= 2 + h \\ &= 2 + 1 \\ &= 3 = m_{P_1P_2}. \end{aligned}$

Problem 3. Given $t$ and $h$ , evaluate the gradient of $P_t P_{t+h}$ .

Verify your answer in Problem 2.

(Click for Solution)

Solution. Since $P_t(t,t^2)$ and $P_{t+h}(t+h, (t+h)^2)$ , so that the gradient is given by

$\begin{aligned} m_{P_tP_{t+h}} &= \frac{(t+h)^2-t^2}{(t+h)-t} \\ &= \frac{(t^2+2th+h^2)-t^2}{(t+h)-t} \\ &= \frac{2th+h^2}{h} \\ &= \frac{(2t+h)\cdot h}{h} \\ &= 2t+h. \end{aligned}$

This answer agrees with Problem 2 since setting $t = 1$ gives

$\begin{aligned} m_{P_tP_{t+h}} &= 2t + h \\ &= 2 \cdot 1 + h \\ &= 2 + h \\ &= m_{P_1P_{1+h}}. \end{aligned}$

Problem 4. Let $L_t$ denote the line passing through $P_t$ . Show that $L_t$ is tangent to $y = x^2$ if and only if its gradient is $2t$ .

(Click for Solution)

Solution. Denote the gradient of $L_t$ by $m$ . Then

$L_t : y = m(x - t) + t^2.$

When $L_t$ intersects the curve $y = x^2$ ,

$x^2 = m(x-t) + t^2$

so that solving yields

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

Then $x = t$ or $x = m - t$ . For $L_t$ to be a tangent to the curve $y = x^2$ at $P_t$ , we must have both roots equal $t$ , so that

$t = m-t \quad \iff \quad m = 2t,$

as required.

Therefore, by setting $h = 0$ in Problem 3, we obtain the answer in Problem 4: the expression $2t$ . Intuitively, this expression describes the gradient of the tangent at $(t, t^2)$ .

Problem 5. Now consider the graph of $y =x^3$ .

Define the point $Q_t$ on the graph by $(t, t^3)$ .

For any $t$ and $h$ , evaluate the gradient of $Q_t Q_{t+h}$ in terms of $h$ .

(Click for Solution)

Solution. Using the same strategy as per Example 1, since $Q_{t+h}(t+h, (t+h)^3)$ , the gradient is given by

$\begin{aligned} m_{P_tP_{t+h}} &= \frac{(t+h)^3-t^3}{(t+h)-t} = \frac{(t+h)^3-t^3}{ h }. \end{aligned}$

We first expand $(t+h)^3$ :

$\begin{aligned} (t+h)^3 &= (t+h)^2 \cdot (t+h) \\ &= (t^2 + 2th + h^2)(t + h) \\ &= t^3 + (2t^2 + t^2)h + (t+2t)h^2 + h^3 \\ &= t^3 + 3t^2h + 3th^2 + h^3.\end{aligned}$

Therefore,

$\begin{aligned} m_{P_tP_{t+h}} & = \frac{(t+h)^3-t^3}{ h } \\ & = \frac{3t^2h + 3th^2 + h^3}{ h } \\ & = \frac{(3t^2 + 3th + h^2)h}{ h } \\ &= 3t^2 + 3th + h^2. \end{aligned}$

If we set $h = 0$ in the final result, we obtain the expression $3t^2$ . Intuitively, this expression describes the gradient of the tangent at $(t, t^3)$ .

Problem 6. Given that $P_t$ lies on $y = x^n$ , evaluate the gradient of the tangent to $P_t$ . You may freely use the factorisation

$x^n - t^n = (x-t)(x^{n-1} + x^{n-2}t + \cdots + xt^{n-2} + t^{n-1}).$

without proof.

(Click for Solution)

Solution. Denote $f(x) := x^n$ and

$\varphi (x) := x^{n-1} + x^{n-2}t + \cdots + xt^{n-2} + t^{n-1}$

so that $f(x)- f(t) = (x-t) \cdot \varphi (x)$ for brevity (and possible generality).

We follow the approach in Problem 4. Denote the gradient of the tangent $L_t$ to $P_t$ by $m$ . Then

$L_t : y = m(x-t) + f(x).$

When $L_t$ intersects the curve $y = f(x)$ ,

$f(x) = m(x-t) + f(t).$

Solving yields

$\begin{aligned} f(x) - f(t) - m(x-t) &= 0 \\ (x-t)g(x) - m(x-t) &= 0 \\ (x-t)(\varphi (t) - m) &= 0. \end{aligned}$

Hence, $x = t$ or $\varphi (x) = m$ . For $L_t$ to be tangent to $P_t$ , we must have $\varphi (t) = m$ and vice versa. In particular,

$m = \varphi (t) = n \cdot t^{n-1}.$

Remark 1. Problem 6 generalises to other kinds of functions using Carathéodory’s theorem.

—Joel Kindiak, 7 Jan 26, 1542H

January 7, 2026
Pythagoras’ Trophy

Pythagoras’ theorem is the most important idea in geometry, all of mathematics even. But before discussing it, let’s ask a simple question.

Given a unit length $1$ , and a positive number $R$ , how can we construct $\sqrt R$ ?

Recall that $\sqrt R = s$ means that $s^2 = R$ . We can achieve this construction goal using Pythagoras’ theorem.

(A proper construction, as with many other geometric ideas, relies on real analysis.)

Theorem 1 (Pythagoras’ Theorem). Given a right-angled triangle below,

$a^2 + b^2 = c^2.$

We call the longest side with length $c$ the hypotenuse of the triangle.

If $a,b,c$ are positive integers, we call $(a,b,c)$ a Pythagorean triple. The most famous such example would be the $3$ – $4$ – $5$ triple:

$3^2 + 4^2 = 25 = 5^2.$

Proof. Draw three extra identical triangles, so that we get a large square with side length $a + b$ .

We claim that the slanted four-sided shape (i.e. a quadrilateral) in the middle is indeed a square.

Since the right-angled triangles are identical, we take two adjacent triangles and label their angles as follows.

