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Irrational numbers in infinite list of integers

  1. Mar 13, 2010 #1
    Is it safe to assume that the absolute value of sin x is greater than zero for all positive integer values of x? I have no real experience in number theory, and I don't know if you can say that there are no irrational numbers in an infinite list of integers.
  2. jcsd
  3. Mar 13, 2010 #2


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    Can you convince yourself that this conjecture is equivalent to
    Every integer is rational​
  4. Mar 13, 2010 #3
    If sin(x)=0 for positive x, then x=k*pi for some positive integer k. If x is an integer, then x/k=pi should raise a red flag.
  5. Mar 14, 2010 #4
    are you talking of integer degrees or radians?
  6. Mar 14, 2010 #5
    Integer radians. Sorry if this was a super obvious question. I guess I was just thinking too much into it.
  7. Mar 23, 2010 #6
    If you mean to ask whether the intervals [2npi,(2n+1)pi) (where sinx >0) contain integers, the answer is yes ( as pi>1).
  8. Mar 24, 2010 #7
    SineX = 0 only where X = n*pi radians. But n*pi must be an irrational number or 0. Thus SineX is either greater than or less than 0 for any integer angle n in radians where n>0. I think this is what the poser was asking.
    Last edited: Mar 24, 2010
  9. Mar 25, 2010 #8


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    Yes, since any integer, n, can be written as the fraction n/1, all integers are rational numbers. There are no irrational integers. It is true that sin(n) is never 0 for any integer n- but that does NOT mean "greater than 0". sin(4) is negative.
  10. Mar 25, 2010 #9
    While I used the langauge "Sine x" is "greater than or less than 0", the poser asked whether "the absolute value of Sine x is greater than 0 for all integer values of x.". I think you had my language in mind when you overlook the "absolute value" part of the poser's question.
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