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Continuous function

  1. May 30, 2012 #1
    if f : (0, 1]--> R is given by f(x) = 0 if x is irrational, and f(x) = 1/(m+n) if x = m/n in (0, 1] in lowest terms for integers m and n. How can i prove that this function is continuous at 1/√2?
     
  2. jcsd
  3. May 30, 2012 #2
    At x=1/sqrt(2),
    LHL=0 and RHL=0 becoz x+ and x- are both irrational here.
    Also, Lt(x--1/sqrt(2))=0
    x being irrational.

    Hence f(x) is continuous at x=1/sqrt(2)
     
  4. May 30, 2012 #3
    Essentially, you need to use the fact that if p/q is close to sqrt(2), then q must be large. The closer you want it to be, the larger q must be. Therefore, the value of the function for p/q sufficiently close to sqrt(2) must be small. Perhaps you could use something like this: Given Q, let e_Q be the infimum of | (p/q) - sqrt(2)| for q less than or equal to Q. Then e_Q is nonincreasing, is never 0, and tends to zero as Q tends to infinity. You could use that in your proof.
     
  5. May 30, 2012 #4

    Perhaps it' only me but I really can't understand whatever it is you're trying to argue above. Besides this, I think it'd be better

    for all if you write down your piece with LaTeX.

    DonAntonio
     
  6. May 30, 2012 #5

    HallsofIvy

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    What do you mean by "x+" and "x-"??

    What is "Lt"?

     
  7. May 30, 2012 #6

    Clearly, if [itex]\,\{y_n\}\,[/itex] is an irrational seq. s.t. [itex]\,\displaystyle{y_n\to\frac{1}{\sqrt{2}}}\,[/itex] , then [itex]\,\displaystyle{0=f(y_n)=f\left(\frac{1}{\sqrt{2}}\right)=0}[/itex]

    OTOH, if [itex]\,\displaystyle{\left\{x_n=\frac{a_n}{b_n}\right\}}\,[/itex] is a rational seq. s.t. [itex]\,\displaystyle{x_n\to\frac{1}{\sqrt{2}}}\,[/itex] , with [itex]\,(a_n,b_n)=1\,\,\forall n\,[/itex] , then

    Lemma: If [itex]\,\displaystyle{\left\{x_n=\frac{a_n}{b_n}\right\}}\,[/itex] is a rational seq. that converges to an irrational number , then [itex]\,b_n\to \infty[/itex]

    Proof: Exercise (try contradiction and check what happens when an integer seq. converges...)

    DonAntonio
     
  8. May 31, 2012 #7
    Thanks guys for the replies.

    shreyakmath - your argument is not valid as 0- is not defined.
    DonAntonio - it is a good way to proceed, but I've come to a simpler solution (may not be 100 correct).

    Here is the solution:

    We can show that f(x) is continious at x=1/√2 using the ε-δ definition. Let ε>0 and set δ=ε. we first note that |f(x)|≤|x| for all x. Indeed, if x is irrational, f(x)=0, and if x is rational f(x)=1/(m+n); x=m/n.

    Thus, for 0<|x|<δ=ε, we have:

    |f(x)-f(1/√2)|=|f(x)|≤|x|<ε.

    this will then concludes the proof!
     
  9. May 31, 2012 #8


    ¡Muy bien! ¡ Bravo ! Ausgezeichnet ! מצויין ! Excellent ! I really loved it.

    DonAntonio
     
  10. Jun 1, 2012 #9
    I believe there is an error. Your proof demonstrates that if x is close to 0 then f(x) is close to f(1/sqrt(2)).

    You need to show that if x is close to 1/sqrt(2), then f(x) is close to 1/sqrt(2).
     
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