WEBVTT
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Welcome. This is a lesson about proving using induction. So I'll give you several examples
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of how to prove using induction. And I'll sort of tell you the formula or the way that
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you can prove using induction. It's actually a pretty useful thing to do. I'll show you
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in what cases it can be especially useful. Okay. So proving using induction is probably
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the simplest or the most basic form of proof there is. And it's the one that most people
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learn before any other form of proving. So the base case is the first thing you want
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to do. You want to prove if you will when you use induction. You want to show that the
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assumption is true for a very basic trivial case. This is usually where the variable equals
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0 or 1. And sometimes you want to do 2 or 3 just to prove it sort of to yourself but
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also prove that for the very basic for the very first few cases it does work. So you just
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show that base case first. The next thing to do is to pick an arbitrary number which
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we'll call k. You're not actually picking an integer. You're going to say that there's
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some integer k that's picked arbitrarily. And you're going to assume that the conclusion
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is true for this value. You're going to assume what you want to prove is true for k.
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Now you're going to replace n or k or whatever variable you have in your equation and your
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formula. You're going to replace that with k plus 1. And you're going to rewrite the
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left side and the right side of the equation with k plus 1 instead of n or instead of whatever
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variable you had. So you're going to replace both the left and the right side with k plus
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1 which is the integer directly following the one that we assumed it worked with. Then
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we're going to figure out what our goal is which means you're going to simplify the
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right side so you know what you're going to try to get to. You want to know as soon
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as I get the left side to equal whatever this thing is, the right side is, then I'm
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done. Then that showed that the left side equals the right side. And then like I just
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said, you're going to work on the left side trying to get it equal or less than or whatever
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it says. So if it said prove that 2k is less than k squared, then you would say you would
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have to prove that the left side is less than the right side. But sometimes it says prove
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the left side is equal to the right side. But either way, you're trying to work on the
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left side to get it to be equal to or less than or whatever the situation calls for than
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the right side. And the whole trick of the entire proof is to use the induction assumption.
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This is where you use the fact that you have assumed the fact that you have assumed the
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result to be true for the n equals k case. So there is a reason that we assumed that the
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n equals or the the k case is true because we're going to use that assumption that we know
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that's true to prove that the k plus 1 case must be true. And that's all there is to
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it. Once you've proved that, then that shows that if some arbitrary case is true, then
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that means the next case must be true. And if that's true, then you could prove that
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all the way down the line of the integers, then every single case must be true because
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if a certain case is true and you know that case is true, and then that implies that
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the next case is true, then every case must be true.
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All right. Here's a classic example. This is perhaps the most famous proof of them all.
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Prove that 1 plus 2 plus 3 plus all the way up until n, prove that that sum equals
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n times n plus 1 divided by 2. So let me give you an idea of what this is actually saying.
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This means that 1 plus 2 plus 3 plus, let's go until 100. N in this case would be equal
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100 because it's that last number all the way up until 100. That means that this equals
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100 times 100 plus 1 divided by 2. Interesting. Very interesting. Okay. So let's go about
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trying to prove this. First thing to do is the base case. I use several base cases
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just to make sure this is correct. So I use n equals 1, n equals 2, and n equals 3.
