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Hello everyone. In this 
video, we're going to show

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an example of an injective 
and surjective function.

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So let S be the set of real numbers 
without the point minus 1, 

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and let T be the set of real numbers
without the point 2, okay?

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So sometimes you might see this 
written with the set difference notation

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with the minus. Usually 
you'll see it as the

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slash notation, kind of read this is 
the real numbers without negative 1.

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That's sort of - that's how 
you can think about it.

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So S is this set and T is this set,

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show that the function f 
from S to T defined by

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2x plus 1 over x plus 1, I 
guess that's f of x equals…

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[that’ll] make it a little bit clearer,

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is injective and surjective, hence a bijection.

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So the definition of bijective or bijection is

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a function that's injective 
and surjective, for us, okay?

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Solution so…

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how do we...

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how do we solve this? Well we need to 
show that it's injective and surjective,

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so there's two parts of this. So let's 
show that the function is injective first.

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Well what does that mean?
So a function is injective

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provided that if I have two points 
that give me the same output,

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then the original two points must be equal.

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So let's pick two points
in our domain, so let

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x and y be in S

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be such that f of x equals f of y, 

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and then...

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and then what do we have?

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So let’s…

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so if f of x equals f of y…

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we have the following.

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By the way as a little sanity 
check, notice that S is defined

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without the point minus 1 because this 
function doesn't make sense at minus 1.

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So something to keep in mind.

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Okay so let x and y be elements 
of S such that f of x equals f of y,

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then 2x plus 1 over x plus 1 
equals 2y plus 1 over y plus 1.

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That's literally just plugging in x
and y into the function, right?

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Now from here, you're going to 
cross multiply and simplify,

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and then hopefully you get x equals y.

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So let's do that and see 
what we come up with.

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So 2x plus 1, y plus 1

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equals 2y plus 1, x plus 1.

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Now we expand…

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oops.

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Let me wait for it to render properly.

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So here we go with the cross multiplication,
so we're going to expand and simplify now.

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So we’re going to get 2xy

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plus 2x plus y

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plus 1

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[equals] 2xy plus 2y plus x plus 1.

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And then now, once we simplify this,

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let's see so 2xy, 2xy those cancel.

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The x comes over here, that's becomes an x, 
the y comes over here that's just a y, and

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the two 1’s will cancel, so that's exactly

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x equals y.

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Okay.

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So that solves this.

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That solves this part.

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Once it loads.

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But now we need to show surjectivity.

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So how do we show surjectivity? Well 
we need to show that every point

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inside the range, so 
every point inside T here,

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is reachable. So there's 
some point in S that will, 

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when I apply the function to
it, give me the y inside the T.

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So how do I find this x that's going 
to give me the output of y? Well

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this is where something called a 
napkin computation comes into play.

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I call these napkin computations, 
I should call these…

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why do I call them napkin computations?

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Because they’re something you 
should write on a napkin and then

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throw out when you're done.

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Now obviously it's going to be 
very difficult for me to use a napkin

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on a screencast, but I will

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do the following. I will sort 
of write it below, okay, 

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but because of that I’m
going to be very precise,

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right, because it's supposed to be on a napkin. This 
isn't something that you want somebody to see.

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So what I'm going to do…

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how do I do this? Now I'm going to take…

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I'm going to take the 
function and set it equal to y,

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and I'm going to start solving

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for x, okay?

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Some of you might recognize this as finding
the inverse function and that's definitely 

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what we're doing, but I'm going 
to word it a little differently

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and I'm going to think of it a little differently, 
okay? So 2x plus 1 over x plus 1 equals y. 

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Let's go through and
let's try to isolate for x.

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So now I’m going to get 
2x plus 1 if I cross multiply.

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xy plus y.

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I'm going to isolate over 
here so 2x minus xy,

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y minus 1.

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So let’s see where we’re at
now with a little bit of isolation.

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So 2x plus 1 equals xy plus y.

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y minus 1 is equal to 2x minus xy.

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Factor out the x, divide by 2 minus y,

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what's that going to give us?

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y minus 1,

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2 minus y.

