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Math Help - Show that

  1. #1
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    Show that

    For any real number x, let {x} = x −[x] denote the fractional part where [x] is the
    usual floor function.
    (a) Show that there exist infinitely many positive rationals x such that {x}+{x^2} = 0.99.
    (b) Show that there are no positive rational x such that {x} + {x^2} = 1.
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  2. #2
    Senior Member JaneBennet's Avatar
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    (a)

    For any non-negative integer k, observe that (10k+1.3)^2=100k^2+26k+1.69.

    Thus, if x=10k+1.3\quad(k\in\mathbb{Z},\ k\geqslant0), we have \{x\}+\{x^2\}=0.3+0.69=0.99.

    There are infinitely many non-negative integers k and hence infinitely many such rationals x.

    (b)

    Note that for any x\in\mathbb{R}^+,\ \{x\}=0\ \Leftrightarrow\ x\in\mathbb{Z}. Hence, in order to have \{x\}+\{x^2\}=1, we must have x\notin\mathbb{Z}. If x is a positive non-integer rational, x^2 would have more decimal places than x. This would mean that their fractional parts cannot add up to 1. Hence there are no positive rationals x such that \{x\}+\{x^2\}=1,

    Remark: There do however exist positive irrational numbers x with \{x\}+\{x^2\}=1, To see this, note that f(x)=\{x\}+\{x^2\} is continuous on the interval [0.1\,,\,0.9] and f(0.1)<1 while f(0.9)>1; the result follows by the intermediate-value theorem.
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