Results 1 to 5 of 5

Math Help - Is my work correct? Prove the following is bijective.

  1. #1
    Junior Member
    Joined
    Sep 2010
    Posts
    48

    Is my work correct? Prove the following is bijective.

    The problem says:

    Let A and B be sets. Show that there is a bijective map f: A x B --> B x A.

    My professor helped me out and said that the function should be given by f((a,b)) = (b,a) (where a exists in A and b exists in B).

    Here is the work I have so far.

    Proof. (Injection) Let f((a1,b1)) = f((a2,b2)), where (a1,b1) , (a2,b2) exist in A x B. Then, (b1,a1) = (b2,a2), which implies f((b1,a1)) = f((b2,a2)), which implies (a1,b1) = (a2,b2). So, f is injective.

    (Surjection) Fix (b,a) which exists in B x A. Put x = (a,b) and observe that f(x) = f((a,b)) = (b,a). So, f is surjective.

    Since f is both injective and surjective, then f is bijective.

    I feel like my Injection proof is wrong. Do any of you see anything wrong? Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Some times subscripts are troublesome.
    Suppose that f((p,q))=f((r,s)) then (q,p)=(s,r).

    That means q=s~\&~p=r so that (p,q)=(r,s).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2010
    Posts
    48
    Thanks. That makes sense, I just wasn't sure if it was really that simple or not.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398
    Quote Originally Posted by Lprdgecko View Post
    The problem says:

    Let A and B be sets. Show that there is a bijective map f: A x B --> B x A.

    My professor helped me out and said that the function should be given by f((a,b)) = (b,a) (where a exists in A and b exists in B).

    Here is the work I have so far.

    Proof. (Injection) Let f((a1,b1)) = f((a2,b2)), where (a1,b1) , (a2,b2) exist in A x B. Then, (b1,a1) = (b2,a2), which implies b1=b2 and a1=a2,

    not {
    f((b1,a1)) = f((b2,a2))} Actually, this is also implied, but it's not relevant.

    which implies (a1,b1) = (a2,b2). So, f is injective. (In other words: f(p)=f(q) implies p=q.)

    (Surjection) For any (b,a) which exists in B x A, there exists (a,b) in AxB such that f((a,b)) = (b,a). So, f is surjective.

    Since f is both injective and surjective, then f is bijective.

    I feel like my Injection proof is wrong. Do any of you see anything wrong? Thanks in advance
    I made some notes above, in red.

    In general the proof is pretty good - just needs that one significant change.

    Often, when you, the student, define a function, the professor likes you to prove that the function is "well defined" - i.e, p=q, implies f(p)=f(q). In this case, you would show: if (a1,b1)=(a2,b2) in AxB, then f(a1,b1)=f(a2,b2) in BxA. This is often done "indirectly" by showing that f(a_1,b_1)\neq f(a_2,b_2) implies that (a_1,b_1)\neq (a_2,b_2).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    [QUOTE=SammyS;589482]

    Often, when you, the student, define a function, the professor likes you to prove that the function is "well defined" - i.e, p=q, implies f(p)=f(q).
    I don't think that's the case, unless you're doing something with equivalence classes like f:\mathbb{Q}\to\mathbb{Q}:\frac{p}{q}\mapsto pq
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove comnplement function on [n] is bijective
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: June 24th 2010, 08:54 AM
  2. How to prove this function is bijective
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: May 8th 2010, 07:30 AM
  3. Prove about no bijective fuction exist
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: November 8th 2009, 06:31 AM
  4. Replies: 1
    Last Post: April 21st 2009, 10:45 AM
  5. Replies: 2
    Last Post: April 13th 2009, 12:19 PM

Search Tags


/mathhelpforum @mathhelpforum