# Introduction to Math¶

When someone says math, most people think about calculating something with
numbers, e.g. "what is 6 times 7". You don't need to be good at that sort of
things in order to read this tutorial. If we need something like that, we'll
just calculate it using our favorite programming language or calculator, which
is Python in my case, but you can use anything you want. Python calculates
`6 * 7`

a lot faster than you can or I can, and a lot more reliably.

```
>>> 6 * 7
42
```

Instead of boring number stuff, we'll look into formulas written with letters, like this one:

$$\begin{align}(a+b)c = ac + bc\end{align}$$Memorizing these formulas is not the point. Instead, we'll derive these ourselves, prove that they work and do other stuff like that. It's called mathematics.

The above formula says that if you add two numbers and you multiply them by a
third number (we write $ac$ instead of `a*c`

in math, as explained later), you
get the same result as if you had multiplied the first two numbers with the
third number and added the results. Does it actually work?

```
>>> (1 + 2)*3 # left side: a=1, b=2, c=3
9
>>> 1*3 + 2*3 # right side: a=1, b=2, c=3
9
>>> (3 + (-2))*(-5) # left side: a=3, b=-2, c=-5
-5
>>> 3*(-5) + (-2)*(-5) # right side: a=3, b=-2, c=-5
-5
```

It **seems** to work. This is important: now we know for sure that it works for
*some* $a$, $b$ and $c$ values, but we **don't** know for sure if it works for
*all* values. Maybe there are some special $a$, $b$ and $c$ that break it? You
can easily create a for loop that tries this formula with many different values,
but what if you forgot to check for some very special value? Trying out
different values isn't the thing to do it in math, and instead we need things
that you'll learn in this tutorial.

```
<Zaab1t> if n is even
<Zaab1t> is a**n then equal to a**(n/2) * a**(n/2)
<Akuli> yes
<laplous3> on the surface that seems reasonable, but how do you actually
prove something like that?
<laplous3> <- doesn't have a breeze what's involved in proving something
<laplous3> I mean, I suppose picking arbitrary numbers and showing it
works doesn't cut it right
```

## Variable Names and Notation Stuff¶

If you are a programmer, you already know what a variable is. The concept is quite similar in math.

Mathematicians like single-letter variable names. It's bad style in most programming languages, but if you look at this...

$$\begin{align}(a+b)c = ac + bc\end{align}$$...then how would you name the variables, if you wanted to avoid single-letter variable names? Maybe like this:

```
(first_number + second_number)*third_number = first_number*third_number + second_number*third_number
```

It is much longer now, and not any easier to read. The variables don't have a
specific meaning in the formula. Each of them represents a number, and it's hard
to come up with better variable names than the ones above. Also, math was
originally written by hand, and writing things like `third_number`

by hand would
be really annoying.

In math, we can define variables like this:

Let $a=5$.

But in the case of $(a+b)c=ac+bc$ we don't know the values of $a$, $b$ and $c$ yet, so we might define those like this:

Let $a$, $b$ and $c$ be real numbers.

A real number means any number between $-\infty$ and $\infty$, but not $-\infty$ or $\infty$; that is, pretty much any number you can think of, like $2$, $0.1$, $3.14159$, $-5$ or $1000000$. We say "real number" to make a difference between other kinds of numbers; for example, "let $n$ be an integer" means that $n$ can be $2$ or $-5$ or $1000000$, but not $0.1$ or $3.14159$.

Mathematicians are lazy, so usually we don't want to write out "let $x$ be a real number" (think about writing by hand) every time we want to work with numbers. We can do better by abbreviating "real numbers" as "reals"...

Let $a$, $b$ and $c$ be reals.

...but instead, we make it even shorter:

Let $a,b,c \in \mathbb{R}$.

That's a mess! Let me explain how to read it:

- $\mathbb{R}$ is a set of all real numbers. Sets are not like arrays because
some sets contain infinitely many objects, and the order doesn't matter; the
sets $\{a,b\}$ and $\{b,a\}$ are equal, but in programming, the arrays
`[a, b]`

and`[b, a]`

are not. $\mathbb{R}$ is not the only set with a single-letter name, but that's the only one you'll need in this tutorial. - $\in$ means "in", so you should read this like "let a,b,c in reals".
- "Let $a,b,c \in \mathbb{R}$" means "Let $a \in \mathbb{R}$ and $b \in \mathbb{R}$ and $c \in \mathbb{R}$". Copy/pasting is bad, just like in programming.

