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Solving a problem

  1. Aug 3, 2011 #1
    f[x]=x,x is rational, 1-x, x is irrational. prove that f(x) is only continuous at x=1/2.
     
  2. jcsd
  3. Aug 3, 2011 #2
    please give a proper mathematical solution to this
     
  4. Aug 3, 2011 #3

    HallsofIvy

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    Why? It looks like homework to me. You need to try yourself, first, and show us what you have.

    Here's a nudge: for any x, there exist a sequence of rational numbers converging to x and there exist a sequence or irrational numbers converging to x.
     
  5. Aug 4, 2011 #4
    i have already solved the problem in the way you suggested.but i have a problem in solving it using epsilon and those things
     
  6. Aug 4, 2011 #5

    Redbelly98

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    Reminder: members are expected to show their attempt at solving a problem before they receive help.

    ashok vardhan: please see your Private Messages for an important message, if you have not already done so.
     
  7. Aug 13, 2011 #6
    f(x+y)=f(x)+f(y).f is continuous at x=0.prove that f(kx)=kf(x). i have proved it for k os an integer.for k a rational number i assumed it to be of p/q.and i cant proceed further to prove this. would you like to help in this
     
  8. Aug 13, 2011 #7

    HallsofIvy

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    Is this a completely new problem? f(x+ y) is definitely NOT equal to f(x)+ f(y) for the problem you gave before. For example, [itex]1+ \sqrt{2}[/itex] is irrational and so [itex]f(1+ \sqrt{2})= 1- (1+ \sqrt{2})= -\sqrt{2}[/itex] but since 1 is rational and [itex]\sqrt{2}[/itex] is irrational, [itex]f(1)+ f(\sqrt{2})= 1+ (1- \sqrt{2})= 2- \sqrt{2}[/itex].

    You didn't say anything about using [itex]\epsilon[/itex] and [itex]\delta[/itex] in your first post. If you are not allowed to use "[itex]\lim_{x\to a} f(x)= L[/itex] if and only if, for any sequence [itex]\{x_n\}[/itex] that converges to a, the sequence [itex]\{f(x_n)\}[/itex] converges to L", then copy the proof of that theorem, for this particular function.
     
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