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Hello everyone. So in this example we're
going to talk about an induction problem,

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and this happened to 
be one example that

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I skipped from class to 
leave as an exercise, but

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I wanted to go over it in a video.

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And if you've seen it already, then at least you're 
going to get a little bit of extra practice with

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the argument. So

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let f n be the Fibonacci sequence,

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so the first two terms are 
1 and every subsequent

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term is the sum of the previous two, 
that's what this says mathematically.

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now we're supposed to
show that f n is bounded by

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7/4 to the power of n 
for all natural numbers.

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So we're trying to show that each Fibonacci 
number is bounded by 7 over 4, okay?

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The tip-off here to sort of use
induction here is that we're

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trying to show some sentence is 
true for all natural numbers n,

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and the sequence that we 
have is defined recursively.

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So both of these things combined sort of 
suggest let's use Mathematical Induction.

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Now the question is do I use 
Strong Mathematical Induction

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or just the Principle of 
Mathematical Induction?

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And that you sort of - well you can try to
figure out before you write the proof,

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but I like to figure it out as I 
write the proof and if I'm wrong 

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I just go back and modify 
everything to use the other one. 

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You can always use Strong 
Induction if you want

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but, it might save you time to use just 
the Principle of Mathematical Induction.

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They are equivalent of course,

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but that is a non-trivial exercise.

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Okay, so back to the question at hand. So if we're 
going to prove this by induction, we need three things:

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we need a base case, an induction
hypothesis, and then the inductive step. So 

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the induction step…so

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So let's start with the base case.

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So for n equals 1, we have that

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f at 1 equals 1, which is less than

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7 over 4.

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41
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Okay so for the base case, we have 
for n equals 1 we have that f at 1

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equals [1] so let's edit that…

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f at 1 equals 1 which is 
less than 7 over 4. Okay

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so base case is easy.

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Induction hypothesis,

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so assume that

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that f of k

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is less than frac

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7 over 4 to the power of k.

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So our base case is done

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and the induction hypothesis we're 
going to assume it's true for some k.

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Now we're going to do
the induction step.

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So for k plus 1,

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we have…

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f at k plus 1…

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Well...

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okay this is what we'd 
like to write. So let's

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start with what we have. So 
for k plus 1 we have what?

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So for k plus 1 we have f at k plus 1 
we want to write that this is equal to f at

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k plus f at k minus 1.

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Oh I guess maybe I should go back here,
so assume that f k this and this for some -

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I should be a little more precise 
in my induction hypothesis

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So for what k? Well k has to be some natural number,
okay, so the induction hypothesis, let's review that.

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Assume that f k is less than 7 over 4 to 
the k for some k in the natural numbers.

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Now for the induction step, we have…

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we want to write this. We want to write 
f k plus 1 is equal to f k plus f k minus 1,

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but this is valid only when

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k

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plus 1 is at least 3.

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So this is only valid when k is at least 2.

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So this is only valid when k is at least 2,

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then we know we're going to need at least
one more base case because I mean

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here we only have that
k is 1, so k could be 1

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and this is f at 2 which is f at 1 plus f at 0 
and that doesn't really make sense, okay?

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So we're going to need to do at least one 
more base case, so let's add that in now.

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So for n equals 2, watching all my typos,

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we have that f 2 is also equal to 1 and 
that's clearly less than 49 over 16 and that's 

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7/4 squared.

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Okay so now we have
our two base cases, okay?

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So now in our induction hypothesis, 
well we already know that

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1 and 2 are true, so we can
actually additionally suppose here

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that k is greater than or equal to 2. 

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So now this step here, f k plus 1 is

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equal to f k plus f k minus 1, 
well this is valid since now…

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valid since k plus 1 is
greater than or equal to 3.

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Okay, so now we have this, 

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but now, as you've noticed, now
we're going to have to try to use

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just the fact that f k is bounded by 7/4, 

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and that's not enough. We also need 
that f k minus 1 is bounded by 7 over 4

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if we're going to try to make 
any progress with this question.

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So what do we do now? Well

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now we know that we need to use Strong 
Induction, so we're going to go back to the top

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and we're going to fix this up
by using Strong Induction.

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So we proceed by Strong Induction.

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Now we know which one we're using, 
so let's add that into our solution.

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Now our induction
hypothesis is wrong,

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so we need to…

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so we need to account 
for the induction hypothesis.

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So now we're going to 
assume that instead of

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f k is less than 7
over 4 to the k, 

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we're going to assume that f i

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is less than or equal 
to 7 over 4 to the i.

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Now we're going to put for all 1 less 
than or equal to i less than or equal to k

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for some k in the natural numbers 
and k greater than or equal to 2. 

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So now we're assuming this 
holds for all i between 1 and k,

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okay, and we've already 
proven two base cases so

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so there's no loss of generality in 
assuming that k is at least 2 here.

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Okay, now let's go back to the induction step, 

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which iI will magically make reappear.

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Now from here, what are we going to do? 
Well now we can use the induction hypothesis.

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So the induction hypothesis 
says that this is less than…

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So now for k plus 1, we have this 
thing, right? Now remember our goal,

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what's our goal? So maybe I'll 
write down the goal, right?

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So now we want to show 
that f k plus 1 is less than

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7 over 4 to the power k plus 1, okay?

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So now that we're at this step, right,

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usually the tricks in induction are 
either, you know, add 0, add 1...

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or sorry add 0, multiply by 1,

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or factor/expand, okay? In this case, factoring 
seems like the most natural thing to do

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since something is in common. So let's 
try to factor and see where that leads us.

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k minus 1. So the smaller of the two
terms is this 7/4 to the k minus 1.