Real numbers don't have anything to do with things that we can measure in real life. If we say "let $x=2$", we don't mean that $x$ is 2 inches or metres or whatever. It's $2$. This is also true when we draw stuff; it's perfectly normal to draw a rectangle like this:

The rectangle's width is 3 and its height is 2, and therefore its area is 6. It is
*not* 3 inches wide or 2 inches high, and its area is not 6 square inches. If
we *use* the math for something practical, then we need inches and centimeters
and pixels and other stuff, but the math itself is written without that.

A note about $ab$ and `a*b`

: if I had written something like
$a \cdot c + b \cdot c$ you would need to think about that for a while. What was
the operator precedence again? Oh right, multiplication is done first, so we
have $(a \cdot c)+(b \cdot c)$. That's bad because I want to let you focus on
important things instead of notation details, and $ac+bc$ is better; you can see
right away what it means because the multiplied things are closely together.

Finally, here is some boring but important stuff. You probably know what $a < b$ and $a > b$ mean, but we also use these things in math:

- $a \ge b$, or "greater than or equal", means $a > b$ or $a = b$. For example, $2 \ge 1$ and $1 \ge 1$ are true, but $1 > 1$ is false.
- $a \le b$ means $b \ge a$.
- "$x$ is positive" and "$x$ is negative" mean $x > 0$ and $x < 0$, respectively. $0$ is not positive or negative.
- "$x$ is non-negative" or "$x$ is not negative" means $x \ge 0$. This is useful when negative numbers need to be treated specially, but it doesn't matter whether $x$ is zero or positive.
- "$x$ is non-positive" or "$x$ is not positive" means $x \le 0$.

## Proofs and Axioms¶

Here is a proof of $(a+b)c = ac + bc$. The proof works only if $c \in \{1,2,3,...\}$, because otherwise it doesn't make sense to add a number to itself $c$ times. We'll talk about other $c$ values soon.

$$\begin{align}(a+b)c &= \underbrace{(a+b)+(a+b)+...+(a+b)}_{c\text{ times}} \\ &= \underbrace{a+b+a+b+...+a+b}_{ a\text{ repeated }c\text{ times, }b\text{ repeated }c\text{ times}} \\ &= \underbrace{a+a+...+a}_{c\text{ times}} + \underbrace{b+b+...+b}_{c\text{ times}} \\ &= ac + bc\end{align}$$The point with proofs is that you don't believe things because they are in a book, but because the proof convinces you that the formulas actually work like they're supposed to work.

First of all, if you want to prove that $x=y$, the most basic way to do it is to start with $x$, and do some stuff to it until you end up with $y$. That's what we did above.

The above proof relies on a few facts that you probably know already, and we can't do the proof without these "basic facts":

- We can parenthesize $+$'ed things however we want: $(x+y)+z = x+(y+z)$, and because those things are the same, we can simply write the whole thing as just $x+y+z$. In our case, $(a+b)+(a+b)+...+(a+b)=a+b+a+b+...+a+b$.
- We can reorder $+$'ed things however we want: $a+b+a+b=a+a+b+b$. In the simplest form, $x+y=y+x$, and the more complicated ab things can be done by applying the simple $x+y=y+x$ rule multiple times.
- If $x \in \{1,2,3,...\}$, then $xy$ means $y$ added to itself $x$ times.

That's all there is to our proof. But we assumed that $c \in \{1,2,3,...\}$. Let's see if the thing works with a different $c$:

```
>>> (1+2)*0 # left side: a=1, b=2, c=0
0
>>> 1*0 + 2*0 # right side: a=1, b=2, c=0
0
>>> (3+4)*(-3.14) # left side: a=3, b=4, c=-3.14
-21.98
>>> 3*(-3.14) + 4*(-3.14) # right side: a=3, b=4, c=-3.14
-21.98
>>>
```

Again, it *seems* to work and we can *guess* that it will work, but how do we
prove it this time? I thought about this for a while and struggled with it, and
turns out that we can't really do it with the "basic facts" I listed above. The
$(a+b)c=ac+bc$ thing **itself** is yet another "basic fact" that we need to take
for granted in order to get anything done. These "basic facts" are called
**axioms**, and you can google something like "axioms of real numbers" to get a
complete list if you want. There are a few more than the ones we saw here, but
not many.

