[MUSIC] So, I get a number like the 
square root of two. 
And square root of two is irrational. 
So, how am I going to figure out what 
it's approximately equal to? 
Well, it turns out that it's 
approximately, you know, 1.414 and a bit 
more. 
But how would ever figure that out? 
How would I know that that's 
approximately the value of the square 
root of two? 
Let's use the intermediate value theorem 
to try to pull it. 
I want to use the intermediate value 
theorem to approximate the square root of 
two. 
To do that, I'm going to use this 
function, 
f(x) = x^2 - 2. 
Notice something about this function. 
This function is equal to zero when x is 
the square root of 2. 
And how is that? 
f of the square root of 2 is the squared 
of two squared, which is two, minus two, 
which is zero. 
So, if I'm trying to approximate the 
square root of 2, what I'm going to think 
about this, is that I'm going to look for 
a positive value, that I can plug into 
this function to make it a square root of 
zero. 
How am I going to find such a value? 
You know, the other good thing about this 
function is that it's continuous. 
And then, I can plug in some values, like 
f(1). 
Now, what's f(1)? 
It's one squared minus two and that's 
minus one. 
And I can plot that on my graph right 
here, and it's this point right here. 
And take a look at, say, f(2). 
f(2) would be two squared minus two, 
which is two. 
And I can plot that on my graph, right 
here. 
It's 2, 
2. 
Now, f is a continuous function and at 
one, it's negative. 
And at two, it's positive. 
So, it's a continuous function. 
It's negative at one and it's positive at 
two. 
By the intermediate value theorem, there 
must be a point in between when the 
functions value is equal to zero. 
And you can see that on the graph. 
We're looking for that point. 
That is the square root of two where this 
graph crosses the x-axis, 
the positive x-axis. 
And that's the point that we're looking 
for, 
right? 
The intermediate value theorem promises 
me there's a value in between where the 
functions value equals zero. 
And I can see that there must be such a 
point on the graph. 
And now, I know that that value is 
between one and two. 
And I can do better. 
I can cut this interval between one and 
two in half and I get that f(1.5). 
Well, 1.5 squared is 2.25 - 2 is 0.25. 
[SOUND] Alright. 
So, that's, say, this point on my graph. 
Now, what do I know? 
I know that the squared of two, which is 
where this graph crosses the x-axis, is 
between one, where the function is 
negative and 1.5, where the function is 
positive. 
And I can do better, 
alright? 
I can pick some other point. 
Let's be brave and pick 1.4. 
And if I plug in 1.4, well, 14 squared is 
196. 
So, 1.4 squared is 1.96 - 2, 
that makes this -0.4, 
alright? 
This is barely below the x-axis. 
And I can plot that point here. 
And now, I know that the square root of 
two, the positive input to this function, 
which is equal to zero, 
well, it must be between 1.4 and 1.5. 
Because the function's value at 1.4 is 
negative and the function's value at 1.5 
is positive. 
So, this is not so bad. 
I mean, we, we could do this with a few 
calculations on the blackboard. 
Let's, let's bring out some sort of 
computation device so we can try to get 
an even better approximation to the 
square root of two. 
So here, I've got a function. 
The function is f(x) = x^2 - 2. 
And you can see that if I plug zero into 
the function, I get out -2 because 0^2 - 
2 is -2. 
Now, this is a physical function. 
I've got this knob here. And as I spin 
this knob, the function's input value 
changes. 
So, for example, if I spin it up to 
f(05.), f(1/2), that's negative 1.75 and 
that's right because if I take a half 
squared, I get a quarter, and a quarter 
minus two is -1.75. 
Now, I could spin the input up to one. 
f(1) is -1 = because 1^2 - 2 is -1 or 
f(1.5), well, 1.5 squared is 2.25, 2.25 - 
1 is 0.25. 
Now, this is positive or I could wheel it 
back down a little bit. 
And I see that when I get to 1.41, the 
function's output is negative again. 
Then, I could increase the function's 
input a bit. 
Oh, now the function's output is positive 
again, at f(1.145). 
And then, I could decrease the value a 
little bit. 
Oh, 
now, the function's output value is 
negative again and that would increase 
the output value a little bit. 
Oh, 
now, the function's output is positive 
and I could decrease the input a bit. 
