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Rational and irrational numbers

  1. Dec 23, 2014 #1
    Are numbers which tend to an integer for example 1, that is 0.9999... And 1.00....01 irrational?

    [where I got the doubt: There was a question in my textbook--->
    $$f(x) = x^2+ax+1 \text{ if x is rational}$$
    $$f(x) = ax^2+bx+1 \text{ if x is irrational}$$
    If f(x) is continuous at x=1 and 2 find a and b.
    Why I had the doubt: Since if numbers that tend to 1 and 2 are irrational then 1 and 2 are the roots of the equation ##x^2+ax+1=ax^2+bx+1##]
  2. jcsd
  3. Dec 23, 2014 #2


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    Your question makes no sense. 0.999... by which I assume you mean "the limit of the sequence 0.9, 0.99, 0.999, ..." is 1. There is NO such number as [itex]0.000...1[/itex] because you cannot have a number with an "infinite number of 0s ended with a 1"- and infinite sequence has NO end. If you mean, as you should, "the limit of the sequence 0.01, 0.001, 0.0001, ...", then that number is 0. No, 1 and 0 are NOT irrational.

    But I have no idea what you mean by "if numbers that tend to 1 and 2 are irrational". Given any number, whether rational or irrational, there exist sequences of rational number and sequences or irrational number that converge to it.
  4. Dec 23, 2014 #3


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    Every number is the limit of both rationals and irrationals.

    What then does continuity at 1and 2 tell you then about the two expressions for f at 1and 2?
    Last edited: Dec 23, 2014
  5. Dec 23, 2014 #4
    Il rather post the question in the homework forum. Maybe I'm totally wrong. You can close the thread.
  6. Dec 23, 2014 #5


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    Think about my reply.
  7. Dec 24, 2014 #6
    All repeating/terminating decimals are rational.
    Also, as pointed out earlier, 0.99999999999..... is equal to 1.
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