Hello everyone. In this video, we're going to show
an example of an injective and surjective function.
So let S be the set of real numbers without the point minus 1,
and let T be the set of real numbers without the point 2, okay?
So sometimes you might see this written with the set difference notation
with the minus. Usually you'll see it as the
slash notation, kind of read this is the real numbers without negative 1.
That's sort of - that's how you can think about it.
So S is this set and T is this set,
show that the function f from S to T defined by
2x plus 1 over x plus 1, I guess that's f of x equals…
[that’ll] make it a little bit clearer,
is injective and surjective, hence a bijection.
So the definition of bijective or bijection is
a function that's injective and surjective, for us, okay?
Solution so…
how do we...
how do we solve this? Well we need to show that it's injective and surjective,
so there's two parts of this. So let's show that the function is injective first.
Well what does that mean? So a function is injective
provided that if I have two points that give me the same output,
then the original two points must be equal.
So let's pick two points in our domain, so let
x and y be in S
be such that f of x equals f of y,
and then...
and then what do we have?
So let’s…
so if f of x equals f of y…
we have the following.
By the way as a little sanity check, notice that S is defined
without the point minus 1 because this function doesn't make sense at minus 1.
So something to keep in mind.
Okay so let x and y be elements of S such that f of x equals f of y,
then 2x plus 1 over x plus 1 equals 2y plus 1 over y plus 1.
That's literally just plugging in x and y into the function, right?
Now from here, you're going to cross multiply and simplify,
and then hopefully you get x equals y.
So let's do that and see what we come up with.
So 2x plus 1, y plus 1
equals 2y plus 1, x plus 1.
Now we expand…
oops.
Let me wait for it to render properly.
So here we go with the cross multiplication, so we're going to expand and simplify now.
So we’re going to get 2xy
plus 2x plus y
plus 1
[equals] 2xy plus 2y plus x plus 1.
And then now, once we simplify this,
let's see so 2xy, 2xy those cancel.
The x comes over here, that's becomes an x, the y comes over here that's just a y, and
the two 1’s will cancel, so that's exactly
x equals y.
Okay.
So that solves this.
That solves this part.
Once it loads.
But now we need to show surjectivity.
So how do we show surjectivity? Well we need to show that every point
inside the range, so every point inside T here,
is reachable. So there's some point in S that will,
when I apply the function to it, give me the y inside the T.
So how do I find this x that's going to give me the output of y? Well
this is where something called a napkin computation comes into play.
I call these napkin computations, I should call these…
why do I call them napkin computations?
Because they’re something you should write on a napkin and then
throw out when you're done.
Now obviously it's going to be very difficult for me to use a napkin
on a screencast, but I will
do the following. I will sort of write it below, okay,
but because of that I’m going to be very precise,
right, because it's supposed to be on a napkin. This isn't something that you want somebody to see.
So what I'm going to do…
how do I do this? Now I'm going to take…
I'm going to take the function and set it equal to y,
and I'm going to start solving
for x, okay?
Some of you might recognize this as finding the inverse function and that's definitely
what we're doing, but I'm going to word it a little differently
and I'm going to think of it a little differently, okay? So 2x plus 1 over x plus 1 equals y.
Let's go through and let's try to isolate for x.
So now I’m going to get 2x plus 1 if I cross multiply.
xy plus y.
I'm going to isolate over here so 2x minus xy,
y minus 1.
So let’s see where we’re at now with a little bit of isolation.
So 2x plus 1 equals xy plus y.
y minus 1 is equal to 2x minus xy.
Factor out the x, divide by 2 minus y,
what's that going to give us?
y minus 1,
2 minus y.
So there we have it, so this
this x value should be the one that we use…sorry I'm off center…
this x value should be the one we use in order to
get the output y, okay, so if I plug this x
into f, into our function,
I'm going to get the answer y, okay? So that's what I'm going to use.
Okay great, so how do I do this? Well
okay, so now when I write up this proof, I'm going to start
with this value of x, and I'm going to show that we get to the answer.
Let x equal
this value.
Now there's one more thing that we should mention though, right?
So we've written down this value, but we need to make sure that
x is actually in our domain.
So I'm going to hide the napkin computation because I don't need it anymore.
So let x be this value, right? Now notice that
this value is real, right, because y is not 2.
Remember y was taken to be inside T,
T doesn't have the point 2, so x is a real number.
So x is a real number,
and furthermore notice that x is not equal to minus 1.
Okay?
Why is that the case? Why is x not equal to minus 1? Well
we prove this by contradiction,
so suppose it was equal to minus 1, then what would we have?
So let's just look at what I typed. Okay so,
this is going to be a proof by contradiction. So further notice that x does not equal minus 1,
for otherwise, minus 1 is equal to y minus 1 over 2 minus y.
Cross multiplying y minus 2 equals y minus 1, and that
clearly is going to be a problem, right, so cancel the y’s, get minus 2 equals minus 1,
you get minus 1 equals 0, you can get…
you get whatever you want. So I mean let's even do it easier, right,
I’m pretty sure we can all agree that minus 2 is not equal [to] minus 1.
So that's clearly a contradiction, right?
So notice that this was a proof by contradiction, but worded a little subtlety,
so keep that in mind. Remember the “for otherwise” part is
sort of suggesting that we're using a proof by contradiction, right,
and that's what's going on here. So for otherwise I put minus 1 in there,
I've solved and I show that 2 equals minus 1. Okay.
So thus…
so thus x is actually in the domain, and…
and now let's look. So f of x...
f of x is the given function,
so let's do that, plug that in,
and then let's start plugging in our value.
So we know what x is, right? x is equal to this thing
So let's plug this in.
So here we go, f of x is equal to this, let's plug in our value for x,
and remember this should simplify to y so hopefully we do everything correctly, let's do this.
Okay, so what's the numerator going to be? So we're going to bring the 2 inside, so it's going to be
2y minus 2 plus 2 minus y
divided by 2 minus y.
So we're going to find a common denominator as well.
So y minus 1 plus 2 minus y,
2 minus y.
Let's see what we have now.
Okay?
2y minus y that's y, minus 2 plus 2 that's 0, so that's just going to be y,
the 2 minus y here will cancel with the 2 minus y in the denominator’s denominator,
and then y and y will cancel here.
That should clearly be a 1, that's a typo.
So I get y minus 1,
plus 2 minus y, so that gives me 0,
2 minus 1 is 1…
or again 2 minus 1 so I get y over 1 so then that's equal to y.
Okay, and that shows surjectivity.
So we found a point in S that maps to y.
Now notice that the x depended on y, which it almost always should.
And here it does and that's fine.
Notice that, again, our napkin computation well we used it
to find out what the value of x should be, we never mentioned it in our solution, which is fine.
You can also write this back and forth if you want, but I'm not going to get into that in this video,
and again, here we've shown injectivity and surjectivity.
So we started with the injectivity part and then we showed surjectivity,
and that shows that this function is a bijection.
For those of you interested in learning more about these types of examples,
I would check out Mobius Transformations.
This is some subset of them, and
they're pretty interesting. They have a lot of applications,
which I'm not going to get into in this video, but
I will leave it to you to check this out.
Okay so hopefully you got a little sense of how
bijectivity works
and hopefully this video helps, so thank you very much for listening.