## Don’t be a Maths Criminal!

The following is a “proof” that 1 = 2:

Let a and b be any numbers but a must equal b. For example 3 = 3 is a true maths sentence but let’s just keep things in terms of a and b. Given this as a starting point, we can do the following:

Step 1: a = b

Step 2: Multiply the left side by a and the right side by b, which is valid since a = b. This gives a2 = ab

Step 3: Add a2 to both sides: a2 + a2 = a2 + ab

Step 4: Add terms on left side: 2 a2 = a2 + ab

Step 5: Subtract 2ab from both sides: 2 a2 – 2ab = a2 + ab – 2ab

Step 6: Add the like terms on the right side together: 2 a2 – 2ab = a2ab

Step 7: Factor out the 2 on the left side: 2(a2ab) = a2ab

Step 8: Noting that (a2ab)/ (a2ab) = 1, divide both sides by a2ab: 2 = 1

Is this true? Has all our maths training been one big pile of fertiliser?

As much as my jokester half would like to say “yes it has”, there is something wrong with the above “proof”. All the steps are perfectly valid except for the last one. If a = b, then a2ab = 0. So the last step is effectively dividing both sides of the equation by 0.

Dividing by 0 is against maths law because allowing it would make maths inconsistent and all sorts of false equations can result. See what happens if you try to divide any number by 0, even 0 itself, by 0. (Do this in secret as I will not be responsible for getting you out of maths jail.)

Borrowing from an American commercial about drugs:

This is your mind: 2 = 2

This is your mind after dividing by 0: 2 = 1.

## Arithmetic Sequences

This post is about quickly adding a sequential set of numbers from an arithmetic sequence. An arithmetic sequence is a list of numbers that have the same difference between each two adjacent numbers.

For example: 2, 4, 6, 8, 10, … is an arithmetic sequence with a difference of 2 between each two numbers. What if you needed to add the first 20 of these numbers or the next 20 numbers starting with any number in the sequence? Even if you don’t need to do this, the following is very interesting and surprising.

Let’s start with this arithmetic sequence: 7, 14, 21, …, 63, 70. You can see that the difference between the numbers is 7. If you write this sequence down, add the first and last number: 7 + 70 = 77. Now add the second number and the next to the last number: 14 + 63 = 77. What?! Will I get the same number if I keep doing this? It turns out, that you will:

There are 5 pairs of 77’s so 5 × 77 = 385 and this is the same answer you would get if you added all 10 numbers manually.

It turns out that this trick will work no matter what the difference between each number, how many numbers that are to be added, or where you start adding in the sequence. The only condition is that you are using an arithmetic sequence and that the numbers added are sequential, that is you can’t skip numbers.

So a generic sequence addition looks like this:

${a}_{1}\hspace{0.33em}{+}\hspace{0.33em}{a}_{2}\hspace{0.33em}{+}\hspace{0.33em}{a}_{3}\hspace{0.33em}{+}\hspace{0.33em}\cdots\hspace{0.33em}{+}\hspace{0.33em}{a}_{{n}{-}{1}}\hspace{0.33em}{+}\hspace{0.33em}{a}_{n}$

The subscripts just keep track of the order of the terms. The last term, an, means that there are n numbers to be added. In our example, n was 10 and an was 70.

So as in our example, we added the first and last term, then multiplied that by half the numbers to be added, that is n/2. The formula for this trick using the generic sequence is:

${\mathrm{Sum}}\hspace{0.33em}{=}\hspace{0.33em}\frac{n}{2}\hspace{0.33em}\times\hspace{0.33em}{(}{a}_{1}\hspace{0.33em}{+}\hspace{0.33em}{a}_{n}{)}$

This works even if n is odd. What I find fascinating about this formula is that it doesn’t include the difference between each number. It doesn’t matter what the difference between each number is. This formula just needs to know the first and last number and how many numbers are to be added. Isn’t math strange and wonderful!

## Euler’s Initial

In my last post, the irrational number e was used. Let’s define and explain e a bit more in this post.

