1 Introduction to Modulo 7 Arithmetic

Transcription

1 1 Introution to Moulo 7 Arithmti Bor w try our hn t solvin som hr Moulr KnKns, lt s tk los look t on moulr rithmti, mo 7 rithmti. You ll s in this sminr tht rithmti moulo prim is quit irnt rom th ons w t rom omposit numrs. Mo 7 rithmti is it sir us w n lwys solv qutions lik x = or x whn 0 n r ivn. W strt with th ition n multiplition tls or moulo 7 rithmti W r usin n to hlp you unrstn tht ths r not th usul oprtions. Noti w thins out th tl. First, thr is n intity, th numr 1. It stisis 1 x = x 1 = x or vry numr x. Nxt noti tht h non-zro numr hs 1 in its row. For xmpl row hs 1 in th olumn ll 4. In othr wors 4 = 1. This is nothr wy to sy tht n 4 r multiplitiv invrss o h othr. So suppos you h linr qution lik x =. You n multiply on th lt y 4 to t 4 ( x) = 4 = n sin 4 ( x) = (4 ) x = 1 x = x, w hv our solution x =. Wht proprty i w us in th omputtion ov? Th nswr is ssoitivity o. Noti lso tht th numrs 1,, n 4 r squrs, lso ll qurti rsius. Also noti tht just lik in our systm o rl numrs, non-zro numrs tht hv squr roots hv two o thm (rouhly orrsponin to positiv n ntiv squr root). For xmpl 4 hs squr roots n whih r itiv invrss o h othr. Hr s proo tht two numrs with th sm squrs r itiv invrss: x y = s s = 0, so (x y) (x+y) = 0. Th irst tor n t 0, so th son must. Noti tht this rquirs th so-ll zro prout prinipl, nmly tht th prout o two numrs is zro i n only i on o th numrs is zro. This prinipl hols whn th moulr s is prim numr, ut not or omposit numrs. Th two oprtions n r lso ommuttiv. Tht is x y = y x. Do you s vry sy wy to hk this? Bsis ommuttivity, thr is lot mor strutur thn w n or n to isuss. Th 7 ojts w nm 0, 1,..., r tully quivln lsss o intrs unr th inry rltion x y prisly 1

2 whn x n y ir y multipl o 7. So or xmpl th symol zro is rlly th st {0, ±7, ±14,...}. Th st {0, 1,..., } with th two oprtions n is wht is known s Finit Fil or Glois Fil. Finit ils r unmntl in numr o rs o mthmtis n omputr sin, inluin numr thory, lri omtry, Glois thory, init omtry, ryptorphy n oin thory. Division n sutrtion in Moulo 7 Arithmti Nxt w uil th sutrtion n ivision tls or moulo 7 rithmti Look t th ol in th tl. This shows tht sutrtion is littl ountrintuitiv. W hv 3 4 = us 3 = 4. Atully howvr, its not so. W n think o 3 4 s 1, y whih w mn th itiv invrs o 1, whih w writ s. For th tl, w hv = xtly whn =. Th tl n uilt s ollows. = 1. Thus, or xmpl 4 = 4 1 = 4 3 = n 4 = 4 1 = = 3. Noti in tht, unlik stnr KnKn, whr w tk th lrr numr ivi y th smllr on in two-ll, hr oth n r in. This mks ths puzzls muh hrr to solv. W n us ths tls to uil nit sts or ivn trt vlus. For xmpl, suppos w h two-ll 3. Th nit sts r 03, 14,, 3, 04, 1,. 3 A hr Exmpl At this point w ll work our wy throuh hr mo 7 KnKn. Us moulo 7 rithmti to solv th puzzl low. Us th st {0, 1,,,, }. Noti tht w r usin th symols,,, n whn n, to rmin you tht w r lin with irnt oprtions. You n not us ths symols whn you uil your own puzzls.

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4 W n mk som prorss y notiin tht th 0 in h row nnot ppr in. Cn you s why? So th 0 s in rows n lon to th 4 n th s. This ixs th 4 s {0, 4} n th s ithr {0, 3, } or {0,, }. Sin th n ithr {1, 3}, {, } or {4, }, w n limint th st {, }. Th in row hs nit sts {1, } n {4, }, so limint th lttr sin it is inomptil with {0, 4}. Also not tht th 0 o olumn 1 n o only in 1. So w hv , 1, 0, 4 0, 4 3, 3, 4

5 Sly, thr r two istint wys to ill th union o th rows n n th olumns n 7. On wy is ivn hr: ,, 4,, 0,, 3,, 0,, ,, , 4 0, 4 4, 4, S i you n in ontrition thus showin tht this ws th wron hoi or 7.

6 Hr is th othr wy to ix th ottom two rows n lst two olumns ,, 1,,,, 0,, ,, 3 0,, , 4 0, 4 1, 3 1, S i you n inish this up. Look t olumn, n ount. Th must hv n orinry sum o 8, so th (whih is th sm lu s, must hv sum o =. In s you on t s whr to o rom hr, try to ix 1, knowin tht th orinry sum o th ntris in th in row is 8.

7 Th uniqu omplt solution is ivn low