We want to prove that $\gamma = 90^\circ$ . Since the angles in the centre are non-overlapping adjacent angles on a straight line,

$\alpha + \gamma + \beta = 180^\circ.$

On the other hand, since the angles in a triangle sum to $180^\circ$ ,

$\alpha + \beta + 90^\circ = 180^\circ.$

Therefore, using usual calculations,

$\begin{aligned} \alpha + \gamma + \beta &= \alpha + \beta + 90^\circ\\ \gamma &= 90^\circ. \end{aligned}$

Since the hypotenuses are all equal, we know that the quadrilateral has four equal sides. Hence, we have a square in the middle whose area is $c^2$ , so that

$\begin{aligned} (a+b)^2 &= 4 \cdot \frac 12 \cdot a \cdot b + c^2 \\ a^2 + 2ab + b^2 &= 2ab + c^2 \\ a^2 + b^2 &= c^2. \end{aligned}$

There are many applications of Pythagoras theorem that, in my humble opinion, are not worth discussing outside a tuition class. Yet, Pythagoras’ theorem can resolve for us a simple yet non-trivial question: what is a circle?

You might say, “a circle is a non-straight bendy or curvy line that closes in on itself”. This formulation is valid when formalised in the language of algebraic topology and homology theory.

Let’s keep things simple for now.

Definition 1. A circle with centre $P$ and radius $r$ is the unique set of points whose distance from $P$ is $r$ . In particular, we define the unit circle to be a circle with centre $P$ and radius $r$ .

There is one problem with this definition: how do we calculate the distance between two points without requiring any additional measurement?

Lemma 1. The distance $d_{PQ}$ between two points $P(x_1,y_1)$ and $Q(x_2,y_2)$ is given by

$\displaystyle d_{PQ} := \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$

Proof Sketch. Suppose the special case $x_1 < x_2$ and $y_1 < y_2$ .

Construct the right-angled triangle below.

Then the hypotenuse of this right-angled triangle has length $d_{PQ}$ . Applying Pythagoras’ theorem,

$d_{PQ}^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2.$

Taking square roots gives the desired result.

The other cases involving $x_1 \geq x_2$ or $y_1 \geq y_2$ are left as an exerice to the reader.

Call any collection of points a graph.

Theorem 2. The graph $C$ is a circle with centre $(h,k)$ and radius $r$ if and only if it is defined by the equation

$(x-h)^2 + (y-k)^2 = r^2.$

In particular, the unit circle has equation $x^2 + y^2 = 1$ .

Proof. By definition, a point $(x,y)$ belongs to the circle if and only if its distance to $(h,k)$ is $r$ . By Lemma 1, this condition holds if and only if

$\sqrt{(x-h)^2 + (y-k)^2} = r.$

Squaring on both sides, since $r > 0$ , we obtain the equation

$(x-h)^2 + (y-k)^2 = r^2,$

as required.

Once we can properly discuss circles, all sorts of interesting mathematics arise. But we will revisit circles later on.

For now, let’s answer the question we opened with: constructing $\sqrt R$ geometrically.

Theorem 3. Fix $R > 0$ . Define $r := (R+1)/2$ . Define the line $L$ by $x = r-1$ and the circle $C$ to be a circle with centre $O(0,0)$ and radius $r$ .

Then $AB = \sqrt {R}$ .

Proof. By Theorem 2, $C$ has equation

$x^2 + y^2 = r^2.$

To obtain the coordinates of $B$ , we need to solve the equations

$x = r-1\quad \text{and}\quad x^2 + y^2 = r^2,$

and require $y > 0$ . We will use the substitution method as follows—substitute the equation $x = r-1$ into the circle equation:

$\begin{aligned} (r-1)^2 + y^2 &= r^2 \\ (r^2 - 2r + 1) + y^2 &= r^2 \\ - 2r + 1 + y^2 &= 0 \\ -(R+1) + 1 + y^2 &= 0 \\ -R + y^2 &= 0 \\ y^2 &= R. \end{aligned}$

Since $y > 0$ , by definition, $y = \sqrt{R}$ . Since $A(r-1, 0)$ and $B(r-1, \sqrt R)$ , by Lemma 1,

$\begin{aligned} AB &= \sqrt{((r-1) - (r-1))^2 + (\sqrt R - 0)^2} = \sqrt{0 + R} = \sqrt{R}. \end{aligned}$

There is a lot more to circles than this one example, but for now, we return to triangles in the next post. In particular, the isosceles triangle will be a central tool in our geometric expedition.

—Joel Kindiak, 10 Oct 25, 2026H

December 31, 2025
Baby Proving
Here’s our question today: what is the sum of angles in a triangle?

Previously, we have seen how linear equations $ax + by = c$ can represent lines. If $b \neq 0$ , then

$\displaystyle y = -\frac ab \cdot x + \frac cb,$

so that the line has gradient $-a/b$ and $y$ -intercept $(0, c/b)$ .
- If $b = 0$ , then $ax = c$ . In this case, if $a \neq 0$ , then $x = c/a$ , and the line is vertical with $x$ -intercept $(c/a, 0)$ .
We can make a few more basic observations about our lines below, properly constructed here, which uses many tools from linear algebra.

Firstly, we use the word “angle” to describe the amount of separation between two lines. The simplest case occurs when the two lines are at right angles (i.e. ortho-gonal) to each other. We use the symbol $\square$ to denote angles that are right, and in this case write $\ell_1 \perp \ell_2$ :

By convention, we declare one right angle to equal $90^{\circ}$ (i.e. $90$ degrees), so that the sum of adjacent non-overlapping angles on a straight line add up to $180^{\circ}$ .

Definition 1. An angle $\theta$ is said to be
- right if $\theta = 90^\circ$ ,
- acute if $0^\circ < \theta < 90^\circ$ (i.e. smaller than a right angle),
- obtuse if $90^\circ < \theta < 180^\circ$ (i.e. larger than a right angle).
For example, the angle $\alpha$ in the diagram below is acute, while the angle $\beta$ is obtuse.

Eventually, we will consider angles that are reflex (i.e. if $180^\circ < \theta < 360^\circ$ ).

Remark 1. Usually, the Greek letter ‘theta’ $\theta$ denotes an angle. When there are multiple angles, we use the Greek letters ‘alpha’ $\alpha$ , ‘beta’ $\beta$ , ‘gamma’ $\gamma$ , and so on.

Remark 2. We can formalise the idea of angles using calculus (i.e. the study of rates of change), discussed elsewhere. The choice of the number $180$ has a rich human history, and its use is strengthened by its factors $1,2,3,4,5,6$ .

Consider two lines $\ell_1,\ell_2$ :
- We call two non-identical lines $\ell_1,\ell_2$ parallel if they don’t intersect and write $\ell_1 \parallel \ell_2$ .
- In this case, we will use the decorations $>$ , $\gg$ , etc to denote parallel lines.
- We call the pair of angles $\alpha, \beta$ a pair of interior angles if $\alpha + \beta \leq 180^{\circ}$ .
We make the key observation that assuming a flat screen, $\ell_1 \parallel \ell_2$ if and only if any pair of interior angles sum to $180^{\circ}$ . This result is known as the parallel postulate in Euclidean geometry.