If you deposit $1 in the bank which pays 100% interest once each year (I need to find this bank!), then at the end of 1 year, you will have the original$1 plus 100% of that which is another $1, for a total of$2. Now bear with me, but an equivalent expression that will give me this same answer is

${\left({{1}\hspace{0.33em}{+}\hspace{0.33em}\frac{1}{1}}\right)}^{1}\hspace{0.33em}{=}\hspace{0.33em}{(}{2}{)}^{1}\hspace{0.33em}{=}\hspace{0.33em}{2}$

What happens if this generous bank computes part of the interest during the year and that interest is added to your original $1 to be included in subsequent interest calculations? When banks do this, this is called compounding interest. That is the interest made compounds, or is added to, the original investment to get even more interest. Now suppose the bank computes the interest twice per year. In 6 months, the bank will compute your interest but since only half the year has gone by, only half the interest, or 50%, is used. So in 6 months you have$1.50. At the end of the year, 50% interest is again computed but on $1.50 now. This gives a total of$2.25 which is better than the $2 you would get if the bank didn’t compound the interest semi-annually. Now the fractional equivalent of 50% is 1/2, so again bear with me, but an algebraic equivalent expression that computes the amount you will have at the end of 1 year when the bank compounds semi-annually is ${\left({{1}\hspace{0.33em}{+}\hspace{0.33em}\frac{1}{2}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{(}{1}{.}{5}{)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{25}$ Suppose the bank compounds interest four times a year (quarterly)? Again, the interest used each quarter will only be 1/4 of the annual 100% interest but the interest will compound each quarter. At the end of the first quarter, you will have$1.25. This is now the amount to be used at the end of the second quarter, and so on. The algebraic equivalent expression to compute what you have after the entire year is

${\left({{1}\hspace{0.33em}{+}\hspace{0.33em}\frac{1}{4}}\right)}^{4}\hspace{0.33em}{=}\hspace{0.33em}{(}{1}{.}{25}{)}^{4}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{44}$

This is better still! But do you notice the pattern in the algebraic expressions? The denominator of the fraction in the brackets and the exponent (power) are the same as the number of times interest is compounded during the year. So in general, if interest is compounded n times per year, the amount you will have at the end of the year is

${\left({{1}\hspace{0.33em}{+}\hspace{0.33em}\frac{1}{n}}\right)}^{n}$

You may have also noticed that the larger n is, the more money you have. Well if you’re not greedy enough, let’s find a bank that compounds daily. If your dollar is compounded daily, at the end of 1 year you will have

${\left({{1}\hspace{0.33em}{+}\hspace{0.33em}\frac{1}{365}}\right)}^{365}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{71}$

That’s great but you may have thought that would be a larger amount. The problem is that though the power over the bracket stuff is increasing which has the effect of increasing the amount, the fraction part in the brackets is getting smaller, making the stuff in the brackets closer to 1. Raising 1 to a power is just 1. So there are two competing forces here, one that increases the value of the expression and one that decreases it.

Now we can compound more frequently than daily. We can compound half-daily, etc. What happens to the expression as n increases to infinity, ∞?

Well maths does have a process for that, it’s called limits. Let me just show that and then explain it:

$\mathop{\lim}\limits_{{n}\rightarrow\infty}{\left({{1}\hspace{0.33em}{+}\hspace{0.33em}\frac{1}{n}}\right)}^{n}$

This is read: “What is the limit of this expression as n goes to infinity?”. Now you can get an approximate answer to this by putting larger and larger numbers in for n on your calculator. It turns out, that there is no exact answer that can be written in decimal numbers because the answer to the above is an irrational number like 𝜋. This was proven to be the case in 1737 by Leonhard Euler. Because of his work with this number, it is given the symbol e in his honour.

It turns out that to 50 decimal places e = 2.71828182845904523536028747135266249775724709369995…

So you see that the most you can make with your dollar is \$2.72.

This number was first calculated by Jacob Bernoulli in 1683 to solve the very problem about interest we just went through. But Euler did a lot more work with it.

e is a very important number in calculus, probability, finance, and the interesting world of complex numbers.

## Inverse Operations

I said a little about operations that are inverses to each other when I talked about square roots. Operations (or functions on numbers – I may use operations and functions interchangeably) are inverses of each other if those operations, when performed sequentially on a number, result in the original number.

