It's all about undoing the mess

In this article, we are going to examine the general principle of equation solving, in particular, the algorithmic aspect. For simplicity, we will just focus on the case of single variable equation.

Before we dive into that discussion, let’s begin by looking at the following example:

$\displaystyle 3x+5=11\qquad \Rightarrow\qquad 3x=6\qquad\Rightarrow\qquad x=2$

The first equation can be read this way:

Begin with x → Multiply by 3 → Add 5 → The value is 11

To understand what the solution is, we need to “unpack” x and isolate it on one side of the equation. How? By going in the opposite direction and undoing each operation!

Begin with 11 → Subtract 5 → Divide by 3 → The value is x

The above example, though as simple as it looks, is enough to demonstrate the fundamental principle of solving an equation.

Typically, we start off with an equation where the variable is often the object that’s acted upon by some operations, and the process of solving usually culminates with the variable as the subject via some mathematical manipulations.

Simply put, it’s all about inverting the “object-subject” role of the variable.

Also, the complexity of the algorithm depends heavily on the mathematical operations involved. Our example above is rather simplistic as it only involves the four basic operations: addition, subtraction, multiplication, and division.

Suppose we have the quadratic equation

$\displaystyle 2x^2-8x+1=0$

which involves the squaring process. To undo this operation, one could expect that the solution will entail the process of taking the square root (more on this in future article).

At any rate, it’s time to start investigating some common mathematical operations to facilitate our understanding in the equation-solving business. Why don’t we start modest?

The four basic operations

In the usual number system, numbers can “communicate” with each other through four basic means: addition, subtraction, multiplication, and division.

What we’re interested in this time are their inverse relationships, which is the key to understanding how the transition from one to another works.

Addition vs Subtraction

Question: What number does 50 – 29 evaluate to?

Answer: 21

But why 21? It’s because 21 is that number such that when we add 29 to it, we get back 50.

$\displaystyle \_\_\_\_ =50-29\qquad\Leftrightarrow\qquad \_\_\_\_ +29=50$

Mathematically speaking, the two equations above are equivalent. In other words, they are two sides of the same coin in the sense that a subtraction problem can be rephrased as an addition problem.

By the way, this is exactly how people in the olden days can swiftly determine the change in a transaction. Say your groceries cost $29, and you pay using a $50 note. Instead of calculating the difference between $50 and $29, the cashier could just add on to $29 the amount of money that it takes to make up $50, which is $21.

When you try to work out the change before the cashier does so

Multiplication vs Division

Question: What number does 18 ÷ 3 evaluate to?

Answer: 6

And why 6? Because 6 is the number such that when we multiply it by 3, we get 18.

$\displaystyle \_\_\_\_ =18\div 3\qquad\Leftrightarrow\qquad \_\_\_\_ \times 3=18$

Therefore, we can similarly reformulate a division problem into a multiplication one.

The roles of 0 and 1

The numbers 0 and 1 deserve special attention because they each play a specific role in the number system:

Any number added by 0 stays the same, i.e. a + 0 = a.
Any number multiplied by 1 stays the same, i.e. a × 1 = a.

For these reasons, we call 0 the additive identity and 1 the multiplicative identity. Hence, 0 is to addition what 1 is to multiplication.

Okay, we give them cool names, so what’s the deal? The fact is, they are often the invisible heroes that help undo the mess in the process of equation solving. How so?

Let’s solve the example equation 3x + 5 = 11 again, but this time, we’ll be a bit more pedantic.

First, we subtract 5 from both sides to undo “addition by 5”:

$\displaystyle 3x+5-5 =11-5\qquad\Leftrightarrow\qquad 3x+0=6\qquad\Leftrightarrow\qquad 3x=6$

See how the part “+ 0” vanishes (since 0 is the additive identity ) which results in a cleaner equation?

Next, we divide both sides by 3 to undo “multiplication by 3”:

$\displaystyle 3x \div 3 =6\div 3 \qquad\Leftrightarrow\qquad 1x=2\qquad\Leftrightarrow\qquad x=2$

Again, notice how we can discard the coefficient “1” (since 1 is the multiplicative identity), making x the only subject appearing on the left.

