Results 1 to 9 of 9
Like Tree3Thanks
  • 1 Post By SlipEternal
  • 1 Post By SlipEternal
  • 1 Post By SlipEternal

Thread: proving a bijection

  1. #1
    Member
    Joined
    Mar 2017
    From
    Annabay
    Posts
    198

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    2,706
    Thanks
    1033

    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 $x = \dfrac{1}{y}$).
    Last edited by SlipEternal; Aug 23rd 2017 at 06:50 PM.
    Thanks from bee77
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2017
    From
    Annabay
    Posts
    198

    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2017
    From
    Annabay
    Posts
    198

    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 ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,267
    Thanks
    2641
    Awards
    1

    Re: proving a bijection

    Quote Originally Posted by bee77 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2010
    Posts
    2,706
    Thanks
    1033

    Re: proving a bijection

    Quote Originally Posted by bee77 View Post
    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.
    Thanks from bee77
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Mar 2017
    From
    Annabay
    Posts
    198

    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
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Mar 2017
    From
    Annabay
    Posts
    198

    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
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Nov 2010
    Posts
    2,706
    Thanks
    1033

    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, $f(x)=\dfrac{1}{x}$
    Thanks from bee77
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Bijection
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: Mar 27th 2011, 11:01 AM
  2. Proving a function is a bijection
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Apr 21st 2010, 12:39 PM
  3. Bijection
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Sep 15th 2009, 12:55 PM
  4. Bijection
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Sep 11th 2009, 04:28 AM
  5. Replies: 1
    Last Post: Nov 9th 2008, 12:31 PM

/mathhelpforum @mathhelpforum