For example, consider the operations of adding, then subtracting 4 to/from a number. Let’s use 5:

5 + 4 = 9, 9 – 4 = 5, the original number. More directly,

5 + 4 – 4 = 5 + 0 = 5

So no surprise, addition and subtraction are inverse operations. It doesn’t matter what the starting number is or the number you are adding or subtracting. In general, if you start with a number x, + y and – y are inverse operations:

x + yy = x

The order of these operations does not matter – I can first subtract 4 then add 4. Multiplication and division are also inverse operations. For example,

5 × 2 = 10, 10 ÷ 2 = 5 or 5 × 2 ÷ 2 = 5 × 1 = 5

Other less familiar inverse operations are squaring a number and taking a number’s square root. For example, I can square 4 then take it square root or take its square root the square the result and I will get 4 back in either direction:

$\begin{array}{l} {\sqrt{{4}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{16}\hspace{0.33em}{=}\hspace{0.33em}{4}}\\ {{\left({\sqrt{4}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{2}^{2}\hspace{0.33em}{=}\hspace{0.33em}{4}} \end{array}$

In general,

$\begin{array}{l} {\sqrt{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{x}}\\ {{\left({\sqrt{x}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{x}} \end{array}$

There are other ones like this as well: cube and cube roots, raise to the fourth power and fourth roots, etc.

Another less well known one, and one that confuses many of my students, is exponential and logarithm operations.

When x is the base and a number, say 2, is in the exponent, as in x², this is a power operation. However, when x is the exponent, and a number called e is in the base, this is called an exponential operation, .

I won’t spend much time on e, but it is an irrational number like 𝜋 and there are several reasons why it is the ‘base of choice’. So if the operation is to take a number and make it the power of e, you will generate a number. Many calculators have an ‘.’ key where you can do this. So for example, e² = 7.389 … . What inverse operation can I perform on this number to get the ‘2’ back? This operation is called the log operation.

Now logs can use any base but again, the base e is most commonly used in math. The notation for taking the log of a number is logₑx which is often shortened to ln x. The shorthand ‘ln’ stands for ‘natural log’ as e is a natural base to use in math (topic for another post). If you have an ‘ ‘ key on your calculator, you probably also have an ‘logₑx‘ or ‘ln x‘ key as well.

So what does ‘ln x‘ mean? This operation is asking the mathematical question: “What number do I need to raise e to, so I get x?”. Can you see how this is the opposite of ‘eˣ ‘ where I actually raise e to the x power? So, ln() is asking the maths question: “What power do I need to raise e to, so that I get ?”. Isn’t it obvious that x is the number I have to raise e to, to get ? So and ln x are inverse operations. If you take the number we generated before, 7.389 … , and hit the ln x key on your calculator, you will get the original ‘2’ back.

There are many other inverse functions. In trigonometry, the sin x key in your calculator will give the sine of the angle x. Sometimes 𝜃 is used instead of x. The inverse sine, sometimes called the arcsine, is the inverse of this. The notation you may see on your calculator for this operation is ‘ASIN x‘ or ‘SIN ̅¹x‘. SIN ̅¹x is asking the maths question: ” What angle has x as its sine?”. So SIN ̅¹[SIN(x)] is x.

There are inverse functions associated with the other trig functions as well. You will usually see inverse function keys as the same key on a calculator, you just need to hit another key first to activate the inverse definition for the key.

## How do you Rate?

This post will be about the relationship between distance, speed (or rate) and time, as well as the manipulation of units.

I think most of you are comfortable solving this problem:

If you are travelling at 60 km/hr, how far will you have gone after 2 hours? So if you travel 60 km every hour, after 2 hours you will have travelled 60 × 2 = 120 km. The 60 km/hr is a rate (or speed as normal people would call it) and the 2 hours is the time. The result of multiplying these two things gives a distance. So in equation form, the relationship between these three things is d = rt, where d is distance, r is the rate or speed, and t is time. Now this equation is in the form that allows solving for distance. But we could just as well use this equation to solve for an unknown rate or speed. If the problem was: you travel 120 km at a constant speed for 2 hours. How fast were you travelling? So the unknown thing here is rate. If I take the above equation and divide both sides by t, I get r = d/t. So for this problem, 120km/2hr = 60 km/hr. I can similarly solve for an unknown time.

Now notice how the units work out. in the original problem, I am multiplying a rate times a time. That is, the “hr” cancel to leave just “km” just as if they were variables:

$\frac{\mathrm{km}}{\rlap{/}{\mathrm{h}}\rlap{/}{\mathrm{r}}}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{\mathrm{h}}\rlap{/}{\mathrm{r}}\hspace{0.33em}{=}\hspace{0.33em}{\mathrm{km}}$

In the second problem we are dividing a distance, km, by a time, hr. That is, km/hr and that is the unit of the answer, a rate. Now let’s do a problem where time is the unknown.