We can also use our construction to conclude Playfair’s axiom:

Lemma 1 (Playfair’s Axiom). Let $\ell$ be a line and $P$ be a point that does not lie on $\ell$ . Then there exists a unique line $\ell'$ that passes through $P$ and is parallel to $\ell$ .

Proof. See this post.

Going forward, we will take advantage of these basic observations to derive several foundational tools for geometric inference, and resolve the sum-of-angles question that we started with.

Theorem 1. Given two non-parallel lines, the vertically opposite angles $\alpha, \beta$ are equal to each other.

Proof. Construct the angle $\gamma$ below.

Since adjacent non-overlapping angles on a straight line sum to $180^\circ$ ,

$\displaystyle \alpha + \gamma = 180^\circ \quad \text{and} \quad \beta + \gamma = 180^\circ.$

Therefore, since we can cancel the $\gamma$ term on both sides (this is called the cancellation law),

$\begin{aligned}\alpha + \gamma &= \beta + \gamma \\ \alpha &= \beta. \end{aligned}$

Theorem 2. The two lines $\ell_1,\ell_2$ below are parallel if and only if the alternate angles $\alpha, \beta$ are equal to each other.

Proof. Construct the angle $\gamma$ below.

Since adjacent non-overlapping angles on a straight line sum to $180^\circ$ ,

$\displaystyle \alpha + \gamma = 180^\circ.$

By the parallel postulate,

$\ell_1 \parallel \ell_2 \quad \iff \quad \beta + \gamma = 180^\circ.$

Using the cancellation law in Theorem 1,

$\beta + \gamma = 180^\circ \quad \iff \quad \alpha + \gamma = \beta + \gamma \quad \iff \quad \alpha = \beta.$

Theorem 3. The two lines $\ell_1,\ell_2$ below are parallel if and only if the corresponding angles $\alpha, \beta$ are equal to each other.

Proof. Construct the angle $\gamma$ below.

Then $\beta, \gamma$ are vertically opposite. By Theorem 1,

$\displaystyle \beta = \gamma.$

Furthermore, $\alpha, \gamma$ are alternate angles. By Theorem 2,

$\ell_1 \parallel \ell_2 \quad \iff \quad \alpha = \gamma \quad\iff \quad \alpha = \beta.$

Remark 2. In most school problems, you would be hand-given the parallel lines $\ell_1, \ell_2$ , and then, if needed, prove the angle equality result (e.g. $\alpha = \beta$ ). You can use the following presentation for school assessments (e.g. homework, tests, examinations, etc):

$\begin{aligned} \alpha &= \gamma \quad (\text{alt}\ \angle,\ \ell_1 \parallel \ell_2) \\ &= \beta \quad (\text{vert opp}\ \angle). \end{aligned}$

At least, this working got me full credit when I was a secondary school student.

Theorem 4. The angles in a triangle sum to $180^\circ$ .

Proof. Consider the triangle below without loss of generality.

We need to prove that $\alpha + \beta + \gamma = 180^\circ$ .

Use Playfair’s axiom to construct a line $\ell$ parallel to the base $b$ of the triangle that passes through the opposite point (i.e. vertex) of the triangle:

Construct the angles $\alpha'$ and $\beta'$ .

Since $\alpha', \gamma, \beta'$ are non-overlapping adjacent angles on a straight line,

$\alpha' + \gamma + \beta' = 180^\circ.$

We also observe that $(\alpha, \alpha')$ and $(\beta, \beta')$ are pairs of alternate angles. Since $b \parallel \ell$ , by Theorem 2,

$\alpha = \alpha' \quad \text{and}\quad \beta = \beta'.$

Therefore, using the usual rules of addition,

$\begin{aligned} \alpha + \beta + \gamma &= \alpha' + \beta' + \gamma \\ &= \alpha' + \gamma + \beta' \\ &= 180^\circ. \end{aligned}$

While there are many results that can and should be proven using these basic results, we will delay them to connect their usefulness with other geometric objects. For example:
- The fact that a triangle is isosceles if and only if its base angles are equal is a consequence of our discussion on congruent triangles.
- Two angles in the same segment in a circle must equal each other, and we will explore this idea when considering circles in more depth.
Rather than explore these results all in one post, we will smatter them throughout our discussion in connection with other parts of geometry.

For now, we will use Theorem 4 to prove one of the most important results in secondary school (and hence higher-level) mathematics—Pythagoras’ theorem.

—Joel Kindiak, 6 Oct 25, 2100H
December 24, 2025
Baby Linear Algebra
Previously, we tried to earn some money selling fidget spinners. To do that, we will sell $x$ fidget spinners, where $x$ is given by the $x$ -coordinate of the intersection point of the two graphs below.

Remark 1. In economics terms,
- the graph of $y = 2x + 1$ is called the supply curve,
- the graph of $y = -\frac 12 x + 11$ is called the demand curve.
In more mathy terms, we need to find a pair $(x, y)$ of numbers that satisfy both equations:

$\left\{ \begin{aligned} y &= 2x+1, \\ y &= \textstyle -\frac 12 x + 11. \end{aligned} \right.$

We call this process solving a system of simultaneous linear equations.

Example 1. Solve the system of simultaneous linear equations:

$\left\{ \begin{aligned} y &= 2x+1, \\ y &= \textstyle -\frac 12 x + 11. \end{aligned} \right.$

Solution. If $(x,y)$ satisfies both equations, then the $y$ terms on the left-hand side equal. Subtracting the first equation by the second:

$\begin{aligned} y - y &= (2x+ 1) - \left(\textstyle -\frac 12 x + 11 \right) \\ 0 &= 2x + 1 + \textstyle \frac 12 x - 11 \\ &= \left( 2x + \textstyle \frac 12 x \right) + 1 - 11 \\ &= \textstyle \frac 52 x - 10. \end{aligned}$

Therefore, we just need to solve for $x$ :

$\begin{aligned} \textstyle \frac 52 x - 10 &= 0 \\ \textstyle \frac 52 x &= 10 \\ x &= \textstyle \frac 25 \cdot 10 \\ &= \textstyle \frac 25 \cdot 5 \cdot 2 \\ &= 2 \cdot 2 = 4. \end{aligned}$

Hence, $x = 4$ . What is its matching $y$ -value?