## Proof Techniques¶

If you want to prove that $\green x=\blue y$, you can try to simplify $\green x$ and get $\blue y$, or you can simplify $\blue y$ and try to get $\green x$. Alternatively, if you simplify $\green x$ and get a third thing $\maroon z$, you can try to simplify $\blue y$; if you also get the same $\maroon z$ from that, you know that $\green x=\maroon z$ and $\blue y=\maroon z$, and so $\green x=\blue y$. For example, to prove this...

$$\begin{align}(1+2) \cdot 3 = 1 \cdot 3 + 2 \cdot 3\end{align}$$...without the $(a+b)c$ axiom, you can do this:

$$\begin{align}\green{(1+2) \cdot 3} = 3 \cdot 3 = \maroon9 \\ \blue{1 \cdot 3 + 2 \cdot 3} = 3 + 6 = \maroon9\end{align}$$If you want to prove that something is **false**, all you need to do is to give
an example of the thing not working. For example, if you want to prove that this
is false...

For all $a,b,c \in \mathbb{R}$, $(a+b)c=a+bc$.

...all you need to do is to plug in some values:

```
>>> (1+2)*3
9
>>> 1 + 2*3
7
```

Here is the proof that the above statement is false:

$(1+2) \cdot 3 = 3 \cdot 3 = 9 \ne 7 = 1+6 = 1+(2 \cdot 3)$

We found such $a,b,c$ that $(a+b)c \ne a+bc$.

So far we have talked about proving that something is true for **all** $a$, $b$
and $c$ values, but what if you need to prove that there is **some** number or
combination of numbers that satisfies a thing? For example, what if you want to
prove this statement?

There is a number $x \in \mathbb{R}$ so that $2x = 6$.

Here is the proof:

$2 \cdot 3 = 6$, so $3$ is the number.

Well, that was easy.

Here's yet another powerful technique: **proof by contradiction**. If you want
to prove something, you can assume that what you want to prove is *false*, and
see what happens; if you end up with nonsense like $1=2$, the assumption was
wrong, and what you're trying to prove must then be correct.

For example, let's prove this statement...

There are no positive reals $a$ and $b$ so that $a+b=0$.

...like this:

Let $a,b \in \mathbb{R}$ and $a,b > 0$. Let's assume $a+b=0$. Substracting $b$ on both sides gives $a+b-b=0-b$; that is, $a=-b$. $b$ is positive, so $-b$ is negative, and $a$ is also negative because it's $-b$. This is not possible because we just said that $a$ is positive. Our assumption gave us nonsense, so it must be false. This means that there are no positive $a$ and $b$ so that $a+b=0$, which is what we wanted.

Of course, you don't need to write down that much text every time. This is enough:

Let $a,b \in \mathbb{R}$ and $a,b > 0$. If $a+b=0$, then $a = -b < 0$, which is a contradiction.

## Simplification Techniques¶

In this tutorial, division is usually not written as $a/b$; it is almost always $\frac a b$, and later you'll find that much nicer to work with.

Let $a,b,c,d,e,x,y,z \in \mathbb{R}$. The things I'm about to show can be proved with axioms of reals.

For every $a$, there is a number $-a$ such that $a + (-a) = 0$. If $a$ is positive, then $-a$ is negative, and vice versa. On a number line, this looks like flipping to the other side. In this picture, $a$ is positive and $b$ is negative:

Substraction $a-b$ actually means $a + (-b)$.

A similar thing happens with division: for any real number $a$ except $0$ (we'll talk about dividing by $0$ soon), $\frac 1 a$ is a number that gives 1 when you multiply it by $a$:

$$\begin{align}a + (-a) &= 0 \\ a \cdot \frac 1 a &= 1, \qquad a \ne 0\end{align}$$As you can see, adding and multiplying rules look similar, but some of them are slightly different.

If you add together many things, you can put them to any order you want. The same thing works with multiplying.

$$\begin{align}a+b+c+d+e &= e+d+a+c+b \\ abcde &= edacb\end{align}$$If we combine this with the $a-b=a+(-b)$ thing, we'll notice that we can order $+\text{something}$ and $-\text{something}$ things however we want. For example:

$$\begin{align}a-b-c &= a + (-b) + (-c) \\ &= (-c) + a + (-b) \\ &= -c+a-b\end{align}$$It works the same way with more things too:

$$\begin{align}\red{a} \blue{+b} \green{-c} \maroon{+d} \color{black}{-e} = \color{black}{-e} \maroon{+d} \red{+a} \green{-c} \blue{+b}\end{align}$$Note that $a = +a$.