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So there we have it, so this

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this x value should be the one
that we use…sorry I'm off center…

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this x value should be the 
one we use in order to

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get the output y, okay, so if I plug this x

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into f, into our function,

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I'm going to get the answer y, okay? 
So that's what I'm going to use.

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Okay great, so how do I do this? Well

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okay, so now when I write up 
this proof, I'm going to start

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with this value of x, and I'm going 
to show that we get to the answer.

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Let x equal

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this value.

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Now there's one more thing that 
we should mention though, right?

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So we've written down this value, 
but we need to make sure that

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x is actually in our domain.

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So I'm going to hide the napkin computation 
because I don't need it anymore.

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So let x be this value, 
right? Now notice that

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this value is real, right, 
because y is not 2.

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Remember y was taken to be inside T,

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T doesn't have the point 2, 
so x is a real number.

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So x is a real number,

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and furthermore notice that 
x is not equal to minus 1.

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Okay?

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Why is that the case? Why is 
x not equal to minus 1? Well

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we prove this by contradiction,

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so suppose it was equal to minus 
1, then what would we have?

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So let's just look at
what I typed. Okay so,

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this is going to be a proof by contradiction. So 
further notice that x does not equal minus 1,

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for otherwise, minus 1 is equal 
to y minus 1 over 2 minus y.

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Cross multiplying y minus 2 
equals y minus 1, and that

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clearly is going to be a problem, right, so 
cancel the y’s, get minus 2 equals minus 1,

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you get minus 1 equals 0, you can get…

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you get whatever you want. So I 
mean let's even do it easier, right,

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I’m pretty sure we can all agree that 
minus 2 is not equal [to] minus 1.

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So that's clearly a contradiction, right?

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So notice that this was a proof by 
contradiction, but worded a little subtlety,

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so keep that in mind. Remember
the “for otherwise” part is

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sort of suggesting that we're using 
a proof by contradiction, right,

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and that's what's going on here. So 
for otherwise I put minus 1 in there, 

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I've solved and I show that 
2 equals minus 1. Okay.

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So thus…

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so thus x is actually in the domain, and…

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and now let's look. So f of x...

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f of x is the given function,

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so let's do that, plug that in,

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and then let's start plugging in our value.

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So we know what x is, 
right? x is equal to this thing

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So let's plug this in.

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So here we go, f of x is equal to
this, let's plug in our value for x,

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and remember this should simplify to y so
hopefully we do everything correctly, let's do this.

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Okay, so what's the numerator going to be? So 
we're going to bring the 2 inside, so it's going to be

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2y minus 2 plus 2 minus y

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divided by 2 minus y.

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So we're going to find a 
common denominator as well.

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So y minus 1 plus 2 minus y,

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2 minus y.

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Let's see what we have now.

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Okay?

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2y minus y that's y, minus 2 plus 2 
that's 0, so that's just going to be y,

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the 2 minus y here will cancel with the 2 
minus y in the denominator’s denominator,

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and then y and y will cancel here. 

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That should clearly be a 1, that's a typo.

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So I get y minus 1,

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plus 2 minus y, so that gives me 0,

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2 minus 1 is 1…

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or again 2 minus 1 so I get y 
over 1 so then that's equal to y.

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Okay, and that shows surjectivity.

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So we found a point in S that maps to y.

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Now notice that the x depended on 
y, which it almost always should.

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And here it does and that's fine.

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Notice that, again, our napkin 
computation well we used it

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to find out what the value of x should be, we 
never mentioned it in our solution, which is fine.

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You can also write this back and forth if you want, 
but I'm not going to get into that in this video, 

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and again, here we've shown 
injectivity and surjectivity.

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So we started with the injectivity 
part and then we showed surjectivity,

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and that shows that this 
function is a bijection. 

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For those of you interested in learning 
more about these types of examples,

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I would check out Mobius Transformations.

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This is some subset of them, and

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they're pretty interesting. They
have a lot of applications,

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which I'm not going to 
get into in this video, but

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I will leave it to you to check this out.

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Okay so hopefully you 
got a little sense of how

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bijectivity works

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and hopefully this video helps, so 
thank you very much for listening.