Since this $x$ -value must satisfy both equations, we can back-substitute $x =4$ into either equation:
- Supply curve: $y = 2 \cdot 4 + 1 = 9$ .
- Demand curve: $y = -\frac 12 \cdot 4 + 11 = 9$ .
Therefore, our solution is $x = 4$ and $y = 9$ .
- In ordered-pair notation, the solution is $(x, y) = (4, 9)$ .
Whether we calculated the intersection point $(4, 9)$ , or read it off from the graph, we obtained the same answer. This observation is massive—it connects pure symbols with familiar pictures!

This connection is the simplest building block behind linear algebra—which uses calculations to represent “straight-ish” objects like lines and planes (see Theorem 1 below). Its humble beginnings arise from what we just did—solving linear equations simultaneously.
- In higher-level mathematics, the elimination method we used is given the fancier title Gaussian elimination.
- There is an alternate (and more general) strategy to solve simultaneous equations, called the substitution method. We will explore this method when investigating quadratic curves.
Example 2. Let $L$ denote the line given by the equation $x + 2y = 3$ . Determine the intersection point between $L$ and the $x$ -axis, called the $x$ -intercept of $L$ .

Solution. Let $M$ denote the $x$ -axis. Since $M$ is horizontal and passes through the point $(0, 0)$ , it has equation $y = 0$ . Therefore we need to solve the system of simultaneous linear equations

$\left\{ \begin{aligned} x+2y &= 3, \\ y &=0. \end{aligned} \right.$

Multiplying the second equation by $2$ :

$\left\{ \begin{aligned} x+2y &= 3, \\ 2y &=0. \end{aligned} \right.$

Subtracting the first equation by the second:

$\begin{aligned} (x+2y) - ( 2y) &= 3 - 0 \\ x &=3. \end{aligned}$

Back-substituting $x = 3$ into the second equation, since there are no $x$ terms, we just have $y = 0$ . Therefore, $L$ has an $x$ -intercept $(3, 0)$ .

Example 3. Using the definition of $L$ in Example 2, determine the $y$ -intercept of $L$ .

Solution. Let $M$ denote the $y$ -axis. Since $M$ is vertical and passes through the point $(0, 0)$ , it has equation $x = 0$ . Therefore we need to solve the system of simultaneous linear equations

$\left\{ \begin{aligned} x+2y &= 3, \\ x &=0. \end{aligned} \right.$

Subtracting the second equation from the first:

$\begin{aligned} \textstyle (x+2y) - x &= 3 - 0 \\ (x-x) + 2y &= 3 \\ 2y &= 3 \\ y &= \textstyle \frac 32. \end{aligned}$

Back-substituting $y = \frac 32$ into the second equation, since there are no $y$ terms, we just have $x = 0$ . Therefore, $L$ has a $y$ -intercept $(0, 3/2)$ .

Our examples aren’t special; here’s the more general theory:

Theorem 1. If $ad - bc \neq 0$ , then there exists a unique pair $(x,y)$ of real numbers that solves the system of simultaneous linear equations

$\left\{ \begin{aligned} ax+by &= p, \\ cx +dy &= q. \end{aligned} \right.$

In fact, $\displaystyle x = \frac{1}{ad-bc} (dp - bq)$ , $\displaystyle y = \frac{1}{ad-bc} (-cp + aq)$ .

Proof Sketch. Multiply the first equation by $d$ and the second equation by $b$ :

$\left\{ \begin{aligned} adx+bdy &= dp, \\ bcx +bdy &= bq. \end{aligned} \right.$

Subtract the first equation by the second:

$\begin{aligned} (ad- bc)x &=dp - bq. \end{aligned}$

Assuming $ad - bc \neq 0$ , we can obtain a unique value for $x$ , then back-substitute to obtain a unique value for $y$ .

Corollary 1. If $a$ and $b$ are both nonzero, then the line $ax+by = p$ has $x$ -intercept $(p/a,0)$ and $y$ -intercept $(0, p/b)$ .

Proof. Use Theorem 1:
- Set $(c,d,q) = (0,1,0)$ to obtain the $x$ -intercept.
- Set $(c,d,q) = (1,0,0)$ to obtain the $y$ -intercept.
Theorem 1 gives us a taste of the theory of linear algebra, and our discussion on supply and demand curves illustrate one (over-simplified) real-world use case of said theory. We can summarise our findings with a simple cliche:

X marks the spot!

Next time, we will discuss some simple ideas about lines and angles, which can be described using lines, and be precisely formulated using—you guessed it—linear algebra. 90% of my blog was dedicated to ensure that what we do at the O-Level really is legitimate, and thankfully, I can confidently say that it is!

—Joel Kindiak, 4 Oct 25, 1516H
December 17, 2025
Basic Business Math
Let’s try to make some money.

You want to earn money by selling fidget spinners, but two questions arise:
1. How much revenue can you earn by selling the fidget spinners?
2. How much cost would you incur by obtaining the fidget spinners in the first place?
Your profit is then defined by

$(\text{profit}) = (\text{revenue}) - (\text{cost}).$

Let’s first investigate your potential costs.

Example 1. Suppose you buy fidget spinners at a unit cost of $\$2$ per fidget spinner. Make the following definitions:
- $x$ represents the number of fidget spinners you obtain.
- $\$ y$ represents the total cost of buying $x$ fidget spinners.
What would be the relationship between $x$ and $y$ ?

Solution. Using basic counting,

$(\text{total cost}) = (\text{unit cost}) \times (\text{number of fidget spinners}).$

Hence, $y = 2 \times x$ . For clarity, we suppress the $\times$ symbol and write $y = 2x$ .

Example 2. In Example 1, if you can spend a maximum of $\$30$ , what is the maximum number of fidget spinners you can buy?

Solution. Since the maximum cost is capped at $\$30$ , the maximum number of fidget spinners is captured by setting $y = 30$ :

$\displaystyle 30 = 2x.$

We need to determine the number that $x$ represents (i.e. the value of $x$ ). Recalling that $\frac 12 \cdot 2 = 1$ and $1 \cdot x = x$ ,

$\begin{aligned} \frac 12 \cdot 30 &= \frac 12 \cdot 2x \\ \frac 12 \cdot 2 \cdot 15 &= \frac 12 \cdot 2 \cdot x \\ 1 \cdot 15 &= 1 \cdot x \\ 15 &= x. \end{aligned}$

Therefore, $x = 15$ is the maximum possible number of fidget spinners.