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For n equals 1, that's just 1 on the left side, counting up to 1, and then that should
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equal 1 times 1 plus 1 divided by 2. I just plugged 1 in in this formula, and I get 1
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equals 2 divided by 2. That's true. Okay. That's good. For n equals 2, 1 plus 2 equals 2 times
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2 plus 1 all over 2. So that means that 3 equals 6 divided by 2. Yep, that's good. And
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then for n equals 3, 1 plus 2 plus 3 equals 3 times 3 plus 1 divided by 2, 6 equals 12
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to 5 divided by 2. Yep, we're good. Okay. And you can try the 100th example if you'd like
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to, but I assure you that is also true. Okay. So now let's pick an arbitrary integer
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k, and we have to say k is less than or equal to n, and I'll assume that 1 plus 2 plus 3
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plus up to k is equal to k times k plus 1 divided by 2. Okay. So all I'm doing right now
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is assuming that's true for this kth case. That must be true. Okay. Now we'll take the
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k plus 1 case. So if we do 1 plus 2 plus 3 all the way up to k plus 1, that should equal
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k plus 1 times k plus 1 plus 1 divided by 2. So again, all I did was I substitute this
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k plus 1, and wherever I saw n up here, knows 3 spots, I put a k plus 1. All right. So let's
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clean up this right side. So this right side is actually going to be k plus 1 times k plus
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2 divided by 2. That's what we want. That's our goal for the left side. We want that left
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side to equal that now. So it does 1 plus 2 plus 3 all the way up to k plus 1 does that
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equal k plus 1 times k plus 2 divided by 2. Well, let's see. Here's the big, here's
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the big step here. I'm going to break this up into 1 plus 2 plus 3 all the way up to
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k plus k plus 1. So I'm just going to include that term before the k plus 1, which would
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be k. Well, here's the idea. We know that 1 plus 2 plus 3 all the way up to k. We know
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that is k times k plus 1 over 2. We already knew that because that's what we assumed to
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be true. This is our induction assumption that's right up here. We assumed that was true
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so we can use the fact that we know that that must be true or we assume it to be true
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anyway. So we'll use that. So I'll replace that whole part right here with k times k
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plus 1 divided by 2. All right. And then we still have this k plus 1 hanging around.
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Okay, so we have to add these two things together. I'll find a common denominator. That's
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2. Okay. Then you can do this a couple different ways. You could multiply both of them out
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combined and then factor. But I think it's a little easier if we just notice that they
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both have a k plus 1 in common. Okay. So we have a k times a k plus 1 and we have a 2 times
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a k plus 1. That means we have a k plus 2 times k plus 1. Okay. And it's all over 2 still.
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It's whole time we've had a denominator of 2. So, well, that's what we had. This k plus
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1 times k plus 2 divided by 2. That equals k plus 1 times k plus 2 divided by 2. So we're
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all set. We got what we wanted. And therefore, because we proved the theorem was true for
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all k up to n, or sorry, for all numbers up to k, which is less than or equal to n. And
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we proved that if it's true for k, then it must be true for k plus 1. Then that means
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it must be true for all integers bigger than or equal to the smallest base k's found
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to be true. In this case, it was 1. So, for all integers 1 and bigger, this theorem
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or this formula must be true. All right. And so it's proved. We did it. All right. Let's
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try another one. Prove that 2 to the n is less than n factorial for n is bigger than
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or equal to 4. So this is a little different of a k's than the last one we just saw.
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This time, we want to prove something less than something else. And we're not starting
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at 1. We're starting at 4. Okay. So now our base k's is we have to start at 4, at least,
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at least start at 4. So we'll do the 4, the 4 case. So I just put 4 in where I see
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n. So 2 to the 4th is less than 4 factorial is that true? Well, 2 to the 4th is 16. And
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4 factorial is 24. So 16 is less than 24. Yep. We're good. I just do 2 to the 5th, or
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I'm sorry, the n equals 5 case just for good measure. So 2 to the 5th less than 5 factorial.
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Yeah. 32 is less than 120. And you can see the gap is getting bigger and bigger. So
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you can tell that factorials are growing much more rapidly, which makes sense because
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all we're doing on the left side is multiplying by 2 each time. And what we're doing on the
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right side is multiplying by an ever-growing number, 5 and then 6 and then 7 on the right
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side. So of course, you know this is true, but let's prove it using induction. So we'll
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pick an arbitrary integer k. So k is less than or equal to n. And we're going to assume
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that 2 to the k is less than k factorial. Again, we're just going to assume this for the
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k case. So therefore, for n equals k plus 1, we'll put k plus 1 in where we see n. And
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what do we want? We want k plus this to be less than k plus 1 factorial. And I broke that
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down a little bit. I said that that was the same thing as k factorial times k plus 1.
