There’s a famous (and probably apocryphal) story about the mathematician Carl Friedrich Gauss that goes something like this:

Gauss was 9 years old, and sitting in his math class. He was a genius even at this young age, and as such was incredibly bored in his class and would always goof off and get into trouble. One day his teacher wanted to punish him for goofing off, and told him that if he was so smart, why didn’t he go sit in the corner and add up all the integers from 1 to 100? Gauss went and sat in the corner, but didn’t pick up his pencil. The teacher confronted him, saying “Carl! Why aren’t you working? I suppose you’ve figured it out already, have you?” Gauss responded with “Yes – it’s 5,050.” The teacher didn’t believe him and spent the next ten minutes or so adding everything up by hand, only to find that Gauss was right!

So how did Gauss find the answer so fast? What did he see that his teacher didn’t? The answer is simple, really – it’s all about pattern recognition. Let’s look at the problem more closely.

1 + 2 + 3 + 4 + 5 +…+ 95 + 96 + 97 + 98 + 99 + 100 = ?

Now it’s true that adding all that up by hand would take forever, but we don’t really need to add it all up by hand. Look at this series from each end simultaneously instead of just left to right. You’ll see that we can think of this series as a set of pairs of numbers, each of which adds up to 101:

1 + 2 + 3 + 4 + 5 +…+ 95 + 96 + 97 + 98 + 99 + 100 = ?

1 + 100 = 101
2 + 99 = 101
3 + 98 = 101
4 + 97 = 101
5 + 96 = 101

and so on right through to the middle, where:

48 + 53 = 101
49 + 52 = 101
50 + 51 = 101

So we’ve got a total of 50 pairs, each of which adds up to 101. Since all our chunks are the same size, we can take a shortcut and simply multiply the size of the pair (101) by the number of pairs (50). Which is really easy to do in your head, since it’s just (100 x 50) plus (1 x 50), or 5000 + 50 = 5,050.

This reasoning can actually be extrapolated to work with any arithmetic series, and in fact is how we get the formula for the sum of an arithmetic series. Check it out:

Each pair added up to the same number, so we could actually use the mathematical expression for any one of those pairs in our formula. Since the first and last term are the ones most often known, let’s use those. Remember, the last term in the series is written as an, where n is the number of terms in the series. So that 101 will be represented by:

The first term (a1) + the last term (an), or (a1 + an)

Check out the first term in each of our pairs above. They range from 1 to 50, so there are 50 terms – exactly half the number of terms in the series. Which makes sense, since we’re pairing up the terms and that by definition gives us half as many pairs as terms.

So 50 in our example is represented by one-half of n, or n/2.

And what are we doing with these two bits of information? Multiplying them together. So our final formula would be:

Σ = (n/2)(a1 + an)

Now, sometimes you see this formula written as n[(a1 + an)/2],

or the number of terms times the average of the first and last terms. Practically that is exactly the same thing as the way we wrote it first, it’s just written a little differently. But they both have the same value, so if it’s easier for you to remember it as the average times the number of terms, do so.

This formula works for any arithmetic series, so the next time you come up against one, remember Gauss and his pairs of terms!

There’s no such thing as the square root of a negative number. Right?

Since squaring a number is defined as multiplying it by itself, and multiplying a negative times a negative gives a positive, all squares should be positive. Right?

So any number you want to take the square root of should be positive to begin with. Right?

So what if it’s not?

What do you do if you’re chugging through a problem and suddenly find yourself confronted with

x = √(-9)

It seems like to finish this problem we’ll need to take the square root of a negative number – but we can’t, so what do we do? Drop the sign and hope nobody notices? Mark it as ‘undefined’ like dividing by zero? Give up? Cry?

Well, actually, we don’t have to do any of that, because we’ve got an imaginary friend to help us.

