Proof function is bijective check please

[tylerc1991]

Homework Statement

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.

Homework Equations

$g_{ab} = ax + b$.

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 $<br /> g_{ab} = ax + b = f_{ab}$, where $a \neq 0$.

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

Therefore, $G$ is surjective.

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

 

[micromass]

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

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

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

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?

 

[tylerc1991]

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?

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 $a = c$.

 

[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 )

 

[tylerc1991]

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.

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 I am 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.

(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 )

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

 

[micromass]

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 I am 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

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.

 

[tylerc1991]

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

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?

 

[micromass]

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?

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...