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**Sven****Member**- Registered: 2014-05-02
- Posts: 18

I was reading a book on elementary Algebra when I came across an article regarding recurring decimals, in it was a paragraph to find how many digits would recur in the decimal(Or I think it was for this reason).I could not understand the paragraph, so I'll just give the matter as it is:

Let n/d be any real fraction irreducible to lower terms.And suppose it were required to find how many decimal places to reduce it to before same terms return again.To determine this, suppose that 10n > d if not & if 100n or 1000n > d then we begin by reducing 10n or 100n / d to less terms or fraction n/d.

Therefore same number will return in quotient when n is again remainder.Suppose when this happens we have added s zeros & that q is the integral part of quotient, then abstracting from the point we have n x 10^s/d = q + n/d, therefore q = n x (10^s - 1)/d . Now for q to be an integer, we must find least integer number for 's' such that n x (10^s - 1)/d or 10^s - 1 / d is an integer, for this we consider 3 cases:

a: If d is factor 10, 100, 1000 etc. & so there will be no circulating fraction.

b: If d is odd & not a factor of 10 or its power, in which case value of 's' may rise to d - 1 but frequently it is less.

c: d is even but not factor of 10 etc. but nevertheless has a common divisor with one of those powers of 10, which can only be a number of the form 2^c, so that d/2^c = c, so the period will be same for fraction n/d but will not commence before figure represented by c.

I do not understand the meaning of this.

God Bless

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