With multiplication and division, you can also reorder things however you want, as long as things under the horizontal line are being divided, and things above it are being multiplied.

$$\begin{align}a \cdot \frac{b}{c} \cdot \frac{d}{e} = \frac{abd}{ce}\end{align}$$The $a$ goes to top because $\frac b c=b\frac 1 c$, and so $a\frac b c = ab\frac 1 c=\frac{ab}{c}$.

In the above math, $c$ and $e$ must not be $0$ because you can't divide by zero.

```
<laplous3> when you divide by 0 you get theelous3
<laplous3> it's a sort of math black hole
```

Division $\frac x y$ is essentially "I want a number that gives $x$ when I multiply it by $y$":

$$\begin{align}\frac{x}{y}y = x\end{align}$$Division by zero is saying "I want a number that gives $x$ when I multiply it by zero". If you multiply a number by zero, you always get zero:

$$\begin{align}0z = 0 \qquad \text{for all } z \in \mathbb{R}\end{align}$$So, saying $z = \frac x 0$ means $0z = x$, which *never* works if $x \ne 0$.
However, if $x=0$, it works for *any* $z$. In other words, it seems like
$\frac 0 0$ can be anything. That doesn't really make sense either, but later
we'll see that if we are getting $\frac 0 0$, we don't know what the answer
*should* be based on just that, and we need to go back and look at where the
$\frac 0 0$ came from. Usually there's a way to do whatever we're trying to do
so that we can avoid $\frac 0 0$, but if you get $\frac x 0$ with $x \ne 0$,
you probably did something wrong.

Sometimes things *cancel* out nicely:

We're adding the $x$ and taking it away, so after all the $x$ goes away completely, no matter what the value of $x$ is. It works similarly with multiplication and division:

$$\begin{align}\frac{ \red{x} y }{ \red{x} } = y\end{align}$$$$\begin{align}\frac{ ab\red{c}d\blue{e} }{ \red{c}\blue{e}xy} = \frac{abd}{xy}\end{align}$$Sometimes it's useful to use this canceling rule right-to-left and add more stuff:

$$\begin{align}\frac{ a\blue c }{ b\blue c } &= \frac a b \\ \frac a b &= \frac{ a\blue c }{ b\blue c }\end{align}$$The "backwards canceling" thing also works with $+$ and $-$; sometimes it's useful to look at $y$ as $y+x-x$ or something like that.

Be aware of mixing adding and multiplying rules. They don't mix very well, for example:

$$\begin{align}\frac{a+b}{2} c = \frac{a+bc}{2} \qquad \red{\text{wrong!}}\end{align}$$The problem is that with multiplying and dividing, we need to treat $a+b$ as one thing; if we had $ab$ on the top instead of $a+b$, then $a$ and $b$ would be multiplied with the other stuff as we've seen before, but now the $a+b$ as a whole is being multiplied with the other stuff. We need to do this instead:

$$\begin{align}\frac{a+b}{2} c = \frac{(a+b)}{2} c = \frac{(a+b)c}{2}\end{align}$$Of course, you can skip the middle step if you like:

$$\begin{align}\frac{a+b}{2} c = \frac{(a+b)c}{2}\end{align}$$The same thing happens with $a-b$ instead of $a+b$.

Also, when doing the multiplication-division canceling, you can only cancel
*multipliers*; that is, things that we're multiplying everything by:

The problem is that $2+b$ is not $2 \cdot \text{something}$, so $2$ is not a
multiplier here and it can't be cancelled. However, if we have $2b$ instead of
$b$, we can use $(a+b)c=ac+bc$ backwards with $c=2$ and *then* cancel:

If you want to simplify $\frac{2+b}{2}$, recall that $\frac x y = x\frac 1 y$:

$$\begin{align}\frac{\blue{2+b}}{2} = \blue{(2+b)}\frac{1}{2} = \blue2\cdot\frac 1 2 + \blue b\cdot\frac 1 2 = 1 + \frac b 2\end{align}$$Or use one of these handy rules that do the same thing...

$$\begin{align}\frac{\blue{a+b}}{c} &= \frac{\blue a}c + \frac{\blue b}c \\ \frac{\blue{a-b}}{c} &= \frac{\blue a}c - \frac{\blue b}c\end{align}$$...like this:

$$\begin{align}\frac{\blue{2+b}}{2} = \frac{\blue2}2 + \frac{\blue b}2 = 1 + \frac b 2\end{align}$$Note that this does *not* work if $a + b$ is on the bottom: $\frac{c}{a+b}$
doesn't simplify nicely in any way because it's $c\frac{1}{a+b}$, not
$(a+b)\frac c 1$ or something.