Meet i.

i is a mathematical constant, whose sole definition is that i^2 = -1. Or, in other words, i = √(-1). i is an imaginary number – people used to think taking the square root of a negative number was impossible, so they called such results imaginary. i is known as the “imaginary unit” or the “unit” imaginary number, and he functions very similarly to the number 1 in the realm of real numbers.

Because he’s imaginary, i can be a bit difficult to wrap your head around. Just remember that he’s a constant, like 3 or 12 or even π. Unlike other special constants like pi or e, though, we have no real way to articulate his value. We can say that pi is roughly equivalent to 3.14…, and that e is roughly 2.718…, but what is i? i is just i. i is the square root of negative one, and that’s the only way we can really describe it, since he’s not a real number. We just have to accept that “the square root of negative one” would theoretically have a concrete value, and assign it a special symbol like we did with the other special constants.

So let’s put our imaginary friend i to work on our earlier problem of

x = √(-9)

i functions much as 1 does for real numbers, so we can rewrite that equation as

x = √(9 * -1)

Which we can split up into

x = √(9) * √(-1)

And now we know how to deal with that second term – we just use our imaginary friend! √(-1) = i, so

x = 3 * i

Which we can rewrite as simply x = 3i.

Imaginary numbers all include the symbol i, since they’re all essentially multiples of the imaginary unit. So when math hands you an impossibility, just grab your imaginary friend and jump into the realm of the unreal!

This week’s Math Journey builds on the material in The Function Machine. If you have not yet read that journey, I suggest you do so now.

In The Function Machine we discussed why graphing a function is possible at all on a conceptual level – essentially, since every x value of a function has a corresponding y value, we can plot those corresponding values as an ordered pair on a coordinate plane. Plot enough pairs and a pattern begins to emerge; we join the points into a continuous line as an indication that there are actually an infinite number of pairs when you account for all real numbers as possible x values.

But plotting point after point is a tedious and time-consuming process. Wouldn’t it be great if there was a quick way to tell what the graph was going to look like, and to be able to sketch it after plotting just a few carefully-chosen points?

Well, there is! Mathematicians look for an assortment of clues that help to determine the shape of a function’s graph from the equation itself – and it’s those clues that we’ll be talking about today. They come in four basic flavors: the power, the sign, the co-efficient, and the constant.

The Power

Let’s start with our old standby from the previous journey: y = x + 4. When we talk about “power” in this context, we’re referring specifically to the highest exponent on an x value. The highest power in this problem is one; there are no exponents so the x is simply raised to the first power. This means that for every value of y, there is exactly one corresponding value of x. If x is 1, y is 5. If x is 2, y is 6, and so on. For every given increase in x, there is a proportional increase in y, in this case it’s 1 to 1. And that means that this graph is a straight line. Easy enough, right?

Well, let’s throw a bit of a wrench into the works here, shall we? Your new function is y = x^2. Now, if I turned the machine around backwards and told you that y was 4, what would you give me for x? You might give me 2, right? 2 squared is 4. But hang on, there’s more than one thing you can square to get 4.

Not seeing it?

How about negative 2? When you square a negative number it goes positive, right? So your x value could just as easily have been – 2 as positive 2. And the same thing would have been true for any value of y, right – the corresponding x value could be either the square root of y or the negative square root of y. So in this case, there is more than one corresponding x value for any given value of y – in fact there’s exactly 2 corresponding x values for each y (with the exception of 0, of course). That means that this graph is NOT a straight line.

Turns out, it’s actually a parabola. All functions with x^2 as their highest power (known as quadratic functions) graph out as parabolas. The specific parameters of each parabola are determined by the other categories of clues, but the power tells us that this graph will be some kind of parabola. In the same way, the powers of higher-power functions also tell us the type of shape they will graph; third power functions (ones with a cube as their highest power) will form hyperbolas, and so on. This holds true with functions that include radicals as well; the type of power indicates the rough shape of the graph.