Remark 1. Suppose you had $\$31$ instead. Then the equation

$2x = 31$

produces the value $x = 15.5$ .
- We may be tempted to reject the answer $x = 15.5$ , since $0.5$ of a fidget spinner doesn’t arise in real life.
- However, if $x$ refers to the number of fidget spinners measured in groups of $10$ , then the answer $x = 15.5$ corresponds to $15.5 \times 10 = 155$ fidget spinners—a perfectly reasonable solution!
Therefore, in general, we can accept non-whole number values of $x$ .

Example 3. If you need to pay a fixed delivery fee of $\$1$ for your fidget spinners, what would your total cost, in terms of $x$ , now be?

Solution. By including the delivery fee,

$\displaystyle y = 2x + 1.$

For example, if $x = 8$ , then $y = 2 \cdot 8 + 1 = 17$ . Here, we write

$a \cdot b + c \equiv (a \cdot b) + c$

using the usual order of operations.

We can picture this relationship using a graph.
- The horizontal right arrow, called the $x$ -axis, represents the different possible $x$ -values that we can substitute into the expression.
- The vertical up arrow, called the $y$ -axis, represents the different possible $y$ -values that we can obtain.
If we allowed $x$ to take non-integer values, we will recover a richer picture. For example, if $x = \frac 12$ , then $y = 2 \cdot \frac 12 + 1 = 2$ . Repeating this process, we get a striking picture:

We recover a straight line!

Example 4. Calculate the change in cost incurred by obtaining one additional fidget spinner. This change is called the gradient of the graph.

Solution. Let $x_0$ be any fixed amount of fidget spinner bought. The cost of buying this number of fidget spinners is

$y_0 = 2x_0 + 1.$

By incrementing the number of fidget spinners, we want to buy $x_1 := x_0 + 1$ fidget spinners and incur a new cost $y_1$ given by

$\begin{aligned} y_1 &= 2x_1 + 1 \\ &= 2(x_0 + 1) + 1 \\ &= 2x_0 + 3. \end{aligned}$

Hence, the required change in cost is

$\begin{aligned} y_1 - y_0 &= (2x_0 + 3) - (2x_0 + 1) \\ &= 2x_0 + 3 - 2x_0 - 1 \\ &= (2x_0 - 2x_0) + (3 - 1) \\ &= 0 + 2 = 2. \end{aligned}$

Example 5. Using $y = 2x + 1$ , calculate the fixed cost that you would incur.

Solution. Setting $x = 0$ , the fixed cost is given by

$y = 2 \cdot 0 + 1 = 1.$

Definition 1. A graph is called a (non-vertical) straight line with:
- gradient $m$ ,
- $y$ -intercept $c$ ,
if it is continuously drawn using the equation

$y = mx + c.$

The value $m$ describes the steepness of the line:
- if $m > 0$ , then the line is upward-sloping,
- if $m = 0$ , then the line is horizontal,
- if $m < 0$ , then the line is downward-sloping.
Furthermore,
- the more positive the gradient, the steeper the increase,
- the more negative the gradient, the steeper the decrease.
The value $c$ fixes the position of the line:
- if $c$ is positive, then the $y$ -intercept of the line is positive.
- if $c$ is negative, then the $y$ -intercept of the line is negative.
- if $c$ is zero, then the line passes .
If $c$ is positive (resp. zero or negative), then the $y$ -intercept of the line is positive (resp. zero or negative).

In short,
- $m$ describes the steepness of the line,
- $c$ fixes the position of the line.
Definition 2. Define a point as a position in space described by a pair of real numbers, denoted $(x_0, y_0)$ , called the coordinates of the point.

Lemma 1. If two points $(x_1,y_1)$ and $(x_2,y_2)$ with $x_1 \neq x_2$ lie on a non-vertical straight line with gradient $m$ , then

$\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}.$

Proof Sketch. Let the non-vertical straight line have equation $y = mx + c$ . Substituting the two pairs of values gives us the equations

$y_1 = mx_1 + c,\quad y_2 = mx_2 + c.$

We leave it as an exercise to check that

$\displaystyle \frac{y_2 - y_1}{x_2 - x_1} = m.$

Lemma 2. The equation of a non-vertical straight line passing through the point $(x_0, y_0)$ with gradient $m$ is given by

$y - y_0 = m(x- x_0).$

Proof Sketch. Using Lemma 1, any point $(x, y)$ that lies on the straight line must satisfy the equation

$\displaystyle \frac{y - y_0}{x - x_0} = m.$

Hence, $y-y_1 = m(x-x_0)$ .

Example 6. You discover that:
- $10$ people are willing to pay $\$6$ per fidget spinner,
- $16$ people are willing to pay $\$3$ per fidget spinner.
Let $\$ y$ represent the price that $x$ people are willing to buy (of course, one fidget spinner per person).

Stating your assumptions, determine a reasonable relationship between $x$ and $y$ .

Solution. We can picture the two situations in the graph below.

By assuming a straight-line (i.e. linear) graph, we will first apply Lemma 1 using the points $(10, 6)$ and $(16, 3)$ to calculate its gradient:

$\displaystyle m = \frac{3 - 6}{16 - 10} = \frac{-(6-3)}{6} = \frac{-3}{6} = -\frac 12.$

Since the straight line passes through $(10, 6)$ , by Lemma 2, it has the equation

$\begin{aligned} y - 6 &= -\frac 12 \cdot (x-10) \\ &= -\frac 12 \cdot x-\frac 12 \cdot (-10) \\ &= -\frac 12 \cdot x - (-5) \\ &= -\frac 12 \cdot x + 5. \end{aligned}$

Hence, $y = -\frac 12 x + 5 + 6 \Rightarrow y = -\frac 12 x + 11$ .

Definition 3. A graph is a vertical line with $x$ -intercept $c$ if it is continuously drawn using the equation $x = c$ .

Example 7. The line $y = 0$ corresponds to the $x$ -axis and the line $x = 0$ corresponds to the $y$ -axis.

Definition 4. A graph is called a straight line if it can be continuously drawn using the linear equation $ax + by = c$ , where $a, b$ are not both zero.

Theorem 1. A graph is a straight line if and only if it is either a vertical line or a non-vertical straight line.

Proof Sketch. We notice either $b = 0$ or $b \neq 0$ :
- Using Definition 3, $b = 0$ holds if and only if the graph is a vertical line.
- Using Definition 1, $b \neq 0$ holds if and only if the graph is a non-vertical straight line.
The complete proof using the definitions is left as a good exercise in algebraic manipulation.

Remark 2. This blog post is inspired by ideas in business and economics:
- Example 3 pictures the supply curve: how much you are willing to pay in order to sell $x$ fidget spinners.
- Example 6 pictures the demand curve: how much people are willing to pay in order to buy $x$ fidget spinners.
Let’s now draw Example 3 and Example 6 on the same graph.

