# (Real Analysis) Show the function is Bijection

1. Aug 28, 2010

### phillyolly

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

The problem and my attempt are attached.

I am unable to solve the function for x.

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

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• ###### IMG_0142.jpg
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2. Aug 28, 2010

### annoymage

hmm, if you want to show injection, you need to show for all the values in the domain, now you only showing $$-1\neq1$$ how about $$-1\neq-1.1$$?

so if you want to show an injection first see the definition of injection

whenever $$x\neq y \Rightarrow f(x) \neq f(y)$$

or the contrapositive $$f(x)=f(y) \Rightarrow x=y$$

normally i use the second one, so, for any for any x,y in R $$f(x)=f(y)\Rightarrow....... \Rightarrow x=y$$

that will complete the injection

Last edited: Aug 28, 2010
3. Aug 28, 2010

### annoymage

hmm i just learn the latex, i don't know if there's a bug or its just me who did it wrong, i'm sorry, if i convey the different thing, try to refresh it ;P and sorry for my english

4. Aug 29, 2010

### Staff: Mentor

I don't understand what you're trying to do in 1). You say "Injection proof -1 $\neq$1, and then go on to set f(-1) equal to f(1). ???

The idea here is the if f(a) = f(b), and f is one-to-one (i.e., f is an injection), then a = b.

The function is one-to-one.

To solve for x in the equation y = x/sqrt(x^2 + 1), square both sides of the equation, then move all terms in x to one side, and then factor.

It's a little tricky, because squaring both sides introduces extraneous solutions, so the equation for x does not seem to be a function. You will need to simplify the equation somewhat to make x a one-to-one function of y.

5. Aug 29, 2010

The definition for a bijection I usually use is:

Given $$f: A \rightarrow B,$$ f is a bijection if and only if

EDIT (thanks annoymage): $$\forall y \in B , \exists !x \in A \ni f(x) = y.$$

"For all y in B, there exists a unique x in A such that f(x) = y."

Last edited: Aug 29, 2010
6. Aug 29, 2010

### annoymage

hey, aren't that the definition of a function?

7. Aug 29, 2010

Yep, I have it backwards. Sorry, I'm a bit sleepy. It should be:

$$\forall y \in B , \exists !x \in A \ni f(x) = y.$$

8. Aug 29, 2010

### annoymage

but that's the definition of a surjection

9. Aug 29, 2010

### Staff: Mentor

annoymage, I think you're right.

10. Aug 29, 2010

nope, the unique x part makes it a bijection. It's subtle but it's there! A surjection only guarantees there exists an x in A. A bijection tells us it's a unique x.

11. Aug 29, 2010

### annoymage

i see i see

12. Aug 29, 2010

The process is still of course the same:
Solve for x in terms of y and show that's it's a unique solution.

13. Aug 29, 2010

### phillyolly

Based on your feedback. This is my bijection thing.

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14. Aug 29, 2010