1. ## proving a bijection

Prove that that f : R \ {0} ->R \ {0} : R -> 1/x is a bijection.

I understand that with a bijection it is both an injective function and Surjective.

and a bijective function provides a one to one correspondence between the elements .

so would it be true to say all Real numbers besides 0 are in the set

and then I prove that it is both surjective and bijective?

2. ## Re: proving a bijection

To prove that it is an injection: For any $x,y \in \mathbb{R}\setminus \{ 0 \}$: Show that $f(x) = f(y) \Longrightarrow x=y$. (This follows trivially).

To prove that it is a surjection: For any $y \in \mathbb{R} \setminus \{ 0 \}$: Show that $\exists x \in \mathbb{R} \setminus \{ 0 \}$ such that $y = f(x)$. (Hint: let $x = \dfrac{1}{y}$).

3. ## Re: proving a bijection

let x be any real number besides 0 ...
the rule for injective functions are that set R of real numbers is (-∞ ,∞ ) -> (-1/∞,1/∞)
injective
f(-∞) = f(-1/∞ ) for all Real numbers
Surjective
f(-∞) = f(-1/∞ ) for all real numbers if we sub into the rational number....
I think this is wrong but the concept of trying to prove surjective and injective to be Bijective is there ...
if I invert the function......f (-∞...∞\{0})-> {-1/∞...1/∞\{0}) to
f^-1 {-1/∞...1∞)->f (-∞...∞)
wouldn't that be bijection?

any help would be good ...I had a go but I'm pretty sure I'm wrong

4. ## Re: proving a bijection

Thanks slip eternal ,
so for the y = f(x) would I just sub 1/y into x as you mentioned and say that R->1/1/y from the original R -> 1/x
making it 1 ÷ 1/y = 1 x y/1 = y
there fore proving y ∈ r\{0} following the surjective function definition for y ∈ Y there is an x ∈ X such that y =f(x)
I think I was typing the way I tried it at the same time you posted ...is this correct ?

5. ## Re: proving a bijection

Originally Posted by bee77
let x be any real number besides 0 ...
the rule for injective functions are that set R of real numbers is (-∞ ,∞ ) -> (-1/∞,1/∞)
injective
f(-∞) = f(-1/∞ ) for all Real numbers
Surjective
f(-∞) = f(-1/∞ ) for all real numbers if we sub into the rational number....
I think this is wrong but the concept of trying to prove surjective and injective to be Bijective is there ...
if I invert the function......f (-∞...∞\{0})-> {-1/∞...1/∞\{0}) to
f^-1 {-1/∞...1∞)->f (-∞...∞)
wouldn't that be bijection? any help would be good ...I had a go but I'm pretty sure I'm wrong
Let me be very blunt with you. If any student in one of my classes wrote $\color{red}{\Large\frac{1}{\infty}}$ on a paper, that paper would fail. Not only that, would require that student to drop out of the class. Not only is it grossly wrong, it is stupid.

6. ## Re: proving a bijection

Originally Posted by bee77
Thanks slip eternal ,
so for the y = f(x) would I just sub 1/y into x as you mentioned and say that R->1/1/y from the original R -> 1/x
making it 1 ÷ 1/y = 1 x y/1 = y
there fore proving y ∈ r\{0} following the surjective function definition for y ∈ Y there is an x ∈ X such that y =f(x)
I think I was typing the way I tried it at the same time you posted ...is this correct ?
Here is the complete proof:

Given $f: \mathbb{R} \setminus \{ 0 \} \to \mathbb{R} \setminus \{ 0 \}$ defined as $f(x) = \dfrac{1}{x}$. We want to prove this is a bijection.

Proof that $f$ is an injection:
Let $x,y \in \mathbb{R} \setminus \{ 0 \}$ such that $f(x) = f(y)$. This means $\dfrac{1}{x} = \dfrac{1}{y}$, and by cross multiplying, we get $x=y$. So, the function is an injection.

Proof that $f$ is a surjection:
Let $y \in \mathbb{R} \setminus \{ 0 \}$ be any number. We want to find $x \in \mathbb{R} \setminus \{ 0 \}$ such that $f(x) = y$. Let $x = \dfrac{1}{y}$. Then $f(x) = f\left( \dfrac{1}{y} \right) = \dfrac{1}{ \tfrac{1}{y} } = y$. Therefore, $f$ is a surjection.

Since $f$ is both an injection and a surjection, it is a bijection. Q.E.D.

7. ## Re: proving a bijection

oh I meant it to be real numbers...point taken ...I was thinking in terms of limits and had a brain malfunction...sorry my mind wanders with this stuff

8. ## Re: proving a bijection

Thanks Slip Eternal do those principles apply to all bijection proofs ie :x=y for injection ?
and x =1/y for subjection ? (or what ever function letters are being used ie f(g) or f(h) etc ) ...
thanks

9. ## Re: proving a bijection

For any function $f: D \to R$, to prove it is an injection you must show that for any $a,b \in D$ such that $f(a)=f(b)$ then $a=b$.
To show $f$ is a surjection, you must show that for all $b\in R, \exists a\in D$ with $f(a)=b$. It will not always be $a=\dfrac{1}{b}$. That only happened because in your problem, $f(x)=\dfrac{1}{x}$