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This remember factorial just means multiplying all the way down from whatever number is being
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factorial. So k plus 1 factorial is just k plus 1 times k times k minus 1 times k minus
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2. So another way I could write that is k plus 1 times k factorial. That is the same
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thing. So I'm going to break up this left side into 2 times 2 to the k. Now you should
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be able to see why this is an advantageous for me because we know something about 2 to
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the k. What do we know about that? We know 2 to the k is less than k factorial because
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we assume that 2 to the k is less than k factorial right up here. So we can use that again. And
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we have to use that. That's the whole idea of induction. So we'll assume 2 to the k
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is less than k factorial which means this thing which was the left side. Actually this is
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what we started off with. As I said it's the same thing as that. And that is less than
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this thing. If k is greater than or equal to 4 again it can only be for the constraints
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that we put on it. We said it was only for k bigger than or equal to 4. So this
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is only true for k bigger than or equal to 4. So if that's less than 2 times k factorial
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would that be true or would that be less than what we want? Would that be less than k
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plus 1 times k factorial which again is what we want? Well yes it is. As long as k is
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bigger than or equal to 2 because this is the exact same thing. And what's being multiplied
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it here? You're multiplying by a 2. Here you're multiplying by a k plus 1. Well if k is
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bigger than or equal to 2 then k plus 1 is going to be bigger than 2. So yeah that's
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true. So this thing we started off with is less than the thing we wanted it to be less
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than. Only fk is bigger than or equal to 4 though. And this is kind of a big deal right
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here. This was only if k is bigger than or equal to 2 which means if our original thing
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had been only if k were bigger than or equal to 1 we could no longer say that. It would
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have to be bigger than or equal to 2 because of this step. But because the first thing
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was bigger than or equal to 4 and this is bigger than or equal to 2 it's okay that's
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fine as long as this is at least or at most bigger than or equal to 4. And that's fine.
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Alright and so therefore again because we proved it for k and we proved that if k is
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true then k plus 1 is true then it must be true for everything bigger than or equal
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to 4 the first case that we saw. Okay here's another one that's kind of interesting.
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It's a little different than again the things we've seen before. We want to prove that
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5 to the n minus 1 is divisible by 4 if n is bigger than or equal to 1. So what does
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it mean for 5 to the n minus 1 to be divisible by 4? Well it means 4 can go into it right
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and even number of times 4 multiplies into the number. Okay so it's 5 to the first minus
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1 divisible by 4 well 5 to the first minus 1 is 4 and yes 4 is divisible by 4 so that's
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good. How about 5 to the second minus 1? 5 to the second is 25 25 minus 1 is 24. Yep 24
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is divisible by 4 so that works also. So our true base case is work. Okay we'll pick
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an arbitrary k and assume 5 to the k minus 1 is divisible by 4 and then for n equals
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k minus 1 we would put k or sorry for n equals k plus 1 we'll put k plus 1 in where n was
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and what do we want we want divisible by 4 so that's all we're looking for so it's
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kind of a different goal that we're looking for we're just looking for divisible by 4.
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So I'll break this up again into 5 times 5 to the k minus 1 because again we know something
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about 5 to the k minus 1. However we don't quite have 5 to the k minus 1 yet we have 5
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times 5 to the k minus 1. However if I break up 5 times 5 to the k and to 4 times 5 to
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the k plus 5 to the k so these are the same thing. I just broke it up into 4 5 to the
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k's and 1 5 to the k just broke it up and why would I do something like that or where
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would you have the insight to do something like that? Well it's just because we wanted
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5 to the k minus 1 we wanted this right here and so just do anything you can to get it
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anything you can to get this thing because we know that that's divisible by 4 so if
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we can get that we can use our induction assumption so now we have it so we know that
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this thing is divisible by 4 we already knew that. So again being divisible by 4 just means
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it's 4 times something. So here all I did was I replaced this thing we know is divisible
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by 4 with 4 times something we don't really care what it's 4 times it could be 4 times
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anything but as long as it's 4 times something that's all we care about. So we have 4 times
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5 to the k plus 4 times something well we can factor out that 4 they both have a 4
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in common so it's 4 times 5 to the k plus something. Okay well we again we don't really
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care what this is it's something whatever it is so it's 4 times some stuff it's 4 times
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something is in there but it's 4 times that so what does that mean? Well that means it's
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divisible by 4 which is exactly what we wanted. Again the definition of being divisible
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by 4 means it's 4 times something and this is divisible by 4 so we have what we want.
