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!

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.

The Plot Thickens

Doing maths looking at equations all day can sometimes get boring. Maths gets a lot more interesting when there are pictures. Most pictures in math involve graphs, so let’s start simple and begin with plotting points on a graph.

You are familiar with the number line:

You already know how to plot a point on this. But this is only a 1-dimensional plot. The most interesting thing you can plot on this is a horizontal line which represent all the numbers between the endpoints of the line. Wouldn’t it be nice if we could plot curves in 2-dimensions! Enter the cartesion coordinate system.

To the number line, let’s add a vertical number line intersecting the horizontal one at 0:

Each of these number lines are called an axis. The horizontal one is called the x-axis and the vertical one is called the y-axis. Now you can plot a point on any of these axes, but the strength of this system is that you can plot points anywhere on the surface that the coordinate system is on. But to plot a point in 2-dimensions, you need 2 numbers.

The typical way to indicate a point in maths is by using brackets. For example: (2, 1), (0, -7), (-4,3), (-1.5, -2.75). By convention, the first number is the x coordinate and the second number is the y coordinate, so the general point is (x, y). To plot a point, say (2, 1), you first go along the x axis 2 units to the right since 2 is positive, then go up 1 unit. That’s where the point (2, 1) is. I’ve plotted several other points below:

The point (0, 0) where the axes meet is called the origin. Note that positive x values are to the right of the origin and positive y values are above the origin. Negative values are to the left and below respectively.

Now this is cool but gets quickly boring just plotting points. The interesting things happen when we plot a set of points that satisfy an equation. I’ll get into that in my next post.

Algebra, Subscripts

In my tutoring travels, I notice that some students get confused when they see subscripts, for example \[{x}_{1}\]. As you know, there are only 26 letters in the alphabet. This is almost always enough to represent variables in algebra, but if a formula indicates a pattern, then this is difficult to do using just letters.A subscript is just a way of showing different unknowns using the same letter. \[{x}_{1}\] is a different unknown than \[{x}_{2}\] but the same letter x is used – only the subscript has changed. The subscript number just indicates an order and is not used in calculations.So for example, the method for finding the average of a set of numbers is to add up all the numbers, then divide by the number of numbers you just added. To show this in a maths formula:Average = \[\frac{{x}_{1}\hspace{0.33em}{+}\hspace{0.33em}{x}_{2}\hspace{0.33em}{+}\hspace{0.33em}{x}_{3}\hspace{0.33em}{+}\hspace{0.33em}\cdots\hspace{0.33em}{+}\hspace{0.33em}{x}_{n}}{n}\]Here, the pattern is easy to see. Each number in the set of numbers is given a different subscript. Since the subscript starts at 1 and ends in n, you can immediately see that there are n numbers, which is why the formula shows us dividing by n. The symbol “⋯” is called an ellipsis and indicates that you just follow the indicated pattern until you get to the last number, \[{x}_{n}\]. For any specific set of numbers, you know what n is, but since the formula is to apply for any set of numbers, we need to use the unknown n.Sometimes, the subscript is called an index. So in more complex formulas, you may see  \[{x}_{i}\] to represent any of the unknowns. So we could say that the average is the sum of all  \[{x}_{i}\]‘s divided by n.

 

Algebra, Word Problems

Nothing strikes fear in a math student like the dreaded word problem. I have seen many students who are very good at solving equations but do poorly with word problems. The problem is that they lack the skill to convert english into an equivalent math language. In my last post, I started with converting english phrases into algebraic expressions. Let’s graduate to a full word problem and create the equivalent algebraic equation.

Karen is twice as old as Lori. Three years from now, the sum of their ages will be 42. How old is Karen and Lori?

As I suggested in my last post, let’s break this down. So here we have two unknowns: Karen’s and Lori’s ages. So a good first step is to assign letters to these unknowns.  Let’s let K be Karen’s age and L be Lori’s age. Now the first sentence in the problem has a word that means “=” in math. That word is “is”.  In other word problems, you may see words like “the same as”, “equals”, “was”, “will be”.

The first sentence in the word problem directly converts to an equation since we already assigned letters to the two ages:

Karen is twice as old as Lori: K = 2L

Now there are two unknowns here but that’s OK. We can’t solve anything yet, but there is more information in the word problem. As we read it, write down the equivalent math expressions.

Three years from now, what are their ages three years from now? Well three years from now, Karen will be K + 3 and Lori will be L + 3.

the sum of their ages will be 42. Another equation here because of the words “will be”. So we add the ages of Lori and Karen three years from now to get 42:

(K + 3) + (L + 3) = 42

The brackets are not really needed. I just put them there so you can see that I am adding Karen’s age 3 years from now to Lori’s age 3 years from now.

Now we have the equation but there are two unknowns. You usually cannot solve a single equation  with more than one unknown. But remember the first equation we wrote down: K = 2L? This equation means that algebraically, K is exactly the same as 2L. In the second equation, we can replace the K with 2L:

(K + 3) + (L + 3) = 42: (2+ 3) + (L + 3) = 42

Now we can solve this equation to find what is. I covered solving equations before, so I won’t do a lot of explaining here. I will start by removing the brackets and proceed:

2+ 3 + L + 3 = 42

3L + 6 = 42

3L = 42 – 6 = 36

L = 36/3 = 12

So Lori is 12. What about Karen? Again, look at the things we’ve written down so far. We have K = 2L, that is, Karen is twice as old as Lori. Since we already know Lori’s age, Karen must be 24.

So in most word problems, it will help if you first assign letters to the unknowns, then create expressions and/or equations from each part of the word problem. Have these all together and usually, the equation you need to solve will pop out.