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Irrational Numbers.

  1. Dec 17, 2009 #1
    Irrational Numbers are contained by infinite numerical values?
     
  2. jcsd
  3. Dec 17, 2009 #2
    huh?!i didn't get what u mean:D
     
  4. Dec 17, 2009 #3
    meaning if we would to write a irrational number out , we need a infinite number of digits?
     
  5. Dec 17, 2009 #4
    Do you care? If we write out 1/7 we would require an infinite number of digits.
     
  6. Dec 17, 2009 #5

    Borek

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    Staff: Mentor

    Depends on the base of the number system used. 1/7 is 0.17, 1/3 is 0.13, both require infinite number of digits if they are to be written base 10.

    Edit: do you hate it when you make an idiot out of yourself just because you think in your first language when you should in English? I do. Irrational as it sounds, I was all the time thinking about rational numbers.
     
    Last edited: Dec 17, 2009
  7. Dec 17, 2009 #6
    An irrational number cannot be written as a fraction. (A fraction is a "ratio," so it is considered to be "rational.") Therefore, it has an infinite number of digits after the decimal point. (If there was a finite number of digits after the decimal point, it could be written as a fraction and would therefore NOT be irrational).

    On the other hand, just because there are an infinite number of digits following the decimal point, doesn't mean that the the value is irrational. (0.111111111... can be written as 1/9, so it is rational, whereas [tex]\pi, \: e, \: and \: \sqrt{2}[/tex] are all examples of irrational numbers).
     
  8. Dec 17, 2009 #7

    Mark44

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    1/7 can also be written as 0.06666666...7, and 1/3 can be written as 0.02222222...3, requiring an infinite number of digits in these bases.
     
  9. Dec 17, 2009 #8

    Landau

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    Every irrational number has an infinite non-repeating decimal expansion. Every rational number has either a finite decimal expansion, or an infinite repeating decimal expansion. I.e. for every rational number with an infinite decimal expansion, there is a repetion in the expansion, e.g. the above example 1/9=0.1111..., or 5/12=0.416666...
     
  10. Dec 17, 2009 #9

    Hurkyl

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    I can write [itex]\sqrt{2}[/itex] with two symbols: 2 and [itex]\sqrt{\ }[/itex].

    The decimal expansion of [itex]\sqrt{2}[/itex] has infinitely many digits, though. And so does the decimal expansion of every irrational number, most rational numbers, and even every integer. (don't forget about the infinitely many zeros!)
     
  11. Dec 17, 2009 #10
    That's what I was thinking. We can write "2" without all the zeroes (2.000000...) because by convention we leave them off. What if by convention we left off .4142135623731...? Then things would be different (we'd write sqrt(2) as "1"). The point is, representations of numbers tell you about conventions, not so much about the numbers themselves.
     
  12. Dec 17, 2009 #11
    Last edited by a moderator: Apr 24, 2017
  13. Jan 21, 2010 #12
    hey qntty ! thanks for your help. its greatly appreciated!
     
  14. Jan 22, 2010 #13

    HallsofIvy

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    Several years ago, a poster asked how to prove that a rational number could be written as a fraction! My first reaction was that that is the definition of "rational number" and could not be proven.

    Then I realized that he had been taught "a rational number can be written as a terminating or repeating decimal" as the definition of rational number and now wanted to prove that a number satisfies that definition if and only if it can be written as a fraction, the opposite to the way it is normally done.
     
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