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And therefore again because we proved it was true for k and if it's true for k that's
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true for k plus 1 then it must be true for all k bigger than or equal to 1. Okay and
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I think there's one more example here proving that for all positive integers k k cubed
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plus 2k is a multiple of 3. Okay the base cases the first case 1 cubed plus 2 times 1 is
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a multiple of 3 1 plus 2 is a multiple of 3 so 3 is a multiple of 3 that's true. Okay
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if we do for k equals 2 2 cubed plus 2 times 2 is a plus 4 or 12 and 12 is a multiple
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of 3 so our two base cases are true. Okay again we'll pick an arbitrary number k and we'll
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assume that k cubed plus 2k is a multiple of 3. Okay recall what it means for something
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to be a multiple of 3 it means that it equals 3 times something again. So what we want
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again is a multiple of 3. Okay so we'll put k plus 1 in where we see n. I guess that
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here I should have said for all integers n cubed plus 2n is a multiple of 3. Okay so
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for n equals k plus 1 we'll put k plus 1 in where we see n so k plus 1 cubed plus 2 times
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k plus 1 we want that to be a multiple of 3 that's what we want. Let's see if we can
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get it. Okay so think about if you can figure out what we would want to get to use our
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induction assumption. Okay what might we want? Well I mean if you remember from the
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last ones we want this somewhere on the left side. We want this somewhere over here.
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We want k cubed plus 2k. Well let's see if we can get it. Well I think we're going to
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have to multiply this out. So let's let's multiply it all out. So k plus 1 cubed is k cubed
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plus 3k squared plus 3k plus 1 and if we distribute the 2 we get 2k plus 2k plus 2. Okay
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now we're going to strategically arrange our terms and again like I said we want this
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k cubed plus 2k somewhere. Well here's a k cubed, here's a 2k. Let's put those together
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and then whatever we want to do with the rest of the stuff. Well we can combine the 1 and
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2 to make 3. Then we have a 3k squared, we have a 3k and then as I said we grouped k
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cubed plus 2k over on the left side. So we got what we wanted. This thing, this k cubed
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plus 2k is a multiple of 3 because we assumed it would be. Alright so we have 3 times something
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plus the rest of the stuff that we had, the 3k squared plus 3k plus 3. Okay you might
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see where this is going. We can factor out a 3 out of this stuff on the right. So it's
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3 times k squared plus k plus 1. So it's 3 times something plus 3 times some other stuff.
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So it's 3 times something plus that other stuff. And again it's just 3 times something.
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It doesn't matter. So that means it's divisible by 3, by the definition of being divisible.
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So because we proved it was true for all k, up all numbers up to k and we proved it was
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if it's true for k that's true for k plus 1 and must be true for all integers. Alright
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that's the last example I had. And that's the way most induction problems go. You just
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follow those steps. You might want to write these down or commit them to memory. You just
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do a base case. You pick an arbitrary k. Assume it's true. Replace the variable in the
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problem with k plus 1. Work through it with k plus 1. First figuring out what your goal
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is, what you want to obtain. Work on the left side trying to get it to be true or get
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it to be equal or get to be less than whatever. And then the trick is the induction assumption.
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Somehow get your induction assumption to show up in the problem so you can use it and
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then prove it. That's all there is to it. Alright thank you for watching. Talk to you
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soon.