Why are repeating decimals often self-inversions?Given a reptend r with 2k digits as in1x=0.⋯r1r2⋯r(2k)―,you will often find that ri+r(k+i)≡−1±odb for all i≤k

Step 1That the decimal expansion of 1p repeats after 2k digits means b2k≡1bmodpwhich means bk≡±1bmodpand if it doesn't repeat after k digits that meansbk≡p−1bmodpThus, if d1p=0.r1r2…r2k, then(r1r2⋯rk)×p=bk−(p−1)and (rk+1rk+2⋯r2k)×p=(p−1)×bk−1,so (r1r2⋯rk+rk+1rk+2⋯r2k)×p=p×(bk−1)i.e., r1r2⋯rk+rk+1rk+2⋯r2k=bk−1

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Noelle King

Answered question

2022-03-12

Why are repeating decimals often self-inversions?
Given a reptend r with 2k digits as in

\frac{1}{x} = 0 . \dots \overset{―}{r_{1} r_{2} \dots r_{(2 k)}}

,
you will often find that

r_{i} + r_{(k + i)} \equiv - 1 \pm o d b

for all

i \leq k

Answer & Explanation

Jamiya Bradford

Beginner2022-03-13Added 6 answers

Step 1
That the decimal expansion of

\frac{1}{p}

repeats after 2k digits means

b^{2 k} \equiv 1 b mod p

which means

b^{k} \equiv \pm 1 b mod p

and if it doesn't repeat after k digits that means

b^{k} \equiv p - 1 b mod p

Thus, if

d \frac{1}{p} = 0 . r_{1} r_{2} \dots r_{2 k}

, then

(r_{1} r_{2} \dots r_{k}) \times p = b^{k} - (p - 1)

and

(r_{k + 1} r_{k + 2} \dots r_{2 k}) \times p = (p - 1) \times b^{k} - 1,

(r_{1} r_{2} \dots r_{k} + r_{k + 1} r_{k + 2} \dots r_{2 k}) \times p = p \times (b^{k} - 1)

i.e.,

r_{1} r_{2} \dots r_{k} + r_{k + 1} r_{k + 2} \dots r_{2 k} = b^{k} - 1

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