Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

manjuthottam

could someone explain what equinumerous means in mathematical Theoretical course?!

  • one year ago
  • one year ago

  • This Question is Closed
  1. manjuthottam
    Best Response
    You've already chosen the best response.
    Medals 0

    the definition states " two sets S and T are called equinumerous, and we write S ~ T, if there exists a bijective function from S onto T". Bijective definition is "a function is bijective if it is surjective and injective". I'm asked to "Prove that if (S \ T) is equinumerous to (T \ S), then S is equinumerous to T" how do i do that?

    • one year ago
  2. KingGeorge
    Best Response
    You've already chosen the best response.
    Medals 1

    What exactly do you mean by the S\T and T\S? Are those set minuses?

    • one year ago
  3. manjuthottam
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

    • one year ago
  4. manjuthottam
    Best Response
    You've already chosen the best response.
    Medals 0

    set S without T

    • one year ago
  5. KingGeorge
    Best Response
    You've already chosen the best response.
    Medals 1

    So there are the same number of elements in S that aren't in T as there are elements in T that aren't in S. Let \(U=S\cap T\). So \(S\setminus T=S\setminus (S\cap T)\). Then \(S=S\setminus T +U\). Now let \(f:S\setminus T\to T\setminus S\) such that \(f\) is a bijection. Now define\[g:S\setminus T+U\longrightarrow T\setminus S+U\]by\[ g(s)=\begin{cases} f(s)\qquad s\in S\setminus T \\ s\qquad \;\;\;\;\,s\in U \end{cases}\]

    • one year ago
  6. KingGeorge
    Best Response
    You've already chosen the best response.
    Medals 1

    Note that if we restrict \(g\) to only the domain \(U\), we get a bijective function since \(g(g(s))=s\). I.e., \(g\) has an inverse. Similarly, if we restrict \(g\) to only the domain \(S\setminus T\), then we have a bijective function since \(f\) is bijective. So \(g\) is bijective over its whole domain, and is therefore a bijective function. Finally, since \(S\setminus T+U=S\) and \(T\setminus S+U=T\), \(S\) and \(T\) are equinumerous.

    • one year ago
  7. KingGeorge
    Best Response
    You've already chosen the best response.
    Medals 1

    Did that all make sense?

    • one year ago
  8. manjuthottam
    Best Response
    You've already chosen the best response.
    Medals 0

    oh yes thank you!!

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.