The Sign

Let’s take our quadratic function of y = x^2. When you plot some points it becomes clear that this is a parabola opening upwards; the larger the x values become, the exponentially larger the y value becomes. But what if I made one slight change to this equation?

Y = – (x^2)

Now I’m asking you, essentially, to take each of those y values and invert it. If x is 2 (or negative 2), y would now be negative 4. This holds true for every value of y, so if you plot a few of those sets of points it quickly appears that you’ve just flipped the parabola upside down. And indeed, the sign on the highest-power x value dictates which direction the graph will be facing (at least in terms of up-and-down; the side-to-side graphs are usually dictated by higher powers in the first place or by radicals or other more complex types of functions). If we were dealing with a straight line, the negative sign would indicate that the line travels downward as it moves to the right, rather than upward. Y = -x, for example, is a line with a negative slope, which means it moves down and to the right rather than up and to the right as y = x does. If you graphed both of those line functions, they’d come out to be mirror images of each other. So the sign on the highest-power x value dictates direction.

The Co-Efficient

When we talk about a co-efficient in math, we’re generally referring to the number that is multiplied by a variable. Take, for example, the function y = 3(x^2). How would this differ from our original y = x^2?

Well, let’s follow the problem through. With a co-efficient, each time we get the square we’ll need to multiply it by 3 before it becomes the y value. This will mean that each y value is quite a bit larger than the y value in our original problem. The curve will be quite a bit steeper, since using 2 for x will give us 12 for y instead of 4. So with a co-efficient above 1, the graph will show up steeper/skinnier/more closed. With a co-efficient that is a fraction, however, the graph will show up shallower or more open. Think about y = (1/3)(x^2). With 2 for x, you’d now end up with 4/3 for y; even less than with the original problem. So the co-efficient tells us how steep or sharp the progression of the curve is. Higher numbers mean sharper curves, while smaller fractions mean more gentle progressions.

The Constant

The constant is my favorite clue. A constant is a number that does not involve a variable. In our original y = x + 4, that +4 is the constant. That constant is the y-intercept – the value at which x is 0. If x were 0, all terms with x’s in them would become zeros and all you’d have left would be the constant. So with a quick look at the constant you can figure out one of your points with no work at all. But here’s the really fun part. Since it doesn’t involve a variable, the constant doesn’t actually change the shape of the curve itself. What it does do is move it around the plane. Take a look at y = x^2 versus y = x^2 + 4. That +4 on the end simply means that every y value you normally would have gotten is now 4 places higher on the graph. The whole curve has been lifted up four places on the graph. If it were a negative 4 – you guessed it – it would have moved down four places.

So the natural next question is: what if you want to move it right or left on the plane? Well, that involves getting a second co-efficient into play. Let’s change our equation to x^2 + 2x + 4. That 2x will shift the graph horizontally – but it’s a little bit more complicated than you might think. The signs here are actually reversed – adding 2x moves the graph to the left, and subtracting it moves the graph to the right. Also, it’s not a one-to-one ratio; in fact the ratio varies depending on the equation itself. Remember, too, that the constant is still the y-intercept, so if you get sideways transposition involved the center won’t necessarily be cleanly at an easily-discernible value anymore; but the curve will still cross the y-axis at 4. Combining those two pieces of information, along with the power, sign, and leading co-efficient to tell you the shape of the curve, will get you well on your way to knowing what the graph looks like.

Remember back at the beginning when I told you that using these clues would allow you to plot just a few points and sketch the graph more quickly? Well, here’s how we put it all together. Let’s take a new equation:

y = 3(x^2) + 5x – 2

What can we tell about the graph from the clues presented here?

First, the power. This is a quadratic function, which means we’re dealing with a parabola. The leading sign is positive, so it’ll open upward. The leading co-efficient is 3, which is greater than 1, so it’ll be a sharper, steeper curve, 3 times steeper than the basic parabola. We’re adding 5x, so the graph will be transposed to the left, and the y-intercept is at – 2. We’d still need to work out and plot a couple of points (personally, I’d factor the quadratic to find the x-intercepts and work from there – more on that next time), but now we have a better idea of what the graph would look like – and we can see all of that just from the equation alone!