What is the maximum number of fidget spinners that would be worth selling?

Obviously, when the two graphs intersect!

By using our eyeballs (i.e. inspection), the point of intersection has coordinates $(4, 9)$ . However, we will explore this idea more systematically next time without inspection by solving simultaneous equations.

—Joel Kindiak, 27 Sept 25, 1450H
December 10, 2025
The Number Line
The number line helps us picture positive numbers, negative numbers, and even any point in between: we call these points real numbers, and represent them using the symbol $\mathbb R$ :

All familiar numbers belong on this line:
- Natural numbers: $1, 2, 3, \dots$
- Integers: $0, -1, -2, \dots$
- Fractions: $\frac 12, \frac 13, \frac 23, \dots$
- Negative fractions: $-\frac 12, -\frac 13, -\frac 23, \dots$
A number is called a rational number if it is either $0$ or a fraction or a negative fraction. Other numbers like

$\displaystyle \pi \approx 3 + \frac 1{10} + \frac 4{100} + \frac 2{1000} =: 3.142$

or

$\displaystyle \sqrt 2 \approx 1 + \frac 4{10} + \frac 1{100} + \frac 4{1000} =: 1.414$

are also real numbers that are not rational numbers:
- $\pi$ calculates the length of the circumference of a circle with unit radius.
- $\sqrt 2$ calculates the length of the diagonal of a unit square.
Using this point of view (pun intended) we can accept these rules for adding real numbers:

$\begin{aligned} a + b &= b + a, \\ a + (b + c) &= (a + b) + c, \\ a + 0 &= 0 + a = a, \\ a + (-a) &= (-a) + a = 0.\end{aligned}$

When positive, real numbers can also be used to describe areas.

Each positive real number length $b > 0$ matches a rectangle with base $b$ and height $1$ . If $b = 0$ , the rectangle becomes a line, which has $0$ area, giving the trivial equations

$b \times 1 = b,\quad 0 \times 1 = 0.$

We can therefore use different rectangles with the same area to recover many correct equations.

In particular, we obtain the following area properties for positive real numbers:
- Commutativity: $b \times h = h \times b$
- Multiplicative identity: $b \times 1 = 1 \times b = b$
Furthermore, we can use different rectangles with the same area to construct different real numbers. For example, we can use a square with area $4$ to construct the real number $2$ .

Likewise, we can use a square with area $2$ to construct the real number approximately equal to $1.414$ , which we represent using the symbol $\sqrt 2$ .

Remark 1. We will revisit this example when discussing geometry.

For this reason, we call this number the “square root” of $2$ : it is the root number, that when used to construct a square, produces a square with area $2$ . Replacing $2$ with any positive number $a$ ,

$(\sqrt a)^2 \equiv \sqrt a \times \sqrt a = a.$

Write $a \times b$ to mean the area of the rectangle with height $a$ and base $b$ :

Since the two smaller rectangles combine into a larger rectangle with height $a$ and base $(b+c)$ , we recover the “rainbow method” to expand expressions (i.e. distributivity):

$\begin{aligned} a \times (b+c) &= (a \times b) + (a \times c), \\ (a + b) \times c &= (a \times c) + (b \times c), \end{aligned}$

where the second identity follows from the first:

$\begin{aligned}(a + b) \times c &= c \times (a + b) \\ &= (c \times a) + (c \times b) \\ &= (a \times c) + (b \times c). \end{aligned}$

Example 1. Evaluate $(a+b) \times (c + d)$ . In particular, evaluate $(a+b)^2$ .

Solution. By using the “rainbow method”,

$\begin{aligned}(a+b) \times (c + d) &= (a \times (c + d))+(b \times (c + d)) \\ &= (a \times c) + (a \times d)+(b \times c) + (b \times d). \end{aligned}$

Setting $a = c$ and $b = d$ ,

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

Remark 2. Many students will sadly make the rookie mistake $(a+b)^2 = a^2 + b^2$ .

Example 2. Explain why it is not true in general that $\displaystyle \sqrt{a+b} = \sqrt{a} + \sqrt{b}$ .

Solution. Consider the case $a = 1$ and $b = 1$ . Then

$\displaystyle \sqrt{a+b} = \sqrt{1+1} = \sqrt 2 \approx 1.414 \neq 2 = 1 + 1 = \sqrt a + \sqrt b.$

We call this violation of a proposed general pattern a counterexample.

We can do better. Consider the rectangle below with base $b > 0$ , height $h > 0$ , and area $1$ .

Since there is only one rectangle with base $b$ and area $1$ , there is exactly one positive number $h$ such that

$b \times h = 1.$

We define $\displaystyle \frac 1b := h$ , and inverse property:

$\displaystyle b \times \frac 1b = \frac 1b \times b = 1.$

Hence, we define real number division by

$\displaystyle \frac nd := n \times \frac 1d,\quad d \neq 0.$

Remark 3. If $b = 0$ , then no matter what height $h$ we have, the area of the corresponding rectangle will be $b \times h = 0 \times h = 0$ . Hence, the quantity $\displaystyle \frac 10$ is undefined. In other words: division by $0$ is illegal.

Example 3. Given $c \neq 0$ , show that $\displaystyle \frac ac +\frac bc = \frac {a+b}{c}$ .

Proof. Using real number division,

$\begin{aligned} \frac ac +\frac bc &= a \times \frac 1c + b \times \frac 1c \\ &= (a + b) \times \frac 1c \\ &= \frac{a+b}{c}.\end{aligned}$

If we want to multiply many times, we can bump things up by one dimension. Consider the volume of a cuboid with base $b$ , width $w$ , and height $h$ .

