Question

For a < b, construct a bijection from (a, b] to (0, 1].

Answer 1

\[
\begin{array}{rcl}
a<&x&\leq b \\
0<&x-a&\leq b-a\\
0<&\frac{x-a}{b-a}&\leq 1
\end{array}
\]

Answer 2

Is that a bijection? And should it be interms of a and b as well as x? I really don't understand this, I'm sorry.

Answer 3

Well, \(x\in(a,b]\) and \(f(x)=(x-a)/(b-a)\in (0,1]\).

Answer 4

A bijection is a function, so there needs to be an input variable.

Answer 5

Since \(a\) and \(b\) are not given, the bijection has to be in terms of them.

Answer 6

The question says create "a" bijection, does it definitely need to be in terms of them? I guess it would still, huh?

Answer 7

Since \(b\) is in the codomain, how can you get \(f(x)=b\)?

Answer 8

Whoops, let me rephrase that..

Answer 9

How can get get \(f(x)=1\)? You don't even know what \(a\) and \(b\) are, so you're going to have go provide some \(x\) in terms of \(a\) and \(b\). If that is the case, you'll need \(f(x)\) to have \(a\) and \(b\) in it to cancel them out.

Answer 10

Okay that definitely makes sense. Thank you so much for being patient with me.

Answer 11

We know \(g(x)=x\) is a bijection.

Answer 12

There should be a property stating that if \(g(x)\) is a bijection then so is \(g(x)+c\) and some property saying that if \(g(x)\) is a bijection then so is \(cg(x)\) where \(c\) is not \(0\).

Answer 13

Yea there definitely is somewhere..been a while since I saw that stuff but it sonuds familiar. Thanks