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A sequence of numbers is a list of numbers, whether infinite or finite. The individual numbers in a sequence are called terms. If you can wrap your head around a sequence, that may be referred to as "coming to terms with infinity." Or maybe not.
The terms in sequences are separated by commas. Finite sequences have a defined last term. If the sequence is infinite, "..." is written at the end to show that the sequence continues forever. Or that the sequence is lost in thought and trailing off.
For example, 1, 2, 3, 4, 5,... is a sequence. The 1st term is 1, the second term is 2, and so on. This guy's infinite.
And 2, 3, 4 is also a sequence. No little dots means this baby is three and done, but it's still a list of numbers, so it's still a sequence.
Even 0, 0, 0,... is a sequence where every term is 0. And what a scintillating sequence it is.
Because a sequence is a pattern, you can usually find any term of the sequence as long as you're given enough terms to figure out where it's going. If our sequence is all the even natural numbers, for example, we can figure out that the 350th term of the sequence would be 700, since 2 × 350 = 700. Beats writing out the first 350 terms by hand. Unless you're in prison and looking for activities to pass the time.
Sequences also come in handy when we're writing algebraic proofs about different types of numbers. We'll get into that over the next few pages.
Here are some more fun sequences.
Like we said before, a sequence is just a list of numbers, and it can help us majorly prove some things about rational and irrational numbers. For starters, though, let's run through a quick proof showing why an irrational number is irrational.
Our good buddy is happy to step in. How can we prove that he's irrational? Besides pointing to his complicated conspiracy theories and fifty cats, we mean.
Let's start with an overview of the proof, and then we'll get into the guts of it. Remember to sterilize your hands first.
Overview of the proof:
This is a proof by contradiction. And yet, this is not a proof by contradiction. See what we did there?
We'll start out with the assumption that is rational. This means the statement (*) below has to be true. We'll then show that our assumption also implies the statement (*) has to be false. Since no statement can be both true and false, we must have made a bad assumption to start out with. This means can't be rational after all. Ready for us to prove ourselves wrong? Here we go...
Guts of the proof:
If were rational, we could write as a fraction . Since any fraction has an equivalent fraction in lowest terms, we can assume is in lowest terms (i.e. it can't be reduced). Someone must have already come by and simplified it for us. We'll have to send them a card.
(*) and there is no whole number that divides both a and b.
; therefore, by multiplying both sides by b, we get:
Let's show that a has to be even. Trust us on this one.
First we square both sides of the equation:
Okay, check this out: b is an integer, right? It would have to be, since we assumed at the beginning that is a rational number, which means a and b were both integers. So b2 is also an integer, 'cause we're not gonna get a fraction or decimal when we square an integer. So since , we know a2 is 2 times some integer b2. That means a2 is an even number; 2 times any integer is an even number. This means a must also be even (think about it—whenever you square an even number, the result is always even). So a = 2 × n for some number n.
Now that that's settled, let's show that b has to be even. We do have a point, and we are getting to it. Bear with us.
Substitute 2 × n for a in the equation :
Again, square both sides:
Divide both sides by 2:
This means b must be even, for the same reasons that a had to be even. Come on, b, that's not cool—come up with your own reasons.
But if both a and b are even, then 2 would divide both a and b, which means the fraction isn't in lowest terms. Uh-oh. This is a contradiction: you're saying there's no whole number that divides both a and b, but 2 divides both a and b? Our fraction is reducible and not reducible? It's okay if this gives you a bit of a headache; it's hard to imagine things that are impossible! Like the idea of anyone enjoying figuring this stuff out for a living.
Since our initial assumption led to a contradiction, our initial assumption must have been false. This means can't be rational.
This is pretty much the opposite of a real-life scenario. Generally, when two people both have even tempers, they can be quite rational.
We've gabbed about rational numbers before, but here's a new spin on an old favorite: do the rational numbers form a sequence? In other words, can we write them in a countable list?
We can't exactly "count" all the rational numbers, since there are infinitely many of them. However, we can put enough of them in a list so that we can project what they'd be. Keep this list and your shopping list separate; it would be a major nuisance if you got home from grocery shopping only to find your canvas bags full of fractions.
