Proof function is bijective check please

1. Sep 12, 2011

tylerc1991

1. The problem statement, all variables and given/known data

Define

$\mathbb{A} = \{ g_{ab} : (a, b) \in \mathbb{R}^2, a \neq 0 \}.$

Define the function $G: \mathbb{R}^2 \to \mathbb{A}$ by the rule

$G\big((a, b)\big) = g_{ab}, \quad (a, b) \in \mathbb{R}^2, a \neq 0.$

Determine whether $G$ is bijective.

2. Relevant equations

$g_{ab} = ax + b$.

3. The attempt at a solution

To prove injectivity, we must show that $G\big((a, b)\big) = G\big((c, d)\big) \implies (a, b) = (c, d)$.

Suppose $G\big((a, b)\big) = G\big((c, d)\big)$ for some $(a, b), (c, d) \in \mathbb{R}^2$.

By the definition of $G\big((a, b)\big)$,

$g_{ab} = G\big((a, b)\big) = G\big((c, d)\big) = g_{cd}.$

By the definition of $g_{ab}$,

$ax + b = g_{ab} = g_{cd} = cx + d.$

This means that $a = c$ and $b = d$, or $(a, b) = (c, d)$.

Therefore, $G$ is injective.

To prove surjectivity, we must show that range$(G) = \mathbb{A}$.

By definition, range$(G) \subseteq \mathbb{A}$.

Choose an element $f_{ab}$ in $\mathbb{A}$.

Then $f_{ab} = ax + b$ such that $a \neq 0$.

Clearly there exists an element $g_{ab}$ in range$(G)$ such that $g_{ab} = ax + b = f_{ab}$, where $a \neq 0$.

Hence, $f_{ab} \in \text{range}(G)$. This means that $\mathbb{A} \subseteq \text{range}(G)$, and hence that range$(G) = \mathbb{A}$.

Therefore, $G$ is surjective.

Because $G$ is injective and surjective, $G$ is bijective by definition.

2. Sep 12, 2011

micromass

All is ok, but I would liike you to explain this a little further:

Why does ax+b=cx+d imply a=c and b=d?? You have an equality of functions, how do you use that to prove the equality of the constants?

3. Sep 12, 2011

tylerc1991

Suppose $ax + b = cx + d$ and $a \neq c, b \neq d$.

Then $ax + b - cx - d = (a - c)x + (b - d) = 0$ (*) for all $x \in \mathbb{R}$.

But $b \neq d \implies b - d \neq 0$ for all $x \in \mathbb{R}$.

This is a contradiction. Hence $b = d$, and it follows from this fact and equation (*) that $a = c$.

4. Sep 12, 2011

micromass

Yes, but can't ax+b=0 for all x without a and b being zero?? Why not??

Try to choose specific values for x to prove this.

(the reason I'm persistent in this is that I was once asked this on an oral exam, I didn't answer it as well as my professor liked )

5. Sep 12, 2011

tylerc1991

Hmm, well if we go with the assumptions that I had in my last post, and say that $x = 0$, then we reach the contradiction I had in the last post.

So suppose $x \neq 0$, then manipulating the function I had written above (sorry im lazy about this)

$x = \frac{-(b - d)}{(a - c)}$. But $x$ can be any real number, so there exists a value of $x$ that does not equal the fraction.

It is a question I took for granted during calculus (when doing partial fractions for instance), we just said well the coefficients must be equal. So I am actually sort of glad that the question went this direction

6. Sep 12, 2011

micromass

OK, that's fine. Let me show you my method anyway:

Say that ax+b=cx+d for all x. Take x=0, then b=d. So we're left with ax=cx. Take x=1, then a=c.

But what if we're given polynomials of higher degree. Or have forbid, series?? Well, here's a method that can be generalized:

Take $ax^2+bx+c=dx^2+ex+f$. Fill in x=0, to obtain c=f. Take the derivate to obtain $2ax+b=2dx+e$. Fill in x=0, to obtain b=e. Take the derivative to obtain 2a=2d, or a=d.

If you're ever taking complex analysis, then this method will show up a few times.

7. Sep 12, 2011

tylerc1991

I've actually taken complex analysis (albeit a VERY elementary introduction). What would you need this algorithm to prove? i.e. what type of problem would you need to show this on?

8. Sep 12, 2011

micromass

For example the proof of the identity theorem uses a very similar approach. That is: if two holomorphic functions are equal on a set with a limit point, then they are equal.

Also, when showing that there exists a unique power series representation for each holomorphic function. This algorithm can be used to prove uniqueness...