Here's another thing that does not work:

$$\begin{align}\frac a b + \frac c d = \frac{a+b}{c+d} \qquad \red{\text{wrong!}}\end{align}$$You can combine fractions with $+$ and $-$ between them only if the bottoms are the same. This is using the $\frac{a+b}{c} = \frac a c + \frac b c$ rule right to left. If you want the bottoms to be the same, you may need to do "canceling backwards":

$$\begin{align}\frac a b + \frac c d = \frac{a\blue d}{b\blue d} + \frac{c\green b}{d\green b}\end{align}$$Now both bottoms are $bd$, so they combine nicely.

$$\begin{align}\frac a b + \frac c d = \frac{a\blue d + c\green b}{bd}\end{align}$$Putting a minus in front of stuff means multiplying by $-1$, so $-x$ is $(-1)x$.
Keep this in mind when you have something like $-(a+b)$; that's *not* $-a+b$
because you need to use the $(a+b)c=ac+bc$ thing with $c=-1$ here:

Of course, there's no need to write down these steps every time; just remember that if you have a $-($, you need to add a $-$ in front of each thing between $($ and $)$.

## Negative Times Negative¶

You have probably been taught that if you multiply a negative number with another negative number, the result will always be a positive number. As usual, I don't want you to believe it just because someone says so, but because there is a proof. To keep things simple, we'll first prove that $(-1)(-1) = 1$, and then using that result, we'll show that it works for any negative numbers.

Let $x = (-1)(-1)$. This is an *equation*; we have two things with an *equal*
sign between them. With an equation, we can divide both sides by anything
nonzero (more details below), so let's do that:

The right side cancels:

$$\begin{align}\frac{x}{(-1)} &= \frac{(-1)\color{blue}{(-1)}}{\color{blue}{(-1)}} = (-1) \\ \frac{x}{-1} &= -1\end{align}$$We can also add and substract things on both sides. Let's add $1$...

$$\begin{align}\frac{x}{-1} + 1 &= -1 + 1 \\ \frac{x}{-1} + 1 &= 0\end{align}$$...and substract the $\frac{x}{-1}$ thing:

$$\begin{align}\color{blue}{\frac{x}{-1}} + 1 - \color{blue}{\frac{x}{-1}} &= 0 - \frac{x}{-1} \\ 1 &= -\frac{x}{-1}\end{align}$$Let's rewrite the prefix $-$ as multiplying by $-1$:

$$\begin{align}1 &= (-1)\frac{x}{-1} = \frac{(-1)x}{(-1)}\end{align}$$Things cancel again, and we get $1=x$.

Now, let $x,y \in \mathbb{R}$ be positive. Then $-x$ and $-y$ are negative:

$$\begin{align}\overbrace{(-x)}^{\text{negative}}\underbrace{(-y)}_{\text{negative}} = ((-1)x)((-1)y) = \overbrace{(-1)(-1)}^1 xy = \underbrace{xy}_{\text{positive}}\end{align}$$The $x$ and $y$ can be any positive reals, so $-x$ and $-y$ can be any negative reals. It works!

Earlier I said that $\color{green}{a} \cdot \frac b c$ simplifies to $\frac{\color{green}{a}b}{c}$, not $\frac{b}{\color{green}{a}c}$. However, if $a=-1$, both are correct:

$$\begin{align}\frac{(-1)b}{c} = \frac{b}{(-1)c} = \frac{-b}{c} = \frac{b}{-c}\end{align}$$Here is a quick proof that $\frac{(-1)b}{c} = \frac{b}{(-1)c}$. Other parts of the above math are more straight-forward. This is one of the cases when "canceling backwards" is useful:

$$\begin{align}\frac{(-1)b}{c} = \frac{\color{blue}{(-1)}(-1)b}{\color{blue}{(-1)}c} = \frac{1b}{(-1)c} = \frac{b}{(-1)c}\end{align}$$## Using Formulas to Make More Formulas¶

Let $a,b,c,x,y \in \mathbb{R}$. Now we know that $ac+bc=(a+b)c$. This is known
as **factoring**. We end up with a bunch of things multiplied together, so the
thing is *factorized*. The *factors* are the things that are multiplied
together; in our case, $(a+b)$ and $c$.

Undoing factoring is known as **expanding**; $(a+b)c=ac+bc$ is a formula that we
can use to expand simple things. But how do we expand something like $(a+b)^2$?
That means $(a+b)(a+b)$, and mathematicians like to use the $x^2$ notation
because copy/pasting and repeating things is bad.

