I\'m having trouble with two theoretical proofs: 1) Let p be a prime integer. Pr

Question

I'm having trouble with two theoretical proofs:
1) Let p be a prime integer. Prove that if[a][b] = [0] in Zp, then either [a] = [0]or [b] = [0].
and
2) Let p be a prime integer. Prove that [1] and[p-1] are the only elements inZp that are their own multiplicativeinverses.
I understand the concepts and can do the work with appliednumbers, but the proofs have me a bit stymied. Thanks inadvance for any assistance.
1) Let p be a prime integer. Prove that if[a][b] = [0] in Zp, then either [a] = [0]or [b] = [0].
and
2) Let p be a prime integer. Prove that [1] and[p-1] are the only elements inZp that are their own multiplicativeinverses.
I understand the concepts and can do the work with appliednumbers, but the proofs have me a bit stymied. Thanks inadvance for any assistance.

Explanation / Answer

QuestionDetails:
1) Let p be a prime integer. Prove that if[a][b] = [0] in Zp, then either [a] = [0]or [b] = [0].
and
ZP CONSISTS OF CONGRUENCE CLASSES
[0];[1];[2]....[P-1]
GIVEN THAT [A][B]=0
================================================================
THE FOLLOWING IS FOR YOUR UNDERSTANDING...
NOT NECESSARY TO PUT THIS IN PROOF.
[A]=A1+PZ
[B]=B1+QZ
WHERE P AND Q ARE INTEGERS AND 0<=A1<P AND 0<=B1<P
HENCE
[A1+PZ][B1+QZ]=0 [MOD Z]
[A1B1+Z(PB1+QA1+PQZ)][MOD Z]=A1B1 [MOD Z] = 0 [MOD Z]
A1B1=0
HENCE A1=0 OR B1=0....
THAT IS [A]=0 OR [B]=0
====================================================================
PROOF
THIS IMPLIES [AB]=0
THAT IS P IS A DIVISOR OF AB
THAT IS P|A OR P|B
SINCE IF A AND B ARE ANY 2 INTEGERS AND P IS A PRIME NUMBER
THEN P|AB IMPLIES P|A OR P|B
THAT IS [A]=0 OR [B]=0

2) Let p be a prime integer. Prove that [1] and[p-1] are the only elements inZp that are their own multiplicativeinverses.
FROM THE ABOVE
LET US TAKE A MODULAR CLASS [X] WHERE ANY ELEMENT IN THE CLASS ISGIVEN BY
X1+PZ WHERE P IS AN INTEGER AND X1=0 OR 1<X1<P-1
[THAT IS WE ARE EXCLUDING [1] and [p-1] ]
TPT
[X1+PZ][X1+QZ] IS NOT EQUAL TO [1]
TPT
[X1*X1+Z{(P+Q)+PQZ}] IS NOT EQUAL TO [1]
TPT
X1*X1 IS NOT EQUAL TO 1[MOD Z]
TST
X1*X1-1 IS NOT EQUAL TO 0[MOD Z]
X1*X1-1=(X1+1)(X1-1)
SINCE P IS PRIME,X1+1 <P AND HENCE X1-1<P [SINCE X1<P-1] ,AND HENCE X1-1<P
OBVIOUSLY P CAN NOT BE PRODUCT OF 2 FACTORS AS THEN IT WILL BECOMECOMPOSITE.
HENCE NO RESIDUE CLASS ...IN 1<X1<P-1 HAS IT SELF ASINVERSE
SIMILARLY X1=0 IS ALSO NOT ITS OWN INVERSE
SINCE THEN OBVIOUSLY [X1][X1]=0 NOT ONE
BUT WHEN X1= 1 , WE CAN EASILY SEE THAT X1*X1 =1 AND HENCE [X] ISITS OWN INVERSE
WHEN X1=P-1, W GET
(P-1)(P-1)=1-2P+P2 = 1 [MOD P] THAT IS
FOR X=P-1 , [X] IS INVERSE OF ITSELF....PROVED



I understand the concepts and can do the work with appliednumbers, but the proofs have me a bit stymied. Thanks inadvance for any assistance.

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