1 / 10

Discovery in Mathematics an example (Click anywhere on the page)

Discovery in Mathematics an example (Click anywhere on the page). Repeating decimal for 1/7 (click screen for the next step). 0. ____________. 1 7 3. 4 0 28 2. 2 0 14 6. 8 0 56 4. 5 0 35 5. 7 0 49 1. 142857…. 7) 1 .000000000000…. 1 _ 1 – x.

ferdinand-joshua
Download Presentation

Discovery in Mathematics an example (Click anywhere on the page)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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)

More Related