Anyway, maybe there is a nice formula for expanding $(a+b)^2$? You can google it or look it up from a book, but don't believe it blindly; instead, you should also find a proof of the formula and read it. Or better yet, you can figure out how to do the thing yourself. Let's do that.

We know that $(a+b)\blue c = a\blue c+b\blue c$ and now we want to do $(a+b)\blue{(a+b)}$. The $\blue c$ can be any number, and $a$ and $b$ are also numbers, so there's nothing wrong with letting $c=a+b$. Let's see what we get from that:

$$\begin{align}(a+b)\blue{(a+b)} = a\blue{(a+b)} + b\blue{(a+b)}\end{align}$$What can we do now? We use the $(a+b)c$ thing again! For $a(a+b)$, we need to let $c=a$, and for $b(a+b)$ we do $c=b$:

$$\begin{align}& a\blue{(a+b)} + b\blue{(a+b)} \\ =& a\blue a + a\blue b + b\blue a + b\blue b\end{align}$$Next, notice that $ab = ba$, so we have $ab+ab$ in the middle. That's $2ab$. With the $x^2$ notation, $aa=a^2$ and $bb=b^2$.

$$\begin{align}(a+b)^2 = a^2 + 2ab + b^2\end{align}$$Tada! It's done now, and this is what you'll find if you google "expand (a+b)^2". Let's make sure that it actually seems to work, just to see if we did any fatal mistakes along the way:

```
>>> (3+4)**2 # (a+b)²
49
>>> 3**2 + 2*3*4 + 4**2 # a² + 2ab + b²
49
```

Figuring this out took quite a while because we had never done anything like this before, but if you want to present your work to someone else, you should usually write it as a concise proof instead:

$$\begin{align}(a+b)^2 &= (a+b)(a+b) \\ &= a(a+b) + b(a+b) \\ &= a^2+ab+ba+b^2 \\ &= a^2+2ab+b^2\end{align}$$Now that we have gotten started with this, we can do more. For example, we know that $a-b=a+(-b)$, so using $(a+b)c=ac+bc$ we get this:

$$\begin{align}(a-b)c &= (a+(-b))c \\ &= ac+(-b)c \\ &= ac-bc\end{align}$$That's all we need for doing a very similar thing for $(a-b)^2$.

$$\begin{align}(a-b)^2 &= (a-b)(a-b) \\ &= a(a-b)-b(a-b) \\ &= a(a-b)+(-b)(a-b) \\ &= a(a-b)+(-b)a-(-b)b \\ &= a^2-ab-ba-(-b)b\end{align}$$Hmm, what can we do to $-(-b)b$? We have two minuses, and we can bring them both to front, so we have $-(-(b^2))$. If you think about this on the number line, every $-$ means flipping to the other side of the number line:

So, $-(b^2)$ is $b^2$ flipped to the other side. If we flip again, we end up right back to where we started:

$$\begin{align}-(-(b^2)) &= +b^2 \\ (a-b)^2 &= a^2-2ab+b^2\end{align}$$Here's another way that's easier to remember: $-x = (-1)x$, so $-(-(b^2)) = (-1)(-1)b^2$, and $(-1)(-1)=1$ because we proved it (see above).

There's another, easier way to prove the $(a-b)^2$ formula, which is using the $(a+b)^2$ formula we already proved with $x-y=x+(-y)$:

$$\begin{align}(a-b)^2 = (a+(-b))^2 = a^2+2a(-b)+(-b)^2\end{align}$$Let's do a similar thing to the $(-b)^2$ as we did last time:

$$\begin{align}(-b)^2=(-b)(-b)=(-1)b(-1)b=(-1)(-1)b^2=1b^2=b^2\end{align}$$If you don't want to do this mess ever again, you can memorize that $(-b)^2=b^2$ for all $b \in \mathbb{R}$, which is what we just proved. We are almost done:

$$\begin{align}(a-b)^2=a^2-2ab+b^2\end{align}$$This happens all the time in math: we have proved a thing or we have an axiom for a thing, and then we use it to prove another thing, and then using that other thing we prove even more stuff, and so on.

$$\begin{align}(a+b)c &= ac+bc \\ \Downarrow \\ (a+b)^2 =&\ a^2+2ab+b^2 \\ \Downarrow \\ (a-b)^2 =&\ a^2-2ab+b^2\end{align}$$There's one more formula I want to show you. I promise, this will be the last one. So far we have looked at $\green{(a+b)(a+b)}$ and $\blue{(a-b)(a-b)}$, but what if we mix them, like $\green{(a+b)}\blue{(a-b)}$?

$$\begin{align}(a+b)(a-b) &= a(a-b) + b(a-b) \\ &= aa-ab+ba+b(-b)\end{align}$$The $-ab$ and $+ba$ cancel because $+ba=+ab$, and $b(-b)=-b^2$:

$$\begin{align}(a+b)(a-b) = a^2 - b^2\end{align}$$Here is an example of using this formula.