Now, the function's output is negative, 
could increase the input a little bit. 
Now, it's positive, could decrease the 
input a little bit. 
Now, it's negative again. 
Now, Look at what happened, 
right? 
What's going on? 
I'm looking at the function f(x) = x^2 - 
2 and I'm walking back and forth across 
the point where this function is equal to 
zero. 
And as I walk to the left and to the 
right, the function is positive and the 
function is negative. 
And I'm getting closer and closer to a, 
to the actual place where the function is 
equal to zero and what is 1.41421356? 
It's awfully close to the square root of 
two. 
And if you think the square root of two 
and square it and subtract two, you get 
zero. 
What we just saw with the knob that I was 
turning is actually an incredibly general 
technique for figuring out where a 
continuous function crosses the x-axis. 
And we can do it very efficiently. 
Let's take a look at how we might do this 
a little bit more carefully. 
Here, I've drawn a graph with some random 
looking continuous function and I've 
labeled some points along the x-axis. 
Imagine that we couldn't see this whole 
graph. 
Just imagine that the only thing you can 
do is interrogate this function. 
You can ask this function at some input 
point, are you positive or negative? 
We're trying to figure out where, 
approximately, this continuous function 
crosses the x-axis. 
We're looking for a zero of this 
continuous function. 
How might the game begin? 
I'll let it begin by asking the function, 
are you positive or negative at zero? 
And the function is positive at zero. 
And then, maybe I'd ask at sixteen. 
At sixteen, is this function positive or 
negative? 
And I'd see the function is negative. 
The function's value is negative at 
sixteen. 
Because it's a continuous function and 
the function's value at zero is positive 
and the function's value at sixteen is 
negative, I know that there's a zero in 
between. 
There's some point in between that if I 
plug it into the function, I get zero 
out. 
Now, I could cut that interval in half. 
I could plug in eight and I could ask the 
function at f(8), are you positive or 
negative? 
And the function f(8), the value of the 
function at eight is positive. 
Now, look. 
I know the function's value at eight is 
positive, the function's value at sixteen 
is negative. 
So, by the Intermediate Value Theorem, 
there must be a value in between, right 
there, where the function's value is 
zero. 
And now, I could take that interval and 
cut it in half again. 
I could take a look at twelve, which is 
exactly between eight and sixteen. 
And I could ask the function, at 
f(12),12), are you positive or negative? 
And the function's value at twelve is 
negative. 
[SOUND] So now, I know by the 
intermediate Value Theorem, the function 
zero zero of this function must land 
between eight and twelve and I could cut 
that interval in half again. 
I could look at ten, and I could ask the 
function, 
well, if I plug in ten, are you positive 
or negative? 
And the function's value with ten is 
negative. 
And now, I know that the zero must land 
between eight and ten, because the 
function's value at eight is positive and 
the function's value at ten is negative, 
so there must be a zero in between, a 
point where the function's value is equal 
to zero. 
Now, I could cut that interval between 
eight and ten in half again. 
I could look at nine. And I could ask the 
function, at nine, are you positive or 
negative? And if I plug in nine, it looks 
like the function is positive. 
And that means that a value that I can 
plug into the function to make the output 
zero lies between nine and ten. 
And this is really quite remarkable. 
I, I, I could cut the interval between 
nine and ten and half again and see if it 
was between nine and 9.5 or 9.5 and ten. 
And I could keep doing this every time 
I'm getting a little bit more precision 
as to the exact value of the input that 
would make the function zero. 
The other really neat thing about this is 
that I could have just tried all of the 
values between zero and sixteen. 
I could have plugged all seventeen of 
those numbers into the function to figure 
this out. But by doing this clever 
cutting in half method, this bisection 
method, I managed to only ask six 
questions, 
right? 
I asked the function's value at zero and 
sixteen, at eight and twelve, at ten and 
at nine, and by only asking six 
questions, I was able to narrow down a 
region where a zero resided. 
Now, that's pretty quite efficient, much 
better than asking seventeen questions. 
So, this is a very general technique for 
figuring out where zeroes of continuous 
functions lie. 
And I encourage you to pick your favorite 
function and see if you can approximate a 
root, a zero of that function. 
[MUSIC]