This journey is heavily inspired by the youtube mathematician Vi Hart, whose videos describing mathematical concepts through doodling in a notebook were the inspiration for much of my mathematical journeys series. I’ll put a link to her video on this topic at the end of the journey, and I highly encourage everyone to go check her out.

Let’s talk exponents.

But to do that, first we should talk about multiplication. Multiplication is a shortcut for adding a bunch of the same number together. If I gave you:

5 + 5 + 5 + 5 + 5 + 5 = ?

You could just add them normally, treating each of those 5’s as a size-5 step along the number line. But since each of these addition steps is the same size, a faster way to figure out the result would be to determine two things: the size of the step, and how many steps we have. Then we can multiply the size of step (in this case, 5) by the number of steps. In this case, we have a total of 6 size-5 steps, so we’d say:

5 * 6 = 30 Size of step (5) times number of steps (6) = total number (30).

Simple enough, right?

Well, exponents are a similar type of shortcut – except this time, it’s actually a shortcut for the shortcut! Exponents are a shortcut for multiplying a bunch of the same number together, in the same way that multiplication is itself a shortcut for adding a bunch of the same number together. So, if we had:

2 * 2 * 2 * 2 * 2 = ?

Again, we could multiply them normally, but that’d take a while. Since all the steps are the same size, we can take a notational shortcut and write this as an exponent. The size of step is the base (2), the number of steps is the exponent (5). So this case could be rewritten as:

2^5

It’s important to note that I said “notational shortcut” up there. This exponent has exactly the same value as that series of * 2’s above it, which in turn has the same value as:

{[(2 + 2) + (2 + 2)] + [(2 + 2) + (2 + 2)] } + {[(2 + 2) + (2 + 2)] + [(2 + 2) + (2 + 2)] }

But that is nearly impossible to read, let alone to work with – it took me several minutes just to work out where all the brackets went! So we use notational shortcuts to express these concepts – which at their heart are just complicated forms of counting – in a way that’s easier to work with. But it’s important to note that even the most complex high-school algebra is just a fancy way of writing some complicated counting. You could take the long way around if you really wanted to, but it’d be much too confusing for everyday manipulations, so we come up with shortcuts.

But back to exponents. Knowing our shortcut makes it relatively simple to figure out the total number when faced with a problem like:

2^5 = x

You just take 2 * 2 * 2 * 2 * 2, and you arrive at 32 as your answer. Again, the base is the size of step, the exponent is the number of steps. The only difference is that these are what Vi Hart calls “times-ish” steps – steps where you’re increasing the value not by adding, but by multiplying.

One of the basic truths of mathematics is that for every operation, there is a way to undo that operation. Want to undo a multiplication? Just divide. Want to undo an addition? Subtraction’s got you covered. (This, incidentally, is sometimes why students find themselves going around and around in an infinite loop within a problem – at some point they’re undoing an operation they’ve done elsewhere in the problem, and getting back to where they started.) To undo an exponent, generally taking a root will have you covered. For example:

We’ve arrived at 32 after 5 “times-ish” steps, but I want to know what size step we took to get there. Or, in math terms:

x^5 = 32

Roots have got you covered here. Just find the 5th root of 32 and you’ll have your answer:

(5th)√32 = 2

But wait! There are actually three different positions in the basic exponent equation, right?

b^p = r b = base, p = power, r = result

We’ve seen how to solve if our variable is the result or the base, but what if the variable IS the exponent? In other words, what if I know the size of step and the number I want to get to, but not how many steps it would take to get there?

2^x = 32

How do we undo THAT one?

This is a case where, because there are three parts to the original operation, we need a second way to undo it. THAT is where logarithms come in. Logarithms are the way to undo exponentiation to solve for the exponent itself.

b^p = r solve for p?

