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The largest rational less than sqrt(10)

  1. Sep 16, 2007 #1
    How do I find the sup of rationals (p/q, where q is even) that is less than sqrt(10)?
  2. jcsd
  3. Sep 16, 2007 #2


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    Here's a hint: while that set does have a supremum, it does not have a maximum.
  4. Sep 16, 2007 #3
    Suppose I gave you a rational number which I claimed to be the closest rational less than sqrt(10). Could you make one that's closer?
  5. Sep 16, 2007 #4
    Suppose you claim x/y is the sup, where y is even (and x must be odd by the way), then I can construct x+1/y+2?
    hmm, seriously I have no idea.
    wait, i got an idea
  6. Sep 16, 2007 #5
    Given the decimal expansion to sqrt(10), you can always truncate it at a point such that it is bigger than any supposed "closest" rational. Basically, this shows that there is no such rational. As Hurkyl hinted, the set of rationals less than sqrt(10) has a supremum -- it's sqrt(10), it does not actually have a maximal element.
  7. Sep 16, 2007 #6
    thanks a lot for all your reply
    I under stand the truncate argument , we know irrationals are infinite decimals but rationals are finite decimals.
    but how to show it rigorously?i was trying to use archimedean principle to construct some contradiction,but didn't work out.
  8. Sep 16, 2007 #7
    Maybe that's what you're looking for:

    If we have a supremum x, then sqrt 10 - x > 0. Since Q has the archimedean property, there exist a rational y such as 10 - x > 1/y. This said, we have 10 - (x + 1/y) > 0. This said, 1/y is rational, and hence so is x + 1/y - contradiction.
  9. Sep 16, 2007 #8


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    Most rationals are not finite decimals. For example, 1/3.
  10. Sep 16, 2007 #9
    sorry, i meant finite decimals or infinitely repeating decimals
  11. Sep 16, 2007 #10
    Brilliant, but I thought archimedean property is defined such that 10-x>1/n where n must be an integer. Also the statement sqrt 10 - x > 0 is quite intuitive.
  12. Sep 16, 2007 #11


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    the supremum of "all rationals less than the irrational number r" or the supremum of "all rational numbers with even denominator (when reduced to lowest terms) less than the irrational number r" is very easy to find- but it's not a rational number.
  13. Sep 16, 2007 #12


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    You can use finite contin ued fractions vs. infinite continued fractions, if you like.
  14. Sep 16, 2007 #13
    Then replace "a rational y" by a "natural y". sqrt 10 - x > 0 is easy to argue for: we know that sqrt 10 > 0. Since 0 belongs to Q, then x can assume values that are lower than sqrt 10, hence a supremum could exist.
  15. Sep 17, 2007 #14


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    It's hard to believe there have been so many posts on this! If you were working only in the rational numbers, then the set of all rationals, p/q, reduced to lowest terms with q even, less than [itex]\sqrt{10}[/itex] does not exist.

    Thought of as subset of real numbers, since the set has an upper bound, it has a supremum: [itex]\sqrt{10}[/itex].
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