Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

hby0214

  • 3 years ago

Show that the set T ={(w,x,y,z)∈R4 such that y=w and x^2 =z^4} is not the graph of any function of w and x.

  • This Question is Closed
  1. helder_edwin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    well u have the correspondence (i don't know if this is the word in english) \[ \large (w,x)\mapsto(y,z) \] such that \(y=w\) and \(z^4=x^2\).

  2. helder_edwin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    do u remember the definition of function?

  3. hby0214
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    A function is a rule that assigns a unique element in Rn to Rm. It is one to one.

  4. helder_edwin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the last part is something else.

  5. helder_edwin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    a function assigns a UNIQUE element to EVERY element of its domain.

  6. hby0214
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok I understand that.

  7. hby0214
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So a certain w and x cannot have two outputs.

  8. helder_edwin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    what does (w,x)=(1,-1) get assigned to?

  9. helder_edwin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes. precisely.

  10. hby0214
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (1,-1) is assigned to (1,1)?

  11. helder_edwin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    just that?

  12. helder_edwin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    let's see: \[ \large y=w=1 \] right?

  13. hby0214
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right

  14. helder_edwin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    BUT \[ \large z^4=x^2=(-1)^2=1\Rightarrow z=\pm\sqrt[4]{1}=\pm1 \]

  15. helder_edwin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so \[ \large (1,-1)\mapsto(1,1) \] and \[ \large (1,-1)\mapsto(1,-1) \]

  16. hby0214
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So does that mean T is not a function?

  17. helder_edwin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes. that's what u were asked to prove.

  18. hby0214
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    T is graph of something, but that something is not a function. right?

  19. helder_edwin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes.

  20. hby0214
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you SO much!

  21. helder_edwin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    u r welcome.

  22. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy