Question list

Prove that if p is an odd positive prime integer that

m = 2 * 3 * . . . * (p-2) will have remainder of 1

when divided by p.

If R is a ring with characteristic 2, and every element is

idempotent, prove that multiplication in R is communitive.

Question # 1

Prove that if p is an odd positive prime integer that

m = 2 * 3 * . . . * (p-2) will have remainder of 1

when divided by p.

There are (p-1) numbers in the set of integers,

{2,3,4,. . . (p-2) }.

It is a theorem in beginning number theory that for any

integer, K, in the set {2,3,4,. . .(p-2)},

there exist another, different, integer, J, in the same set,

such that L = K * J has remainder 1, when divided by p.

Group the integers in the set {2,3,4,. . .(p-2)} into pairs,

such that in each pair the product of the pair has remainder

1 when divided by p.

Another theorem implies that if t1 has remainder 1 when

divied by p, and t2 also has remainder 1 when divided by p,

then (t1 * t2) also has remainder 1 when divided by p.

Multiply all the (p-1) numbers together in the order of the

pairings. Each pair product has remainder 1 when divided by

p.

Thus the product of all the pair products has remainder 1

when divided by p.

m = 2 * 3 * . . . * (p-2) has remainder 1 when divided by p.

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Question # 2

If R is a ring with characteristic 2, and every element is

idempotent, prove that multiplication in R is communitive.

Let s and t be two arbitrary elements of R.

Since every element in R is idempotent,

(s + t)**2 = (s + t)

s**2 + s t + t s + t**2 = s + t

s**2 = s

t**2 = t

s + s t + t s + t = s + t

s t + t s = 0

t s is the unique additive inverse of s t

Since R has characteristic 2, t s = s t.

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