Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

LogicalApple

  • one year ago

Challenge

  • This Question is Closed
  1. LogicalApple
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  2. aacehm
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm pretty sure it's E

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

    What about 0 ?

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

    I think it's just as discontinuous as every other point

  5. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Interesting. I'm going with D. Pathological example, though :)

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

    I think it is D. I am fairly certain of this. If only I could rationalize my opinion!

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

    i support E.

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

    |dw:1356812315918:dw|

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

    It is obviously not A since the function is defined at some points. It cannot be B or C since each of those functions contain infinitely many discontinuous points. The only reason I say D is because both functions share 0 as a vertex which means the function is continuous at 0.

  10. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    The definition of continuity at a point is that there exists a limit as you approach that point, and that the function equals the limit at that point. Both are true for zero. This is an interesting example of a function that is continuous at only one isolated point. Kinda weird sounding, but there you have it.

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

    in theory it's undefined at zero

  12. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Not it isn't, zero is rational.

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

    or you're right sorry

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

    that was silly

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

    hmm.. foolish of me :)

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

    you're right, that's weird

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

    Since 0 is rational, f(x) has a removable discontinuity at 0/3. The point (0, 0) is removable but it defined by the other function. It just so happens that the other function defines f(0) as 0/2 = 0 anyway. This is why I think 0 is continuous.

  18. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Why do you say it has a removable discontinuity?

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

    If we focus only on f(x) = x/3 if x is irrational, then we would expect 0/3 to be discontinuous since 0 is rational and is not defined.

  20. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    But that's not the function. The function is piecewisely defined at every single point. It's never undefined because any point you choose is either rational or irrational.

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

    But piecewise functions are also discontinuous. In this function, each subfunction has infinitely many points of discontinuity. Combined, however, we obtain the set of all real numbers. x/2 is discontinuous for every irrational (the removed discontinuities lay at x/3) x/3 is discontinuous for every rational (the removed discontinuities lay at x/2) We would expect, then, that when x = 0, the removable discontinuity at x/3 would be found at x/2. It happens that 0/3 = 0/2. Thus the removed discontinuity is found at the same location.

  22. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    A removable discontinuity is defined as a point at which a function has a finite limit but the function either does not exist at that point, or it does not equal the limit at that point. Neither of those two things apply here. The function exists at zero, and the limit is zero. It is just good old-fashioned continuous.

  23. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    by the function does not exist, I really mean the function is not well-defined.

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

    You would not agree that f(x) contains two functions containing infinitely many discontinuous points? y = x/2 if x is rational y = x/3 if x is irrational

  25. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    I would completely agree that f(x) has infinitely many irrational points. I disagree that it has a removable discontinuity at 0.

  26. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    I'm sorry, discontinuous, not irrational

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

    Ah, yes. I agree that 0 is continuous. My reasoning was that if we focused only on y = x/3 if x is irrational, that line by itself has a discontinuous point at (0, f(0)). However, since 0 is rational, and f(x) = x/2 when x is rational, then f(0) = 0 anyway. f(x) then is continuous at 0. You already said D didn't you ... we are in agreement :D

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

    My brain probably processed the problem incorrectly but resulted in the correct answer.

  29. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    I think your consideration was logical, but it is just very particular to this problem because there are infinitely many irrational points interspersed between rationals and vice versa. Generally speaking, a function is continuous at a if lim f(x) as x->a = f(a), which is perfectly valid here :)

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

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

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.