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Exploring Repeating Decimals and Prime Numbers in Mathematical Proofs

This exploration delves into the fascinating relationship between repeating decimals and prime numbers, exemplified by the decimal representation of 1/7, which yields a repeating cycle of 142857. It discusses mathematical identities and their implications, including Fermat's Little Theorem, and posits questions about expressing relationships mathematically. The study further investigates how these principles apply to various integer scenarios, particularly within our base-10 system, and speculates on their potential for generalization.

ferdinand-joshua
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Exploring Repeating Decimals and Prime Numbers in Mathematical Proofs

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  1. Discovery in Mathematicsan example (Click anywhere on the page)

  2. Repeating decimal for 1/7(click screen for the next step) 0.____________ 1 73 4 0282 2 0146 8 0564 5 0355 7 0491 142857… 7)1.000000000000…

  3. 1_1 – x = 1 + x + x2 + x3 + · · · 17 = 0.142857142857142857 . . . (Proof involves the identity where x = 10–6) 142857999999 17 = is a repeating decimal with a period of 6. Can this statement be expressed mathematically?

  4. “For which ns does 10n–1 – 1 n integer ?” = 142857999999 17 = 999999 7 142857 (an integer) = 107–1 – 1 7 142857 (an integer) = Is there an interesting question here? ( ) one that generalizes from

  5. (10n–1–1) / n 9 / 2 = 4.599 / 3 = 33999 / 4 = 249.759999 / 5 = 1999.899999 / 6 = 16666.5999999 / 7 = 1428579999999 / 8 = 1249999.87599999999 / 9 = 11111111999999999 / 10 = 99999999.99999999999 / 11 = 90909090999999999999 / 12 = 8333333333.25999999999999 / 13 = 0769230768239999999999999 / 14 = 714285714285.64399999999999999 / 15 = 6666666666666.6999999999999999 / 16 = 62499999999999.99999999999999999 / 17 = 58823529411764799999999999999999 / 18 = 5555555555555555.5999999999999999999 / 19 = 526315789473684219999999999999999999 / 20 = 499999999999999999.9599999999999999999999 / 21 = 4761904761904761904.714999999999999999999999 / 22 = 45454545454545454545.4099999999999999999999999 / 23 = 434782608695652173913

  6. 10n–1 – 1 n 10n–1 – 1 n integer integer = = holds for all prime n except 2 and 5 holds for all non-prime n, except 9 Are we just lucky that we use the base-10 system? What about an–1 – 1 n integer ? = Observations (factors of 10). = 10 – 1. Can we generalize further?

  7. a = 2 (2n–1–1) / n 1 / 2 = 0.53 / 3 = 17 / 4 = 1.7515 / 5 = 331 / 6 = 5.16763 / 7 = 9127 / 8 = 15.875255 / 9 = 28.333511 / 10 = 51.11023 / 11 = 932047 / 12 = 170.5834095 / 13 = 3158191 / 14 = 585.07116383 / 15 = 1092.232767 / 16 = 2047.93865535 / 17 = 3855131071 / 18 = 7281.722262143 / 19 = 13797524287 / 20 = 26214.351048575 / 21 = 49932.1432097151 / 22 = 95325.0454194303 / 23 = 182361

  8. 2n–1 – 1 n 2n–1 – 1 n 2n–1 – 1 n integer integer integer = = = holds for all prime n except 2 (which is a “factor” of a = 2). holds for non-prime n,from 2 to 23 at least. But for n = 341 (a non-prime) we find that Observations

  9. an–1 – 1 n integer = if n is prime and not a factor of a. Conjecture (Guess) 2n–1 – 1 n based on integer observations? = mod(an–1 – 1, n) = 0if n is prime and not a factor of a. This is “Fermat’s Little Theorem”

  10. Usefulness of Fermat’s Little Theorem Test for Primality mod(an–1 – 1, n) = 0almost only if n is prime and not a factor of a. Allows “Public Key Encryption” Pick p = 4099, q = 4111, m = 2 (p and q prime)c = (p – 1)(q – 1)·m + 1 = 33685561c = A·B, A= 2821 (public), B = 11941 (secret)N = p·q = 16850989 (public)x is the secret messageEncrypt: y = mod(xA,N), Decrypt: x = mod(yB,N)

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