All right so in this video we're going to do another proof and this time it's asking us to
prove that a function f is a bijection. This is a really common type of problem that they give
you in a first-semester class of analysis and basically it's a little bit formulaic in that
once you learn how to do it, it's pretty much the same thing no matter what function they give you
and no matter what ranges and domains they give you. The three things that you have to prove to
be able to prove that a function is a bijection is that you have to prove first that a function is
well defined. They're going to give you a function and I've never seen it not be well defined but you
have to just prove it anyway to prove that it is actually a function. Then you have to do another
two things. There's theorem that says that if function is a bijection, it's surjective and
it's also injective. You have to prove that the function is well defined, surjective,
and injective and then you can conclude because it's surjective and injective that it therefore
must be a bijection. We're going to do that and it also doesn't matter which one you do first.
You want to prove that it's well defined first but then if you do injective first or surjective first
it doesn't matter. Let's go ahead and get started. To begin the proof like I said, we have to prove
that the function is well defined. To do that we just have to show that for the given domain of x,
which is in this case is 0 to 1, we have to show that f(x) maps to 2 to 4. We have to show that the
domain and ranges match up to what you expect. And so we're given that x belongs to 0 to 1,
so we're going to say if x belongs to 0 to 1, then we have--and this is just from the
definition of what that means--it goes from 0 to 1. Then what you're going to do is just manipulate
this inequality until it looks like 4 over x plus 1, and that just happens to be f(x), and then you
can prove that f(x) goes from 2 to 4 if you do this correctly. We're eventually going to take
the reciprocal to get 1 over x but first we want to add 1. We're going to add 1 to the inequality,
so that equals that. Then we are ready to take the reciprocal. So 1 over 1 is just 1 and when
you take the reciprocal, you have to flip the signs, so 1 over x plus 1 and then the reciprocal
of 2 is 1/2. Usually I will skip that step and go ahead and rewrite it so that it continues with
the s--with the signs on the same side. I would rewrite that as less than x plus 1, which is less
than 1. And then we're going to multiply by 4 to get the numerator of 4 over x plus 1. That's going
to be 4 times 1/2, which is equal to 2, which is less than or equal to 4 over x plus 1 less
than or equal to 4, and that not coincidentally is equal to f(x). We have therefore proved that f(x)
belongs to 2 to 4. The first part of the three parts that we have to prove is finished. You
can say, thus we have shown that the function is well defined from 0 to 1 to 2 to 4. And the next
thing we're going to do--the next thing I did was prove that it's injective. If you wanted to prove
next that it was surjective, again that's fine. I'm just going to say--I'm just going to say,
next we will show that the function is injective. When you want to prove that function is injective,
you basically follow these steps. You're basically going to show that if you have some variable--some
variables x and y and they're in the same domain, if you have that the functions are equal to each
other, therefore x and y are equal to each other, if you can prove that then you've proved that the
function is injective. We're going to say, suppose some variables x and y belong to the domain,
which in this case is 0 to 1 and f(x) equals f of y, then we have--so f(x) is equal to 4 over x plus
1 and f(y) would be 4 over y plus 1. Now what we want to do is simplify this and manipulate
this so that it can equal x equals 1. We're going to take the reciprocal. That's going to be x plus
1 over 4 equals y plus 1 over 4. We're going to multiply by 4. That equals x plus 1 equals
y plus 1. And we're just going to subtract 1 from both sides. We have x equals y. That was
pretty simple and we went ahead and proved that f, the function, is injective. We're going to say,
thus we have shown that the function is injective and I usually put from 0 to 1 to 2 to 4 because
we don't know technically if the function is well defined through the whole--all the real numbers,
so you want to say, okay we've only proved injectivity on that domain and range. Next we're
going to say, next we will prove that the function is surjective. All right to prove surjectivity I
have another graphic that describes it. I'll place that there. Basically what you're going to do is
you're going to again suppose you have a y and this time it's going to belong to the given range
and you're going to define a variable x naught and you want to show that if you plug in x of naught
into the function it's going to equal y. And therefore you're going to show that x of naught
belongs to the domain and f(x) equals y. Let's go ahead and do it. We're going to say, suppose a or
some y belongs to 2 to 4 and define x of naught. Here we want to do something on scrap paper,
which is not going to be in our proof, and we want to find that if we plug in something to here we're
going to get y. Just the variable y. What I mean by that is that we want to show that if we plug in
something 4 over x plus 1 equal to y. We want to transfer x to be equal to something. We're
going--let me just show you what I mean. We're going to take the reciprocal. That means 1 over
y. That's x plus 1 over 4 multiplied by 4 again, so we get 4 over y, and then subtract by 1. This
is going to be our x of naught and you don't want to include all this work in the actual proof. Just
do that. You're going to see later on that when you plug it in into the function it'll obviously
reverse what you just did and you'll get y. You're going to get, what was it? 4 over y minus 1. Okay
so we have that sentence that sets up the rest of the proof. We're going to say, then we have
by definition again that y is between 2 to 4 and what we're going to do is manipulate that back to
x of naught being equal to 0 to 1. It's similar to what we did before. To get to x of naught,
we're gonna do first the reciprocal. It's going to be 1/2, flip the signs, 1/4, and like I said
I usually skip that step and go ahead and rewrite it back to the signs pointing the same way. Then
we're going to just take--multiply by 4 actually. That's going to be 1 less than 4 over y and we're
going to subtract by 1. Not surprisingly, that gets us back to the domain. We have that this
is x of naught and we've proved that x of naught belongs to the domain 0 to 1. And we're not done
yet. The next thing we have to prove is that if we plug in f of x, this is equal to y eventually.
We're going to say that f of x equals to f over 4 over y minus 1, which is equal to...And we're
going to plug in x of naught into f(x). That's going to be 4 over y minus 1 plus 1. This should
be I guess x naught. Plus 1. That equals to 4 over 4 over y because the 1 and the -1 cancel out. That
equals to 4 times y over 4, which cancels out, which equals to y. That wasn't too difficult
either. We can say, thus we have proved that the function is surjective for any y from--that goes
from 2 to 4. A nd to finish off, we're just going to say by definition because we've proved its
well defined, injective, and surjective. Voila, we have proved bijection. We're going to say,
because a well defined function is bijective if it is--if it is injective and surjective,
by definition we have shown that the function is a bijection from 0 to 1 to 2 to 4 and we're done.