Log(b) r = p

Traditionally, we read the equation above as “The log in base b of r is p.” When you hear the word “log,” think “number of steps to get there.” So you could really read the equation as “The number of steps of size b to get to r is p.”

Most math teachers will give you a funky-looking image with arrows pointing in a circle around the exponent to show you which variable to place where, however those images have always struck me as more confusing than they are helpful. The way I remember which number goes where actually clues off of a later part of the sentence: the phrase “in base b.” Just remember, the base is the original number, the number that says how big the steps are. The logarithm is your way of figuring out the number of steps to a total, so the total should be on the same side of the equation as your base. That leaves the exponent by itself on the other side of the equation, where we can solve for it easily.

So for our example problem from earlier:

2^x = 32

log(2) 32 = x

So now we know what we’re doing. Generally we complete logarithms either by entering them on a scientific calculator or referencing a table, since it is very difficult to calculate the value of a logarithm by hand without using trial and error. Unfortunately this can lead to math teachers focusing on teaching students to simply memorize which numbers go where in the calculator, rather than actually teaching the theory behind what a logarithm IS and why we set it up the way we do. But hopefully you understand a bit more now than you did five minutes ago. Just remember, logarithms are nothing more than your way of undoing an exponent when the exponent itself is unknown.

~

Check out Vi Hart’s video here.

Buckle up readers, it’s Trig time!

Trigonometry can be scary to many students, and in my opinion, a lot
of that is because one of the most confusing concepts in trigonometry
occurs right at the very beginning, in the form of the Unit Circle
and Radians.

Let’s start at the beginning. Give yourself a circle with a radius of 1. Now center that circle on the origin of a coordinate plane, so that the line of the circle itself passes through the points (1,0) (0,1) (-1,0) and (0, -1). Got that?

Now, this circle is referred to as the Unit Circle, because the radius is one unit and it is therefore easier for us to do various manipulations and calculations with it.
Now choose any point on the circle (we’ll call the coordinates of
that point (x,y)), draw the radius to it (which will still be a
length of 1), and drop a line back perpendicular to one of the axes.
Do that and you’ll have a right triangle with the radius as the
hypotenuse, meaning it has a hypotenuse of 1. This becomes very
useful later on for side length calculations involving the
Pythagorean Theorem. But for right now, let’s talk angles.

So in basic geometry you should have learned that there are 360 degrees in a circle. If we want to find out the measure of the angle located at the origin in our right
triangle above (what is usually referred to as a central angle), it
makes sense that we could figure it out based on what portion of the
circle the angle is taking up. If the last slice of pie was taking
up 1/8 of the full pan, we’d know that that angle was 1/8 of 360
degrees, or 45 degrees. That’s why we bother with the circle in the
first place, rather than simply dropping the right triangle randomly
in the middle of a blank piece of paper. It gives us some context to
work with, so that we can think of our angles as parts of a whole.

So what is a Radian?

Well, a Radian is a unit of angle measurement, like degrees. Only this unit is written in terms of the distance around the circumference. So whereas degree measurements
are based on a portion of 360 (360 being the “whole” benchmark),
Radians are based on a portion of the circumference. They become
very handy for finding the measure of an angle when all you know is
the length of the arc that the angle cuts out of the circle. Say you
didn’t know what portion of the pie you had left, but you had a tape
measure and could measure the length of its crust. You’d be left with
what we call an Arc Length, or the distance around the circle that
the angle in question cuts off, and with the power of radians, you
could figure out how much of the pie you had left.

So let’s talk radians, shall we?

The circumference of a circle, you’ll remember from geometry, is 2πr, or 2 pi times the radius. Well, we know the radius of the Unit Circle, it’s 1, remember? That’s the definition of a Unit Circle. That means that the circumference of the Unit Circle (and of our tasty dessert) is simply 2π.