The speed of light is 299,792 km/sec. The average distance from the sun to the earth is 149,597,870 km. I say ‘average’ because the earth’s orbit around the sun is not perfectly circular. So how long does it take for a ray of light to travel from the sun to the earth? Or a more interesting (and morbid) way to ask this is, how long would it take before we knew that the sun exploded?

Going back to our rate equation and solving for t gives t = d/r. So

$\frac{149,597,870\mathrm{km}}{299,792\mathrm{km}/\sec}\hspace{0.33em}\hspace{0.33em}{=}\hspace{0.33em}{499}\sec$

Again, looking at the units, and remembering that when dividing by a fraction, you get an equivalent multiplication problem by multiplying the numerator times the reciprocal of the denominator gives

$\frac{\mathrm{km}}{\mathrm{km}/\sec}\hspace{0.33em}\hspace{0.33em}{=}\hspace{0.33em}\rlap{/}{\mathrm{k}}\rlap{/}{\mathrm{m}}\hspace{0.33em}\times\hspace{0.33em}\frac{\sec}{\rlap{/}{\mathrm{k}}\rlap{/}{\mathrm{m}}}\hspace{0.33em}{=}\hspace{0.33em}\sec$

So ‘sec’ is the appropriate unit for the answer. Let’s convert the answer to minutes. There are 60 sec/min so

$\frac{{499}\hspace{0.33em}\sec}{{60}\hspace{0.33em}\sec{/}\min}\hspace{0.33em}{=}\hspace{0.33em}{499}\hspace{0.33em}\rlap{/}{s}\rlap{/}{e}\rlap{/}{c}\hspace{0.33em}\times\hspace{0.33em}\frac{\min}{60\rlap{/}{s}\rlap{/}{e}\rlap{/}{c}}\hspace{0.33em}{=}\hspace{0.33em}{8}{.}{32}\hspace{0.33em}\min$

So at any moment, you have a little more than 8 minutes to live. What a happy thought!

## Graphs of Trig Equations, Part 3

I know it’s taking a while before I use maths to model a mass on a spring, but that will only make sense by fully describing the graphs of sine equations. Hopefully, this development is interesting in its own right.

Now if you were to plot the daylight length at a certain latitude against days, and if you plotted for a full year, you would see a shape that looks amazingly like the sine graph I showed you in my last post. Except at the equator, the length of a day gets longer in the summer and shorter in the winter. Without actual taking a year to collect the data, I’ve plotted the daylight length in Melbourne Australia against days using the equation

${L}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{63}\sin\left({\frac{360t}{365}}\right)\hspace{0.33em}{+}\hspace{0.33em}{12}{.}{165}$

where L is the length of daylight in hours and t is the number of days after 22 September of any year. Why I chose 22 September is an interesting topic which I may eventually discuss, but it has to do with what are called equinoxes. The plot is below and is almost exactly the plot created if I actually measured the day length each day and plotted these for a year:

Now the shape of this curve is a sine wave but you can see several differences from the standard sine wave explored in my last post:

1. The amplitude is 2.63 instead of 1
2. The wavelength is 365 days instead of 360 degrees
3. The wave is centered at 12.165 instead of the x-axis
4. We are evaluating the sine of time instead of degrees.

Let me explain these differences.

1. Amplitude – The value of sin(x) , regardless of what form x is in, only has values from -1 to +1. So if I multiply the sine by any number, say 2.63, so that I now have y = 2.63 sin(x), then this results in values from 2.63 × (-1) = -2.63 to 2.63 × (+1) = 2.63. This make the amplitude of this new equation 2.63, that is, the number I multiply the sine by. So in general, the amplitude of y = sin(x) is A.
2. Wavelength – Notice that the wavelength 365 is the denominator in
$\sin\left({\frac{360t}{365}}\right)$

The 360 in the numerator is the wavelength of the standard sine wave. The common symbol to represent wavelength is the Greek letter lambda, 𝝀, so in general, when you are taking the sine of something that looks like

$\sin\left({\frac{360t}{\mathit{\lambda}}}\right)$

the denominator, 𝝀, is the wavelength.

Now for those of you who have had exposure to this before, you may have expected to see 2𝜋t in the numerator instead of 360t. This would be the case if we were taking the sine of numbers expressed as radians. But this series of posts is doing everything with calculators in the degree mode. I will explain radians later in a different post.

