(Real Analysis) Show the function is Bijection

[phillyolly]

Homework Statement

The problem and my attempt are attached.

I am unable to solve the function for x.

Homework Equations

The Attempt at a Solution

 

[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

 

[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

 

[Mark44]

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.

 

[Raskolnikov]

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."

 

[annoymage]

The definition for a bijection I usually use is:

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

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

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

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

 

[Raskolnikov]

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

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

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

 

[annoymage]

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

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

but that's the definition of a surjection

 

[Mark44]

annoymage, I think you're right.

 

[Raskolnikov]

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.

 

[annoymage]

i see i see

 

[Raskolnikov]

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

 

[phillyolly]

Based on your feedback. This is my bijection thing.

 

[Raskolnikov]

I would rather you do as Mark44 suggested and solve for x in terms of y.

 

[annoymage]

Based on your feedback. This is my bijection thing.

this complete the injection, and now as mark44 suggested to do for the surjection