So if we wanted to know the measure of that mystery angle at the center of our last piece of pie, we could certainly find it in terms of what portion of 2π the length of that crust was. If it was half of 2π, we’d have half the circle, and so on. In fact, before we come back to our tasty treat, let’s look at some simple examples to give you an idea of what’s going on here.

I just mentioned half the circle, so let’s look at that first. Now, we
know that that angle would be 180 degrees, since it’s a straight line
and half of the full 360, but what portion of the circumference would
it be? What would the length of the curved portion of that
hemisphere be? Well, it’d be half of 2π,
wouldn’t it? So it would be 2π/2, or just a single π.

Let’s look at one-quarter of the circle. We know the angle would be 90
degrees, and the circumference would be one-quarter of 2π,
which is 2π/4,
or π/2. If we look at three-quarters of the circle, similarly we
find that the circumference is ¾ of 2π,
which is 6π/4, or 3π/2.

What we’ve just described are the measures of those angles in radians.
Don’t be confused into thinking that one is arc lengths and the other
is angle measurements; radians are a unit for describing angles in
terms of circumference, so it’s actually still the same thing. It’s
like describing the volume of a container in both cups and liters;
the actual volume hasn’t changed, but the metric by which you’re
measuring it has.

These are all good values to remember, by the way, as they make the
conversion from degrees to radians much easier. Commit these four to
memory – it will come in handy later, I promise!

90 degrees = π/2 radians
180 degrees = π radians
270 degrees = 3π/2 radians
360 degrees = 2π radians

Incidentally, you’ll now see why I harp so much on leaving values as fractions for
as long as possible. Many students like converting to decimals
because the decimal version often looks less intimidating, but a lot
of higher math disciplines (like this one) virtually require you to
be able to work with fractions. Take 3π/2, for example. Plug that
into your calculator, and you’ll get 4.712388980385, but if I handed
you that number on a test, you’d have absolutely no way of knowing
that it’s actually 3π/2 and therefore a 270-degree angle or
three-quarters of the circle, either of which are much more useful
things to know than that random string of numbers. This is
especially important wherever π or other irrational numbers come in,
since they are often non-repeating, non-terminating decimals which
would have to be rounded off and therefore render you an estimation
rather than an exact answer. 3π/2 is exact; 4.712388980385 is not.

But back to our tasty dessert conundrum. Say you measure the crust on
your remaining piece of pie, and it comes out to be π/4. (Don’t ask
me what kind of tape measure would give you that; maybe it was a
hand-me-down from your mathematician grandfather. Problems get
really funky if you try to take the π out of the picture here.) How
do you figure out what the angle is in degrees?

Well, by now you’ve probably figured out that π/4 is the measure of the
angle in radians. So to convert to degrees, what do you do?

The easiest solution is to set it up as a proportion. You remember
those, right?

(π/4)(part) / 2π(whole) = x (part) / 360 (whole)

And solve from there. Cross-multiply:

360π/4 = 2πx

Don’t worry, the π’s cancel each other out. Now do you see why I like
leaving things as fractions?

360/4 = 2x
90 = 2x
x = 45

So that π/4 radians converts to a simple 45 degree angle. And, if
you’re feeling industrious, that means that the piece of pie you have
left is 45/360 or 1/8th of the whole pie. Nifty, huh?

Although, to be fair, you could have figured out how much of the pie you had
without ever converting to degrees in the first place. You simply
know that you have:

(π/4)(part) out of 2π (whole), so

(π/4)/2π = π/4 * 1/2π = 1/(4*2) = 1/8

So 1/8th the total pie. The π’s cancel each other out again. Starting to
make sense to you? Always leave things as fractions!

