≺ Question− algebra ≻ Count the number of ” zero divisor ” of ring , ( Z


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Question Number 175132 by mnjuly1970 last updated on 20/Aug/22

$≺ Q u e s t i o n - a l g e b r a ≻$ $C o u n t t h e n u m b e r o f^{″} z e r o d i v i s o r^{″}$ $o f r i n g, (Z_{45},, \oplus) ◼ m . n$ $- - - - - - - -$
Commented by kaivan.ahmadi last updated on 20/Aug/22
$we must find the number of digits k such that (k,45)≠1. ∅(45)=φ(3^2 ×5)=45(1−(1/3))(1−(1/5)) =45((2/3))((4/5))=24 ⇒number of zero divisor= 44−24=20 or K={3,5,6,9,10,12,15,18,20,21 ,24,25,27,30,33,35,36,39,40,42} for each k∈K ; (45,k)≠1 and ∣K∣=20.$
$w e m u s t f i n d t h e n u m b e r$ $o f d i g i t s k s u c h t h a t$ $(k, 45) \neq 1 .$ $\emptyset (45) = ϕ (3^{2} \times 5) = 45 (1 - \frac{1}{3}) (1 - \frac{1}{5})$ $= 45 (\frac{2}{3}) (\frac{4}{5}) = 24$ $\Rightarrow n u m b e r o f z e r o d i v i s o r =$ $44 - 24 = 20$ $o r K = {3, 5, 6, 9, 10, 12, 15, 18, 20, 21$ $, 24, 25, 27, 30, 33, 35, 36, 39, 40, 42}$ $f o r e a c h k \in K; (45, k) \neq 1 a n d$ $∣ K ∣= 20 .$
Commented by mnjuly1970 last updated on 21/Aug/22

$t h a n k y o u s o m u c h . .$ $i n f a c t$ ${[n - (φ (n) + 1)] = e a s y}$ $= 45 - 25 = 20$
Commented by kaivan.ahmadi last updated on 20/Aug/22

$y e s . a n d s o i f Z_{n} i s a f i e l d$ $i t h a s n o z e r o d i v i s o r .$ $s i n c e n = p = p r i m e a n d$ $p - 1 - ϕ (p) =$ $p - 1 - p (1 - \frac{1}{p}) = p - 1 - p + 1 = 0$
Commented by mnjuly1970 last updated on 21/Aug/22

$z e n d e h b a s h i d . . .$
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