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170
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And what are we left with? 
We’re going to be left with…

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So I factor out this 7/4 to the k minus 1,

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and that’s just 7/4 plus 1,

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and what's that equal to?

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So 7/4 plus 1 that’s 
11/4 plus the first thing.

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And so the second thing is

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11 over 4.

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Now we want to show that 
f k plus 1 is less than

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7/4 to the k plus 1.

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Here we only have a 7/4 
to the k minus 1, so the

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other 2 better come
from this 11 quarters.

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Now...

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if this is going to work, then this 
should be less than 7/4 squared,

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which we've already seen is 49 over 16.

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So maybe now the question 
to ask yourself is 11…

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over- so let's pull up the old paintbrush here.

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So now the question that 
we're going to ask is, “Is

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11 over

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200
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4 less than

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49

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over 16?"

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Question mark.

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And how do we do this? Well if 
we cross multiply, right, we see

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that this is going to 
be true if and only if…

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cross multiply we get 11 times 
16 over 4, we’ll leave the 4 there,

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so 16 the 4 will cancel. 

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So this is going to be true…

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you know let's just write that down.

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So 11 times 16

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over 4

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is less than 49.

217
00:09:19,433 --> 00:09:21,333


218
00:09:21,333 --> 00:09:24,566
And this is true if and only if…

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so if we cancel out a 4 from the 
16, then we’re going to get 44

220
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is less than 49.

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00:09:33,300 --> 00:09:36,933


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So these were all “if and only if’s” so you
can go in either direction of this proof,

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and so we actually have the 
thing that we want to be true.

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Now you don't need to show this, 
so I'm not going to save this

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beautiful masterpiece, but

227
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let's now just write down 
the result that we now want…

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which is that this is…oops… 

230
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strictly less than now

231
00:10:00,100 --> 00:10:04,300


232
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49 over 16.

233
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234
00:10:07,500 --> 00:10:09,800
That's equal to…

235
00:10:09,800 --> 00:10:10,500


236
00:10:10,500 --> 00:10:13,633
now it's just a matter of just crunching 
through, so let’s see the crunch.

237
00:10:13,633 --> 00:10:14,733


238
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Okay so

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00:10:16,500 --> 00:10:19,566
11 over 4 we just saw 
was less than 49 over 16,

240
00:10:19,566 --> 00:10:23,266
that's 7 squared over 4 squared, so I can 
multiply the tops and bottoms and get

241
00:10:23,266 --> 00:10:27,033
7 over 4 to the k plus 1, that's 
exactly what we wanted to show.

242
00:10:27,033 --> 00:10:28,933


243
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So thus,

244
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245
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thus, this holds

246
00:10:36,766 --> 00:10:38,433


247
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and hence the

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249
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the statement

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251
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is true for all n [in the natural numbers]

253
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254
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by Strong Induction.

255
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I probably should capitalize theorems.

256
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257
00:11:00,166 --> 00:11:03,966
Okay so let's back up and just
sort of summarize what we did.

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259
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So our statement that we 
were trying to show here.

260
00:11:08,900 --> 00:11:10,900


261
00:11:10,900 --> 00:11:14,766
The statement we were trying to show is that the 
nth Fibonacci numbers less than 7 quarters to the n. 

262
00:11:14,766 --> 00:11:18,500
We're going to proceed by Strong Induction, 
which we saw by trying to do the proof.

263
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We need two base cases, again 
this was an artifact of the proof.

264
00:11:22,766 --> 00:11:26,333
You could also kind of justify it by the fact 
that we had to go back two previous terms,

265
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so you might want to use two base cases.

266
00:11:28,733 --> 00:11:31,600
Base cases are very easy
to show, just plug in

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1 into both sides of the inequality

268
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and show that it's correct,

269
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or just start - well so really you 
should start from the left-hand side

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and show that it's less 
than the right-hand side.

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Or vice versa, but in this case

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I want to go in this direction. 

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So for n equals 2, we have f 2 equals 1,

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it's less than 49 under 16, 
that's 7 quarters squared.

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Now assume this is true for all…

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so now our induction hypothesis, we're 
going to assume the claim is true for all i

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between 1 and k for some natural number k,

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and we're going to make
sure that k is also at least 2

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since we have two base cases 
that's not a concern for us.

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And the induction step we now want to show that 
f k plus 1 is less than 7 quarters to the k plus 1

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So for k plus 1, we're
just going to plug it in.

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We're going to use the recursive definition 
which is valid since k plus 1 is at least 3.

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Then we use the induction hypothesis,

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or induction hypotheses,

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to show that each of these 
terms is less than 7 quarters 

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to the k or k minus 1 respectively.

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Factor, simplify,

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11 quarters happens to 
be less than 49 over 16,

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we saw that in the old
paintbrush doodle.

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49 over 16 is 7 squared 4
squared, and the reason why we

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picked 49 over over 16 was so 
that we can get the extra two

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values of 7 quarters that we needed.

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00:12:46,900 --> 00:12:49,800
Thus the induction step is true 
and hence the statement is true

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for all natural numbers n by Strong Induction.

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And that's what we wanted to show.

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So hopefully this gives you 
a little bit of [an] idea of 

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when to Strong or Weak
Induction, when to use…

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how many base cases to use,

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and how to go through the 
proof of such an example.

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I apologize this video is very long, but

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induction examples do take a little bit 
of time to explain and get used to

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in the beginning. After a little bit of time

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hopefully this kind of example 
will just be very routine for you,

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but it will take a little bit of practice 
and effort to get to that point.

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So thank you very much for
listening and good luck.