It’s beneficial to get used to working with radians right off the bat, as
most higher math levels will work almost exclusively in radians
rather than degrees. Why? Well, radians are generally more useful
in wider contexts than degrees, partly because they are based on a
distance measurement rather than a somewhat arbitrary 360. You just
have to learn to take a deep breath and not be intimidated by working
with π…it’ll work itself out in the end, I promise. And if you
come up against a problem where the π’s don’t cancel themselves out,
it’s always permissible (and in fact preferable) to leave your answer
in terms of π, where it’s easier for the next mathematician to pick
it up and work with it further.

I hope you’ve enjoyed our mathematical journey, and that radians make a
bit more sense to you now. Stay tuned for more exciting journeys in
the future!

y = f(x)

I can’t tell you how many times I’ve had students come to me profoundly confused about their entire math unit, all because their teachers never fully explained this concept. Teachers throw this equation up on the board without discussion as if it explains everything – which it does, but only if you know what it means. So let’s discuss!

First off, it’s important to remember that this is not just an equation; it’s an indication of a larger concept. We’ll get to that in a minute, but let’s start at the beginning.

Imagine that I have a little machine which I set on the table in front of you and turn on. You place a number in the slot in the top, and the machine begins to hum and churn. After a few moments, a drawer opens at the bottom and you pull out a different number. You can repeat this with any number you like, any number of times.

Now this is a single-purpose machine, which means it has one rule that it uses to transform the starting number into the final number. When you insert a number, it applies the rule to it, and presents the result. For example, say my machine’s rule is that it takes the input number and adds 4 to it. If you give it a six, it will give you a 10. If you give it a 3, it will give you a 7. No matter what number you give it, it will add four and then give you the result. In math terminology, we call machines like this “functions.”

Definition: A “function” is a machine that transforms one value into another value via a set rule.

The notation for a function is

f(x)

where x is the input value. When you see an equation in the form

f(x) = x + 4

That is simply defining the rule of the function. In this case, this is the mathematical notation for our x + 4 machine from earlier. A common question type when learning about functions is as follows:

f(x) = 3x + 6. What is f(5)?

Don’t panic! This question is simply asking you to send a number of their choosing through your machine. In this case, the questioner has handed you a five and a machine with the rule 3x + 6, and wants you to give him his result. You just substitute 5 for the x in the rule and solve it. In our case,

3(5) + 6
15 + 6
21

So the answer is 21.

An important thing to note at this point is that the “f” in f(x) is NOT a variable like x or y. “f” is an OPERATOR, like a plus sign or a division bar. It’s an indication of an operation being performed. You could replace the “f” with literally anything, so long as you define what the rule of the machine is so you know it’s a function. The full notation for this operator is actually “f()”, and you put your starting value inside the parentheses. You’ll see this a lot on the SAT, where they include questions which replace the “f” with a series of nonsense symbols in order to test your understanding of the basic concept. They may even leave out the parentheses to confuse you further. On the SAT, you might see a question like this:

@ is a function such that @x = 2x – 3. What is @6?

Don’t be thrown by the use of an @ sign; they’re just trying to confuse you. Read the rest of the question, and you’ll see that they specifically define it as a new operation, a function with a rule they’ve given you. It’s the same kind of question we just did; they want you to plug the 6 in to the rule. Heck, the operator could be a little picture of an elephant for all you care; what’s important is that they define the rule of the function so that you know what effect it has on the input value.

Now, what about that pesky f(x) = y ?

Okay, bear with me, we’re going on a journey here. You have your machine, and you know that each input number you put into it will present you with a corresponding result number. We’ve been able to pick out individual pairs of corresponding values, but what if we wanted to represent the entirety of this transformation as a graph, something that would show all possible pairs at once, even the nitpicky ones in between integers?

NOTE: For the time being, we’re going to restrict our discussion to only rules of the first power (that means no exponents). Things get a bit more complicated if you’ve got powers in your rule, so we’ll save that for later.