Let $x,y \in \mathbb{R}$. If $x^2 = y^2$, what can we say about $x$ and $y$?

We know that this is true at least if $x=y$ because $x^2=x^2$, or if $x=-y$
because then $x^2=(-y)^2=y^2$. But are there other $x$ and $y$ values that
*also* happen to make this work? We can check it like this: the difference of
two numbers is zero **only if** the two numbers are equal.

This is true **only if** $\green{(x+y)}=0$ or $\blue{(x-y)}=0$;
the **only** ways how multiplying two things can give $0$ are that one of the
things is zero, or they are both zero. So we have two cases:

We are done. There are no other solutions because of all the **only** words I
have bolded; there are more details about this below.

It is kind of surprising that we started with expanding $(a+b)(a-b)$, and we ended up using our result to solve $x^2=y^2$, which doesn't seem to have anything to do with expanding. Things like this often happen in math, and they make math challenging but interesting.

## Or Note¶

If you read the $x^2=y^2$ thing carefully, you notice that there may be some
$x$ and $y$ that have **both** $\green{x+y}=0$ and $\blue{x-y}=0$ at
the same time. It seems like I didn't handle that case at all; I just did
"$\green{x+y}=0$ or $\blue{x-y}=0$". The thing is, I actually
handled that case. Let me explain how.

Let $A$ and $B$ be Booleans. Then "$A$ or $B$" means:

- true, if $A$ is true and $B$ is false
- true, if $A$ is false and $B$ is true
- true, if $A$ is true and $B$ is true (this is the weird thing)
- false, if $A$ is false and $B$ is false.

This is why I didn't need to have a third "$\green{x+y}=0$ and $\blue{x-y}=0$" case above; just "or" was enough. If you haven't seen this "or" thing before, it's kind of surprising, but almost all programming languages do it too:

```
>>> True or True
True
```

This is useful for writing code like this:

```
if ((x is wrong) or (y is wrong)) {
something is wrong
} else {
everything is fine
}
```

If **both** `x`

and `y`

are wrong, something is wrong and we must not
pretend that everything is fine; that's why `True or True == True`

makes
sense. In general, if you don't think about this "or" thing at all, you'll
probably get it right, and that's why `True or True == True`

.

## If and Only If¶

Let $A$ and $B$ be Booleans. Saying "if $A$, then $B$" means that if $A$ is true, then $B$ is also true. That's it. It doesn't say anything about what happens if $A$ is false. Also, if $B$ is true, $A$ can be still false. Here is a list of the possible cases:

"$B$ is true if $A$ is true"

"if $A$, then $B$"

"$A \implies B$"

$A$ | $B$ |
---|---|

true | true |

false | true or false |

For example, if $1 = 2$, then $1 \cdot 0 = 2 \cdot 0$; that is, $0 = 0$. In
other words, we have "if $1=2$, then $0=0$", but we **don't** have "if $0=0$,
then $1=2$". Math is not broken.

Here's a similar table for "if and only if", sometimes written "iff" with double "f":

"$B$ is true if and only if $A$ is true"

"if $A$, and only if $A$, then $B$"

"$A \iff B$"

$A$ | $B$ |
---|---|

true | true |

false | false |

Now $B$ can't be true unless $A$ is also true. "If and only if" guarantees that $A$ and $B$ are either both true, or both false. This is written like $A \iff B$ because you can prove $A \iff B$ by proving that $A \Rightarrow B$ and $A \Leftarrow B$ (exercise for you: why does that work).

**tl;dr:** "$A \iff B$" or "$A$ if and only if $B$" or "$A$ iff $B$" means that
$A$ and $B$ have the same Boolean value, and you can prove that by proving
$A \implies B$ and $B \implies A$.