3. Wave center – Notice that the center of the sine wave is at 12.165 which is the number added to the sine in the daylight length equation. The effect on a graph of adding a number to an equation is to raise or lower it – it does not change shape. So if you can graph and know the shape of y = something, then y = something + 10 will be the same shape, just shifted up 10 units. So adding 12.165 to the sine, doesn’t change its shape, it just changes where it is on the graph.

4. Time – The big change here is that we are no longer finding the sine of an angle. It may appear that we are now taking the sine of numbers in seconds, hours, or days – whatever the units of t are. However, the 360 in the numerator serves the purpose of making the number we are taking the sine of, unitless. That is, 360t/365 does not have any units – it is just a pure number.

Mathematicians/scientists long ago discovered that many periodic physical processes, have motions that follow a sine wave. In fact, when equations were formed that represented the forces on objects that were experiencing periodic motion, the sine of numbers involving time appeared when solving these equations.

And so it is with the length of the day throughout the year. The earth is rotating around the sun and this motion repeats, that is, is periodic. It is no surprise then that the graph of the day length is a sine wave.

I think we are now ready to model a mass on a spring. Let’s do that in my next post.

## Graphs of Trig Equations, Part 2

In my last post, I showed that angles repeat every 360°. So an angle of 45° is the same as 45 + 360 = 405°. I also showed how angles can be negative if a reference line, like the positive x-axis is set up and angles created from that line going in the counter-clockwise direction are positive and going clockwise are negative. And I also showed that for angle 0°, sine 0° = 0 and sine 90° = 1. Please read my last post if needed.

Now without going through the development, it turns out that the sine has values that range from -1 to 1. Angles between 0° an 180° have positive sines and angles between 180° and 360° have negative sines. This repeats as one continues rotating around the x-axis.

Now we have already covered plotting equations so let’s plot the equation

y = sin x

where x is the angle:

So this is what a sine curve looks like. You can see that as you move along the x-axis, the curve moves up and down and repeats itself every 360°. The cosine curve is very similar but it is shifted to the left so that it begins at 1 when x = 0. So you see that the sine equation may prove useful when modelling something that repeats, like a mass on a spring bobbing up and down or a pendulum.

Now to prepare us for the modelling exercise which I will get to eventually, I want to define some characteristics of this sine curve.

First it has an amplitude. Amplitude is how high the curve goes above or below the center-line of the sine curve (or sine wave as it is frequently called). In this case, the center-line is the x-axis and the amplitude is 1 since the maximum extent of the curve is 1 unit above and below the center-line.

The sine wave has wave length. This is the distance between successive peaks (the highest points) or troughs (the lowest points). Lets look at the curve and measure the distance between any successive peaks. There is a peak at x = 90 and the next one is at x = 450. The distance between these two points on the x-axis is 450 – 90 = 360. This is what we expected as we know the sine curve repeats every 360° which is what wavelength means.

Associated with wavelength is something call frequency, but this will not make sense until I do a bit more development and include time in the mix. Stay tuned for the next post!

## Graphs of Trig Equations, Part 1

Happy New Year and welcome to my first post of 2019!

My last post introduced the idea of modelling physical things with math equations. To do this from scratch, requires calculus but seeing the final result is very interesting. So in my last post, I modelled the simple physical event of a ball thrown into the air. Another common example when introducing modelling to students is a mass on a spring. But before I develop this, I want to show what the graphs of some trigonometric equations look like as they will be needed to describe any kind of motion that is cyclic, that is, repeats like a mass on a spring bobbing up and down.

So in a previous post, I defined what sin 𝜃 and cos 𝜃 are in terms of a right triangle. Given the below triangle

the sine and cosine of 𝜃 are defined as$\sin\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{opp}}{\mathrm{hyp}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\cos\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{adj}}{\mathrm{hyp}}$

Let’s look at the sine for now. For very small angles, the opposite side will be small compared to the hypotenuse. Graphically, I think you can see that for an angle of 0°, there would be no opposite side so sin 0° = 0.

In the other extreme, as the angle gets close to 90°, the opposite side is close to the length of the hypotenuse, so the sine approaches 1. In fact,

sin 90° = 1.

