there is a function g which goes from R

to D such that if f(a) = b,

then g(b) = a.

Now the difficulty with this theorem is not that the proof is terribly deep,

but I have to see what it is that has to be proved.

We can, of course,

always write that as a function g from R to D but we have to define it.

Now the definition is,

we take an element B and we look up by this equation, what's its origin,

originally is, under f. That's a and a will then

be the image of g. Now the question is whether this is a good definition.

Well, it wouldn't be a good definition if we

wanted to associate a different element for c also to b.

But that cannot happen because if we have b,

we go to this equation and we know,

that's just since the function f is bijective,

that there's exactly one solution to it.

So for every b we've got exactly one a that fits.

And that a we've put in there has

the image of b on a g. And in this way we define a good function.

And this function is called --- g is called

the inverse of f because if f goes from left to right,

then g goes in the inverse direction,

from right to left.

There are, of course,

a lot of inverse functions.

For instance, if f is the function which adds two to number,

then the inverse of f is the function which subtracts two from a number.

If f is a function that multiplies the number by two,

then the inverse is the function which divides the number two.

If f is the function which squares a number,

then f inverse is the number which takes the square root.

But, wait a second. Here we have to pay

attention because we know

that with the square root there's a problem,

the number x is not allowed to be negative.

So here we have to take care of the definition of f. Now what is the problem?

If we are looking to the graph of f,

which we've just drawn, it looks like that.

We see that it is actually not,

if we define it from R to R,

it is not an injective function because if we take any y,

then f(x) is y has two solutions.

If y is bigger than zero and it has

no solution if y is smaller than zero.

So it's neither injective nor surjective.

You can see that graphically.

If y is bigger than zero,

then we get two solutions,

one there, one there.

If y is smaller than zero,

we've got no solutions at all.

So what should we do here?

Now the way to remedy is always change the domain and the range.

So what we do is, we said, well,

f will be from R+ to R+,

and R+ is here the set for real numbers such that x is larger than zero.

For, in that case,

my domain is only here.

That's now my new domain.

This is when you arrange.

And if I'm looking now at my graph I see that for every value of y,

I've got exactly one solution.

So this is this function up here, is not.

I'm sorry, this function is not bijective.

But the function f from R+ to R+ such that f(x)

is x square is bijective and has as inverse square root.

So the way, that the thing to remember about this example is,

if you are in a situation where you're not injective or not surjective,

you often can remedy the situation by restricting domain and range to

a part of the function which is then bijective and where you can define an inverse.