Since both cuboids have the same volume, we can multiply in either order (i.e. associativity):

$\displaystyle (b \times w) \times h = b \times (w \times h).$

Hence, we can define three-way multiplication by

$a \times b \times c := a \times (b \times c)$

without confusion. For products of even more numbers, though we would find it difficult to imagine objects beyond volume, we don’t need to worry:

$\begin{aligned} a \times (b \times c \times d) &= a \times ((b \times c) \times d) \\ &= (a \times (b \times c)) \times d \\ &= (a \times b \times c) \times d. \end{aligned}$

Example 4. Given $b, d \neq 0$ , show that $\displaystyle (b \times d) \times \left( \frac 1b \times \frac 1d \right) = 1$ . In particular,

$\displaystyle \frac 1{b \times d} = \frac 1b \times \frac 1d.$

Proof. Since we can multiply in any order of “brackets”,

$\begin{aligned}(b \times d) \times \left( \frac 1b \times \frac 1d \right) &= b \times d \times \frac 1b \times \frac 1d \\ &= b \times \frac 1b \times d \times \frac 1d \\ &= 1 \times 1 = 1.\end{aligned}$

Corollary 1. Given $b, d \neq 0$ ,

$\displaystyle \frac ab \times \frac cd = \frac{a \times c}{b \times d},\quad \frac ab + \frac cd = \frac{(a \times d) + (b \times c)}{b \times d}.$

Proof. Using Example 4,

$\begin{aligned} \frac ab \times \frac cd &= \left( a \times \frac 1b \right) \times \left( c \times \frac 1d \right) \\ &= (a \times c) \times \left( \frac 1b \times \frac 1d \right) \\ &= (a \times c) \times \frac 1{b \times d} = \frac{a \times c}{b \times d}. \end{aligned}$

To add two fractions, use Example 3:

$\begin{aligned} \frac ab + \frac cd &= \left(\frac ab \times 1 \right) + \left(\frac cd \times 1 \right) \\ &= \left(\frac ab \times \frac dd \right) + \left(\frac cd \times \frac bb \right) \\ &= \frac{a \times d}{b \times d} + \frac{c \times b}{d \times b} \\ &= \frac{a \times d}{b \times d} + \frac{b \times c}{b \times d} \\ &= \frac{(a \times d) + (b \times c)}{b \times d}. \end{aligned}$

Remark 4. By the principle in Remark 3, you might think that the number $\displaystyle \frac 1{-2}$ is also undefined. After all, how can a rectangle have negative base length? That is because we are using areas as an analogy to describe real number calculations. Nevertheless, the quantity $\displaystyle \frac{1}{-2}$ is still well defined; indeed, using Corollary 1,

$\displaystyle \frac{1}{-2} = \frac{(-1) \times (-1)}{(-1) \times 2} = \frac{-1}{-1} \times \frac{-1}{2} = \frac{-1}{2} = - \frac 1{2},$

and $\frac 12$ is a clearly well-defined positive fraction, so that $-\frac 12$ is a perfectly well-defined rational number. However, trying to define $\displaystyle \frac 10$ will create problems:

$\displaystyle \frac 10 = \frac{\frac 12 \times 2}{\frac 12 \times 0} = \frac {\frac 12 }{\frac 12 } \times \frac 20 = \frac 20 = 2 \times \frac 10$

implies that

$\displaystyle \frac 10 = 0 \quad \Rightarrow \quad 0 \times 0 = 1,$

which is absurd. Therefore, $\displaystyle \frac 10$ remains undefined:

Division by zero is (generally) illegal.

Corollary 2. Given $b, d \neq 0$ ,

$\displaystyle \frac ab - \frac cd = \frac{(a \times d) - (b \times c)}{b \times d}.$

Proof. Recall that $-r = (-1) \times r$ :

$\begin{aligned} - \frac{c}{d} &= (-1) \times \frac{c}{d} = \frac{(-1) \times c}{d} = \frac{-c}{d}. \end{aligned}$

Recall the definition

$s - t = s + (-t) = s + ((-1) \times t).$

Hence, use Corollary 1 to deduce that

$\begin{aligned} \frac ab - \frac cd &= \frac ab + \left( - \frac cd \right) \\ &= \frac ab + \frac {(-1) \times c}d \\ &= \frac{(a \times d) + (b \times (-1) \times c)}{b \times d} \\ &= \frac{(a \times d) + ((-1) \times (b \times c))}{b \times d} \\ &= \frac{(a \times d) - (b \times c)}{b \times d}. \end{aligned}$

Numbers mean nothing unless we can perform meaningful calculations with them. To add two fractions, we first find their (lowest) common denominator, then add them normally.

Example 5. Evaluate $\displaystyle \frac 23 + \frac 45$ without a calculator.

Solution. By using the common denominator $\mathrm{LCM}(3,5) = 3 \times 5= 15$ and Corollary 1,

$\begin{aligned} \frac 23 + \frac 45 &= \frac{(2 \times 5) + (4 \times 3)}{3 \times 5} = \frac{10 + 12}{15} = \frac{22}{15}. \end{aligned}$

Example 6. Evaluate $\displaystyle \frac 67 - \frac 89$ without a calculator.

Solution. Following Example 5 but instead using $\mathrm{LCM}(7,9) = 7 \times 9 = 63$ and Corollary 2,

$\begin{aligned} \frac 67 - \frac 89 &= \frac{(6 \times 9) - (7 \times 8)}{7 \times 9} = \frac{54 - 56}{63} = \frac{-2}{63} = -\frac 2{63}. \end{aligned}$

Example 7. Explain why it is not always true that $\displaystyle \frac 1{a+b} = \frac 1{a} + \frac 1{b}$ .

Solution. Consider the case $a = 1$ and $b = 1$ . Then

$\displaystyle \frac 1{a+b} = \frac 1{1+1} = \frac 12 \neq 2 = 1 + 1 = \frac 1a + \frac 1b.$

Oddly enough, we have derived the important main ideas for calculating using real numbers, so we can extend these ideas to discuss basic algebraic manipulations in the next post.

What remains are some technical remarks about our discussions.

Remark 5. It turns out that for any non-zero real number $a, b$ ,

$\displaystyle \frac 1{a+b} = \frac 1{a} + \frac 1{b}$

is always not true (i.e. false)! Similarly, if $a > 0$ and $b > 0$ , then the statement

$\sqrt{a+b} = \sqrt a + \sqrt b$

is always false.

Remark 6. We have built the basics of real number calculations, and since we will use the letters $x$ and $y$ soon, it gets very messy to write $x \times y$ . To simplify our work, we write $xy := x \times y$ , so that the “rainbow method” looks like

$a(b+c) = ab + ac.$

In particular, Example 1 looks like the following:

$(a+b)^2 = a^2 + 2ab + b^2.$

Sometimes, we use a dot to to clarify that we are doing multiplication:

$x\cdot y \equiv xy \equiv x \times y.$

Hence, we have the identities

$a \cdot (b + c) = a \cdot b + a \cdot c,\quad a^2 = a \cdot a.$

Without brackets, we will always multiply before adding:

$a \cdot b + a \cdot c \equiv (a \cdot b) + (a \cdot c),$

leading to the usual PEMDAS rule:
- parentheses (or brackets),
- exponents (e.g. $a^2$ ),
- multiplication / division,
- and finally addition / subtraction.
In fact, division is defined using multiplication:

$\displaystyle a \div b \equiv \frac ab := a \times \frac 1b,\quad b \neq 0.$

Likewise, subtraction is defined using addition:

$\displaystyle a - b := a + (-b).$

Remark 7. To properly construct the real numbers, beyond the rectangle-area picture, takes a lot more effort. We first need to define the rational numbers, then the real numbers, and finally for completeness, the complex numbers.