If your rule is a first-power rule, then for every input value there’s going to be exactly one output value. Let’s take our original x + 4 machine again. Say I put in a number 2, and it gives me a 6. Now, if I take the 2 and the 6 out of the machine, I’ll have a pair of corresponding numbers that I can plot as a point on a two-dimensional graph (one axis for the starting values, and the other for the results). We’ve been calling our starting value x in the equation, so let’s make the starting value our x coordinate. That would make our resulting value the y coordinate. So our example above gives us the ordered pair (2, 6). Plug in some more x values, and you’ll get more y values and be able to plot more points, and the shape of the graph will begin to appear. In other words, the result of the transformation is the y value for any given value of x.

And THAT can also be written as f(x) = y.

So f(x) = y is not actually an equation as much as an indication of a concept. Basically, we’re saying that we can replace the phrase “f(x)” with the variable “y”, because we know that the number we’d get would be the corresponding y coordinate for the x we started with. So we can rewrite our initial function as:

y = x + 4,

knowing that it’s still a function, just written in a way that’s more conducive to graphing.

By the way, that last equation might look a bit familiar for those of you who’ve worked with slope-intercept format when graphing lines. It’s y = mx + b. In fact, all first-power functions form a line when graphed, and all can be reworked to fit into slope-intercept format (though some get kind of unwieldy when you do). So now we see that there are two simple ways to graph a function: plotting several points and then connecting the dots, or working it into slope-intercept form and graphing it from there.

Functions are a lot of fun when you know what you’re doing, and you can get into all sorts of complex graphs using higher power functions with the same concepts in mind. Just remember: a function is nothing more than a machine that turns x’s into y’s by following a set rule. Hopefully this journey helped you understand functions a bit better, and gave you the desire to get back in there and finish your homework!

Writing a mathematical proof is like explaining math to an alien.  Imagine: you’re sitting at the table doing your math homework when the family alien (for this is the 80s and every family had one) comes to sit beside you.  He asks you what you’re doing and you show him your work.  Now, since he comes from a distant planet, he is unfamiliar with Earth methods and begins to ask lots of questions.  You find yourself stopping after each new mathematical maneuver to explain the reasons why it is possible.  The alien seems to be following along, so you continue in this manner. Eventually, you’ve reached the endpoint of the problem, and the alien can see and understand exactly how you got there.

It’s also like playing an adventure game.  You’re plopped down in a specific place and given nothing but a few select pieces of information about your current surroundings (“This is a pirate ship.  It’s captained by Blackbeard Joe.”)  and an objective (“Rescue the princess.”).  From there, you wind your way around the ship, looking at everything, uncovering more of the map, pocketing the tools which seem useful, and gradually working your way towards figuring out how to achieve your objective.  You backtrack a lot, and you also look forward.  You find out where the clues you have fit into the ones you’re just now unearthing, and soon everything falls into place.  You find Blackbeard Joe, rescue the princess, and even manage to escape with his treasure.

Direct mathematical proofs are a form of deductive reasoning.  You start with what you know, and figure out systematically what you can deduce from it.  Then you take that new piece of information and add it to your list, and repeat the process again.  You move one step at a time from the known to the unknown, until finally the whole thing clicks into place and you see where you were meant to arrive.

Completing an indirect proof, though, is like playing Devil’s Advocate; you prove your point by disproving the opposing one.  Imagine: you’re having a discussion with a friend about a hotly debated topic.  To prove your point, you begin “Let’s suppose that what you’re saying is true.  If that were the case…” You go on to expand upon their starting point until logic brings you to a statement that is blatantly absurd or incorrect.  You use this absurdity to prove to your friend that their opinion, the starting supposition for your argument, is false, thereby proving that your opinion (the opposite one) is correct.  If you are asked to prove indirectly that two lines are not parallel, you begin by supposing they are.  You work logically through the proof from there, and when you reach an inconsistency (such as 3 = 4) or a statement that contradicts one of your givens, you know that your original supposition (that the lines were parallel) is itself false.  The lines are not parallel.  Congratulations, you’ve proved your point.