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Let n belongs to N and given a belongs to Z, let a* represent the congruence class of a modulo n. if a1* = a2* and b1* = b2* for integers a1,a2,b1,b2, then prove that (a1b1)* = (a2b2)*.

if p is a prime, explain why all elements in Z / pZ have multiplicative inverses

I have tried but cant go anywhere with it. need help. thank you.

2006-09-27 10:43:05 · 4 answers · asked by David F 2 in Science & Mathematics Mathematics

4 answers

yes i do. How do family doctors use the scientific method?

2006-09-27 10:46:18 · answer #1 · answered by moni911 1 · 0 3

First, note the following fact: For two integers x and y, x*=y* if and only if x = y + kn for some integer k.

So there is an integer k such that a1 = a2 + kn, and an integer j such that b1 = b2 + jn. Now multiply and see what you get:

a1b1 = ___________.

The second problem is false as stated. It should be something like:

if p is a prime, explain why all elements a* in Z/pZ, with a* not equal to 0*, have multiplicative inverses.

Note that a*=0* if and only if a = 0 + kp = kp, that is, a* = 0* if and only if a is divisible by p. Thus if a* is not equal to 0*, then a and p are relatively prime. That is, the gcd of a and p is 1. Thus there exist integers m and n such that

a X ___ + p X ____ = _____.

Fill in the blanks and use the fact above. The X represents multiplication.

2006-09-27 18:05:10 · answer #2 · answered by the7nt_man 2 · 0 0

1. about why (a1b1)*=(a2b2)* under the given conditions : well you know the following -
a1=k1n+a1*
a2=k2n+a2*
b1=q1n+b1*
b2=q2n+b2*
where k1,k2,q1,q2 belong to N, a1*=a2*, b1*=b2*, and the stars are numbers between 0 and n-1.
so
a1b1 = (k1q1+k1b1*+q1a1*)n + a1*b1*
a2b2 = (k2q2+k2b2*+q2a2*)n + a2*b2*

modulo n, we can neglect whatever multiplies n, and so we get

(a1b1)*=(a1*b1*)*=
=(a2*b2*)*=(a2b2)*
where the middle equality is because a1*=a2*, b1*=b2*.

now about part two... maybe later - I gotta go now.

2006-09-27 18:05:36 · answer #3 · answered by No Mo 2 · 0 0

lol @ moni

well use the following linear congruency
ax ≡ 1 (mod n) [multiplicative inverse]
and what it means for n to be prime
and Z/nZ is the set of equivalence classes
also don't forget to be correct about 0 [0 is NOT coprime to p]

2006-09-27 18:09:24 · answer #4 · answered by xkey 3 · 0 0

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