—Joel Kindiak, 23 Sept 25, 1556H
December 3, 2025
Negative Numbers
What is a negative number?

Let’s say that you started investing $\$100$ . You bought an asset for $\$100$ , and the price changes.
- If the price increases to $\$120$ , then you have gained $\$(120-100) = \$20$ .
- If, however, the price decreases to $\$80$ , then you have lost $\$(100-80) = \$20$ .
A positive number, therefore, represents our idea of an increase in a certain amount. A negative number, therefore, represents our idea of a decrease in a certain amount—i.e. a reverse of an increase.

A useful way to visually think of these ideas is by considering the number line below.

Here, we fix the “no-change” point $0$ :
- We represent a positive change using a rightward arrow $(\rightarrow)$ whose length describes the extent of said change.
- Likewise, we represent a negative change, by flipping the rightarrow $(\rightarrow)$ into a leftward arrow $(\leftarrow)$ whose length describes the extent of said change. We denote points to the left of $0$ with the negative sign prefix $-$ .
It is clear that $5 - 3 = 2$ , since $3 < 5$ , and so we can “take away” a smaller number from a larger number. This idea, however, fails when trying to make sense of $3-5$ .

However, the number line can help us develop some needful ideas.

Example 1. Use the number line above to evaluate the expression $3-5$ .

Solution. We re-draw the number line below.

We can read the expression $3-5$ as follows:
- start at $0$ , then
- move $3$ units rightward $(\rightarrow)$ , and finally,
- move $5$ leftward $(\leftarrow)$ .
The resulting position is $-2$ . Therefore,

$\begin{aligned} 3-5 &= 0 + 3 + (-5) \\ &= -2 \\ &= -(5-3). \end{aligned}$

Since there is nothing special about the numbers $3$ and $5$ in this example, we obtain the following result:

Lemma 1. Given whole numbers $m, n$ , we can define subtraction by

$m - n := m + (-n) = \begin{cases}\phantom{-(} m-n& \text{if}\ m \geq n, \\ -(n-m)& \text{if}\ m < n.\end{cases}$

Proof. Use the number line representation for negative numbers.

Definition 1. A negative number is a number of the form $-n$ , where $n$ is a positive whole number.
- The collection of positive whole numbers, negative numbers, and zero, is together called the collection of integers.
We define addition of integers according to the following rules, where $m,n$ are whole numbers and $m > n$ :

$\begin{aligned} m \phantom{)} + \phantom{(-} n \phantom{)} &= \phantom{-(} m + n, \\ (-m) + \phantom{(-} n \phantom{)} &= -(m - n), \\ m \phantom{)} + (-n) &= \phantom{-(} m - n , \\ (-m) + (-n) &= -(m + n). \end{aligned}$

Furthermore, we define subtraction of integers by $m - n := m + (-n)$ .

For a more complete construction, see this post that uses higher-level mathematics.

Using the number line above, we see that flipping a $3$ unit rightward direction $(\rightarrow)$ gives the same result as directly moving $3$ units leftward $(\leftarrow)$ :

$(-1) \times 3 = -3.$

If we flip a leftward direction $(\leftarrow)$ , say $-3$ , then the negative sign $-$ flips the leftward direction $(\leftarrow)$ into a rightward direction $(\rightarrow)$ :

$-(-3) = (-1) \times (-3) = 3.$

Definition 2. We define multiplication by $-1$ to mean

$(-1) \times n := -n,$

whenever $n$ is an integer.

Example 2. Evaluate $(-1)^2$ .

Solution. By Definition 2, and our prior discussion,

$(-1)^2 = (-1) \times (-1) = -(-1) = 1.$

Consolidating, we have the following sign conventions.

Lemma 2. The following multiplication of negative signs hold:

$\begin{aligned} 1 \phantom{)} \times \phantom{(-} 1 \phantom{)} &= \phantom{-} 1, \\ (-1) \times \phantom{(-} 1 \phantom{)} &= -1, \\ 1 \phantom{)} \times (-1) &= -1, \\ (-1) \times (-1) &= \phantom{-} 1. \end{aligned}$

Example 3. Evaluate the expressions $(-2) \times 3$ , $2 \times (-3)$ , and $(-2) \times (-3)$ .

Solution. Using Definition 2,

$\begin{aligned} (-2) \times 3 &= ((-1) \times 2) \times 3 \\ &= (-1) \times (2 \times 3) \\ &= -(2 \times 3) = -6. \end{aligned}$

Similarly,

$\begin{aligned}2 \times (-3) &= 2 \times ((-1) \times 3) \\ &= (2 \times (-1)) \times 3 \\ &= ((-1) \times 2) \times 3 \\ &= (-2) \times 3 = -6.\end{aligned}$

Finally, using Example 2,

$\begin{aligned} (-2) \times (-3) &= ((-1) \times 2) \times ((-1) \times 3) \\ &= (-1)^2 \times (2 \times 3) \\ &= 1 \times (2 \times 3) \\ &= 2 \times 3 = 6.\end{aligned}$

To shorten our working, we use the double negative rule:

$-(-1) \equiv (-1) \times (-1) = 1.$

Since there is nothing special about the numbers $2$ and $3$ , we obtain the following multiplication rules involving negative signs:

Theorem 1. Given integers $m, n$ , the following multiplication rules hold:

$\begin{aligned} m \phantom{)} \times \phantom{(-} n \phantom{)} &= \phantom{-(} m \times n, \\ (-m) \times \phantom{(-} n \phantom{)} &= -(m \times n), \\ m \phantom{)} \times (-n) &= -(m \times n), \\ (-m) \times (-n) &= \phantom{-(} m \times n. \end{aligned}$

Proof. Use Definition 2 and Lemma 2 and adapt the solution in Example 3.

At last, expressions of the form $m - n$ make sense, whether $m, n$ are positive numbers, negative numbers, or even just zero.

What we have done works for positive and negative fractions as well. We call them rational numbers, since fractions are ratios of numbers. Filling in the holes between the rational numbers gives us the real numbers, which we picture using the number line. More on that in the next post!

—Joel Kindiak, 18 Sept 25, 1641H
November 28, 2025