Negative numbers and fractions

The discussion on the four basic operations is never complete without talking about negative numbers and fractions. We’ll take a look at their respective defining properties and study the benefits of introducing them into the number system.

Question: What’s so special about the number –29?

Answer: It is that number such that when it’s added to 29, we get 0, the additive identity, i.e.

$\displaystyle 29+(-29)=0$

The number –29 is called the additive inverse of 29. More generally, given a number n, we call –n the additive inverse of n, such that when it’s added to n, we get the additive identity 0, i.e.

$\displaystyle n+(-n)=0$

Question: What’s so special about the number 1/3?

Answer: It is that number such that when it’s multiplied to 3, we get 1, the multiplicative identity, i.e.

$\displaystyle 3\times \left(\frac{1}{3}\right)=1$

The number 1/3 is called the multiplicative inverse of 3. More generally, given a nonzero number n (remember, we can’t divide by 0), we call 1/n the multiplicative inverse of n, such that when it’s multiplied to n, we get the multiplicative identity 1, i.e.

$\displaystyle n\times \left(\frac{1}{n}\right)=1$

Essentially, the introduction of negative numbers and fractions allows us to replace every subtraction by addition and every division by multiplication. For instance,

$\displaystyle \begin{aligned} &50-29\qquad &\Leftrightarrow \qquad &50+(-29)\\\\ &18\div 3\qquad&\Leftrightarrow \qquad &18\times\left(\frac{1}{3}\right)\end{aligned}$

Here are two good reasons to do so.

Reason 1: To keep the formulas concise

Imagine coming up with a formula that involves addition. It’s such a nuisance to formulate a variant that deals with the case of subtraction every single time. Fortunately, negative numbers come to our rescue!

For example, recall the binomial identity

$\displaystyle (x+y)^2=x^2+2xy+y^2$

Suppose we need to expand (x – 3)². Although the expression inside the parentheses is a difference of two things, it’s also a sum in disguise, i.e.

$\displaystyle (x-3)^2=\left(x+(-3)\right)^2$

Now we can apply the binomial identity to get

$\displaystyle \left(x+(-3)\right)^2=x^2+2(x)(-3)+(-3)^2=x^2-6x+9$

Hence, by using negative numbers as shown above, the binomial identity can take care of the case that involves subtraction.

Reason 2: Addition and multiplication are better-behaved

Let’s be more precise on what behaviours we are referring to.

Addition and multiplication are commutative (i.e. order doesn’t matter):

$\displaystyle 50+29=29+50,\qquad\qquad 18\times 3=3\times 18$

Addition and multiplication are associative (i.e. grouping doesn’t matter):

$\displaystyle (2+13)+7=2+(13+7),\qquad\qquad (7\times 12)\times 4=7\times (12\times 4)$

In contrast, subtraction and division are ill-behaved in the following sense.

Subtraction and division are not commutative (i.e. order matters):

$\displaystyle 50-29\neq 29-50,\qquad\qquad 18\div 3\neq 3\div 18$

Subtraction and division are not associative (i.e. grouping matters):

$\displaystyle (2-13)-7\neq 2-(13-7),\qquad\qquad (7\div 12)\div 4\neq 7\div (12\div 4)$

Here’s how negative numbers and fractions come to our aid if we wish to reorder or regroup the expression that involves subtraction or division.

$\displaystyle 50-29=50+(-29)=(-29)+50$

$\displaystyle 18\div 3=18\times\frac{1}{3}=\frac{1}{3}\times18$

$\displaystyle (2-13)-7=(2+(-13))+(-7)=2+((-13)+(-7))$

$\displaystyle (7\div 12)\div 4=\left(7\times\frac{1}{12}\right)\times\frac{1}{4}=7\times\left(\frac{1}{12}\times\frac{1}{4}\right)$

To summarize, if we replace subtraction by addition and division by multiplication, we essentially replace the ill-behaved operations by nicely-behaved ones. This empowers us to be more flexible when manipulating equations or expressions.

Now that we have deepen our understanding of the four basic operations, we’re ready to solve a class of equations, called the linear equations. We’ll explore them more fully next time, but until then, stay acute, stay positive. See you!