Hey guys, in this video I'm going to do another real analysis proof. This one says to prove that
h of X is injective and we're given what h of X is here. This h of X is composed of g of X and
also f of X and we're given in the problem that f of X and g of X are injective functions. It's not
too hard to believe that a function that's made up of injective parts is injective but we just have
to prove that formally. Before I go into the proof part, let's go over again what injectivity is and
how you prove it. The definition of injective is that, suppose you have some numbers X1,
X2 belonging to the reals, then you have a function, say f. If you have f of X1 is equal
to f of X2 and this therefore implies that X1 is equal to X2, then that is an injective function.
If this holds true--if you plug in some real number X1 and X2 that are not necessarily the
same but they have to be the same when they're in the function and they're equal to each other,
then that's what an injective function is. We're going to use this definition to prove that h of X
is injective. Let's go ahead and start the proof. We're just going to start with what we just went
over, which is, suppose we have some X1 and some X2 that belong to the reals and let's say that h
of X1 is equal to h of X2. We're going to plug in X1 and X2 into this equation. We have g of
2 f (X1) minus one-third plus f(1) is equal to g 2 f (X2) minus one-third plus f(1). This is
an equation so we can simplify it. We have f(1) on both sides. We can take away that. Now we
have g of 2 f (X1) minus 1/3 is equal to g of 2 f (X2) minus 1/3. Now this is the first line where
we're going to use one of our assumptions. We know that g is an injective function. That means that g
of X1 is equal to g of X2 has to imply that X1 is equal to X2. We can assume because g is injective
that the insides are going to be equal to each other. We can say, by assumption g is injective,
thus we have the equals set to each other. We have 2 f (X1) minus 1/3 is equal to 2 f (X2) minus 1/3.
And again we can simplify by adding 1/3 to both sides. We have 2 f (X1) is equal to 2 f (X2) and
we can divide by 2, so we have f (X1) is equal to f (X2). And again we know that f is an injective
function, so we're going to do the same thing. We can take the insides and assume that those are
equal because f is an injective function. I'm just going to say the same thing, by assumption, thus
we have X1 is equal to X2. Starting from suppose X1, X2 belongs to R and h of X1 equals h of X2.
From that line, we have that X1 is equal to X2. We have proven that h of X has to be injective. For
the last line I'm going to write, because h of X1 is equal to h of X2 implies that X1 is equal to X2
for any X1, X2 in R, by definition h of X must be injective. And we're done with this proof.
*chiptune music*