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    Prove that a real number written as a decimal is equal to a rational number if and only if the decimal eventually recurrs.

    I'm thinking proof by contradiction... I'll have a go, hopefully some other people will be free from their P2 revision to try it out. :cool:
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    I have said first prove that any recurring decimal can be written in the form a/b, where a and b are integers:

    Call the recurring decimal a. It will be in the form:

    0.###...###ė....ġ

    Where x is the number of random digits (shown as #) before the recurrance begins, and y is the number if digits in the set of reccurances, from e to g inclusive.

    Now consider:

    t*10^(x+y)

    This will shift the decimal place up to the point where the first reccurance ends and the second begins (shifting it x+y amount to the left, x to remove the random initial digits from the decimal part, and y removing the first recurrance set). Take t from this number to remove the second and subsequent recurrances, leaving an integer b:

    t*10^(x+y) - t = b

    Factorise,

    t(10^(x+y) - 1) = b
    t = b/(10^(x+y) - 1)

    We know b is an integer, and 10^(x-y) is an integer, as x and y must be integers (as they represent the numbers of digits), therefore 10^(x+y) - 1 is also an integer.
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    If an only if means that we have to prove it in both directions.

    1) If a number is rational it can be written in the form a/b where a and b are both integers and share no common factors. Now, when we do the division it will either eventualy come to a point where it divides nicely, like 3/8 = 0.375 in which case we actually have a reccuring sequence of zeros. The other option is that we will not reach a point where we get zeros, but instead we will continue getting remainders. However, there are only a finite number of possible remainders, in fact there are only (b-1) possible remainders; therefore, there will be a point when we get the same remainder as some time previously, which will be when the decomal recurs again. So, in either case, the decimal expansion of a rational recurs.

    2) If a number can be written as a recuring decimal then it will have an integer part and a decimal part of the form

    x = 0.(A1)(A2)...(An)(B1)(B2)...(Bm) (B1)(B2)...(Bm)...

    That is, there is a starting section with digits given by the As and then a recuring section given by the Bs. For example 1/7 = 0.(14)(285714)(285714)... in which there are some starting digits, 14, and then a recurring section, 285714. With our generic example, with n starting digits and m recuring digits we have

    10^n x = (A1)(A2)...(An).(B1)(B2)...(Bm)(B1)(B2)...(Bm).. .
    10^(n+m) x = (A1)(A2)...(An)(B1)(B2)...(Bm).(B1)(B2)...(Bm)(B1)(B2)...(Bm).. .

    So,

    10^(n+m) x - 10^n x = k (for some integer k)
    x = k/[ 10^(n+m) - 10^n ]

    which is rational. So we have shown that the decimal part of any recuring decimal can be written as a rational, by just adding an integer it will remain a rational therefore any number with a recuring decimal expansion is rational.


    So, the things above show that if a number is rational its decimal is recuring and if a decimal is recurring it is rational; in other words, a number is rational iff its decimal is recurring.
 
 
 
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