Let's start by putting the positive rational numbers into a table:
Now list them, going along the diagonals:
And so on. Eventually, every positive rational number will show up on this list. For example, will show up when we get to row 67, column 1234 of the table. (We used some pretty high-level mathematics to figure that one out.) If we continue forever, we'll have listed all the positive rational numbers. Microsoft Excel goes down quite a ways, but even this program has its limitations.
To list all the rational numbers, start with 0 to make sure it gets on the list. Knowing him, he'll throw a complete hissy-fit if we forget about him. Then list all the positive rational numbers as before, but after each number, list its negative:
Eventually, every rational number will show up on this list.
Since we can put the rational numbers into a list, we say the rational numbers are countable. Yes, that means you, too, zero. Chill.
There's a question mark in the title of this section for a reason. There are so many real numbers, we can't put them all into a list. In fact, we can't even put the real numbers between 0 and 1 into a list. Thought 1 was a small number? Think again. It's a giant among fractions. Sometimes it even tromps through their village, kicking them aside as if they were ants.
In other words, we can't make a complete sequence out of all the real numbers. No matter how hard we try, we'll always be skipping a few terms in our sequence. But talk is cheap, so let's prove it.
Overview of the proof:
Once again, we're going to prove this by contradiction. We'll assume that you have a list of all the real numbers between 0 and 1 (even though we know you're bluffing). Then we'll show that this list can't contain all the real numbers between 0 and 1 after all. This means our assumption was false, so we can't list all the real numbers between 0 and 1. And then we'll take your so-called list and turn it into the authorities so that you can be exposed for the fraud that you are.
Sorry. Sometimes we get carried away with hypotheticals.
Guts of the proof:
Pretend that you could make a list of all the real numbers between 0 and 1. All real numbers can be represented by infinite decimals (a rational number that ends can be thought of as an infinite decimal by sticking infinitely many zeros on the end). Let's write each digit of each infinite decimal like this:
0.d11 d12 d13 ...
0.d21 d22 d23 ...
0.d31 d32 d33 ...
It looks kinda bizarre, but here's how it works: the first digit after the decimal point is written as d11 in the first number. So if the first term in our sequence were 0.164, we'd have d11 = 1, and d12 = 6, and d13 = 4.
This is supposed to be a list of all the real numbers. However, it can't be, because we can find at least one number that didn't make it onto the list. We'll construct, one digit at a time, a real number between 0 and 1 that isn't in this list. The fun bus is leaving the station. Buckle up.
For the first digit of our brand-new imagined decimal, choose any digit other than d11. For the second digit, choose any digit other than d22. For the third digit, choose any digit other than d33, and so on.
You can keep on keepin' on forever! We won't make you, but you can now imagine how it would be very possible to do so. The number that results is a decimal, so it's certainly a real number. It doesn't have any digits before the decimal place, so it's between 0 and 1.
However, it isn't on the list. This number can't be the same as the first number on the list, since it differs from the first number in the first decimal place. This number can't be the same as the second number in the list, since it differs from the second number in the second decimal place. This number can't be on the list at all, since it differs from every number in the list in at least one place!
This argument, known as "Cantor's Diagonal Argument," (if you're guessing this was named after some guy named Cantor, you'd be right) shows that there are too many real numbers to count, even if we only look at real numbers between 0 and 1. This should be a huge relief to abacus makers everywhere.
The reason we can't list the real numbers between 0 and 1 in a sequence is that there are too many irrational numbers. Nice going, guys. Way to ruin it for the team.
Think about it: we can list the rational numbers. If we could also list the irrational numbers, then we could list the real numbers by interweaving the two lists, taking one rational number, one irrational number, one rational number, etc. We might even be able to macramé ourselves an attractive lampshade.
Anyway, our lists would look like this:
rational numbers: r1, r2, r3,...
irrational numbers: s1, s2, s3,...
real numbers: r1, s1, r2, s2, ....
Since we can list the rational numbers but not the real numbers, we must not be able to list the irrational numbers either. What a buzzkill.