## Solving Equations¶

Earlier we proved that there is a real number $x$ so that $2x = 6$ by showing that 3 is that number. In this case, it's easy to find the 3 by trying out different values. But what if we have something more complicated?

$$\begin{align}3x^2+2x+9 = 4x^2+4+5\end{align}$$So, we're trying to find the $x$ values that this equation happens to work for. The basic idea is that if we do something to both sides of an equation, it still works. For example, $1 + 1 = 2$, so $(1+1)^2 = 2^2$.

However, there are things that give you "fake solutions" if you do them to both sides. For example, $x=x+1$ is quite obviously false with any $x$ because the left side is always smaller than the right side, but if you multiply both sides by zero, you get this:

$$\begin{align}0x &= 0(x+1) \\ 0 &= 0\end{align}$$Now it seems to be always true, even though it was supposed to be always false. We got all real numbers as "fake solutions" that seem to work fine, even though they don't actually work at all. This is an if and only if problem, because we showed that if $x=x+1$, then $0=0$. It's perfectly correct, but it's useless because it tells nothing about $x=x+1$.

Usually when dealing with equations, we need "if and only if" things. If we
apply an "if and only if" thing to an equation, we don't introduce any "fake
solutions"; the equation we had before applying the "if and only if" thing is
true *exactly* for the same $x$ values as the equation after the applying.
Obviously, if we apply multiple "if and only if" things, we don't introduce any
"fake solutions" either.

Let $a,x,y \in \mathbb{R}$. Here is a list of "if and only if" things that are commonly done to both sides of an $x=y$ equation:

- Adding a number to both sides: $x=y \iff x+a=y+a$.
- Substracting a number from both sides: $x=y \iff x-a=y-a$.
- Multiplying both sides by something nonzero: if $a \ne 0$, then $x=y \iff xa=ya$ (if $a=0$, then $xa=ya$ is always true).
- Dividing both sides by something nonzero: if $a \ne 0$, then $x=y \iff \frac x a = \frac y a$ (if $a=0$, $\frac x a$ and $\frac y a$ are undefined).
- $xy=0 \iff x=0 \text{ or } y=0$.
- If $x > 0$ and $y > 0$, then $x=y \iff x^2=y^2$.

Most things on this list are quite straight-forward, but maybe the last thing isn't, so let's prove that. Earlier we proved that $x^2=y^2 \iff (x=y\text{ or }x=-y)$, but we have also proved that $x=-y$ never happens if $x$ and $y$ are positive. This means that if $x > 0$ and $y > 0$, then $x^2=y^2 \iff x=y$, which is what we wanted.

## Exercise

Prove that the last thing also works for $x \ge 0$ and $y \ge 0$, not only for $x > 0$ and $y > 0$.

You can also "move" stuff to the other side of an equation. For example, if you have $x+2 = 3$, you can "move" the $2$ to the right side, if you add a $-$ in front of it, so you get $x = 3-2$. What you're actually doing is substracting 2 on both sides, but the moving may be easier to do if you need to do these things a lot. A similar thing works with multiplication and division: $\frac x 2 = 3$ can be solved by "moving" the 2 to the other side, and it gives $x = 3 \cdot 2 = 6$.

## Yet another exercise

Do the $(-1)(-1)=1$ proof by "moving" things to the other side of the equation as much as possible.

Now we know enough stuff to solve our horribly complicated equation. Let's do it.

$$\begin{align}3x^2+2x+9 &= 4x^2+4+5\end{align}$$Simplify $4+5$ to $9$.

$$\begin{align}3x^2+2x+9 &= 4x^2+9\end{align}$$Substract $9$ on both sides to cancel the $+9$.

$$\begin{align}3x^2+2x+9-9 &= 4x^2+9-9 \\ 3x^2+2x &= 4x^2\end{align}$$Substract $3x^2$ on both sides.

$$\begin{align}3x^2+2x-3x^2 &= 4x^2-3x^2 \\ 2x &= x^2\end{align}$$Divide both sides by $x$.

$$\begin{align}2=x\end{align}$$Of course, there's no need to write down that much text between the lines. In fact, you need no text, and much less steps. I would solve the equation like this:

$$\begin{align}3x^2+2x+9 &= 4x^2+\overbrace{4+5}^9 \\ 3x^2+2x &= 4x^2 \\ 2x &= x^2 \\ 2 &= x\end{align}$$Look carefully, we didn't 100% follow the rules. We can divide both sides by something, but only if that something is not zero; in the last step we divided by $x$, and it only works if $x \ne 0$, so we need to handle that case specially. Like this:

If $x=0$, then $2x = 2 \cdot 0 = 0=0^2=x^2$, so $0$ is also a solution.

Answer: $x=0$ or $x=2$