Now angles are periodic in that they repeat every 360°. That is, an angle of 30° is also 30 + 360 = 390°. Another full circle of 360° can be added again to get an equivalent angle 390 + 360 = 750°. Angles can also be negative based on a convention of which direction you move to create the angle. Even with negative angles, multiple of 360° can be added or subtracted to get an equivalent angle whose sine will be the same. The below diagram shows these variations based on angles generated from the positive x-axis:

The angle in red is a positive angle, that is it is formed by going in the counter-clockwise direction from the x-axis. From that angle, you can go 1, 2, 3, etc complete circles to form the same angle. The angle in blue is a negative angle, that is it is formed by going clockwise from the x-axis. One can also go multiple complete circles around this angle to get the same angle. The point is that as you measure angles from 0, either in the positive or negative direction. you eventually repeat the same angles and these same angles will have the same sine value.

In my next post, I will plot the sine values against the angle values and show graphically what “periodic” means.

## Graphing Equations

Well enough statistics, let’s return to some algebra topics. I’ve done a couple of posts on graphing and I would like to return to that.

So if you remember, a graph which we call the  cartesian coordinate system, is a way of plotting points in the form of (x, y) where x is the horizontal axis coordinate and y is the vertical coordinate. Below is a plot of several points on a graph:

But this is rather boring. Much more interesting is the graph of an equation which is a picture of all the x and y values that satisfy the equation. So if I have an equation y = x + 3, the plot of that equation is below with a few points labelled that show that they do indeed solve the equation, that is, make it true:

For example, the point (1, 4) solves this equation because when I substitute in x = 1 and y = 4, I get a true equation:

${y}\hspace{0.33em}{=}\hspace{0.33em}{x}\hspace{0.33em}{+}\hspace{0.33em}{3}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{4}\hspace{0.33em}{=}\hspace{0.33em}{1}\hspace{0.33em}{+}\hspace{0.33em}{3}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{4}\hspace{0.33em}{=}\hspace{0.33em}{4}$

Plots of equations can be many shapes, but the concept is always the same: the plot is a picture of all the points that satisfy the equation. Let’s look at

${y}\hspace{0.33em}{=}\hspace{0.33em}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}$

If you just choose some x values and substitute them in and find the corresponding y values, you can get enough points plotted to show the approximate shape of the graph. For example, if x = 0, then y = -4. So (0, -4) is a point on the graph of this equation. If if x = 2, then y = 0, so (2, 0) is a point on the graph of this equation. Below is the graph of this equation with some points labelled:

Of course, there are an infinite number of points that satisfy this equation like (1.5, -1.75). The point is, this is a picture of all the points that satisfy the equation (within the plot borders of course as the graph goes up forever).

Graphs of equations are very useful in many areas of math, science, and engineering. In my next post, I’ll use a graph to show how it helps visualise a physical process like throwing a ball up in the air.

## The Meaning of Graphs

In my last post, I showed how to plot points on a coordinate system. It is important to remember that a point such as (2, 1) means that for that point, x = 2 and y = 1. For the point (-5, -3), x = -5 and y = -3. The first number is always the x value and the second point is the y value. With that as a background, let’s talk about how to graph an equation.

Now we can solve an equation with one unknown like

${x}\hspace{0.33em}{+}\hspace{0.33em}{3}\hspace{0.33em}{=}\hspace{0.33em}{7}$

Now past posts have talked about how to formally solve an equation like this but I think you can readily see that the solution to this equation is x = 4. That is, if you replace the x with 4, you get a true statement that 4 + 3 = 7. Now this is one equation with one unknown and there is only one solution. But what about y = x + 3 ? Here there are two unknowns, y and x. But you can come up with several solutions. If x = 4, then y = 7. If x = 5, then y = 8. If x = 1, then y = 4. If x = -2, then y = 1. You can see that there are many solutions to this one equation with two unknowns. In fact, there are an infinite number of solutions especially when you consider that fractional numbers are allowed as well. For example, if x = 2.67, then y = 5.67.

Now you see that the solutions are pairs of numbers: an x and a y. So we can think of a solution as a point on a graph. Since any point plotted on a coordinate system is of the form, (x, y), two of the solutions can be shown as (4, 7) and (-2, 1). If I plot these points and others that are solutions to the equation, it appears that all the points that make this equation true are on a line. in fact, they are and I’ve drawn the line over the points:

The main point here is that the graph of this equation is a picture of all the (xy) pairs that satisfy the equation. By looking at this, you can pick out other solutions like (3, 6) or (-3, 0). All the fractional points that satisfy this equation are also on the line. Since there are an infinite number of points that satisfy this equation, the graph is a solid line. Any graph, even curvy ones, are a picture of all the (xy) pairs that satisfy the equation that generated the graph.