Not entirely sure I have my proof right (unless it really is as simple as I thin

Question

Not entirely sure I have my proof right (unless it really is as simple as I think), any help would be appreciated, full proofs get lifesaver. Assume that * is an operation on S with the identity element e and that   x*(y*z)=(x*z)*y for all x,y,z in S. Prove that * is commutative and associative. Not entirely sure I have my proof right (unless it really is as simple as I think), any help would be appreciated, full proofs get lifesaver. Assume that * is an operation on S with the identity element e and that   x*(y*z)=(x*z)*y for all x,y,z in S. Prove that * is commutative and associative.

Explanation / Answer

Abstract Algebra: Prove a binary operation is commutative, associative Question Details Not entirely sure I have my proof right (unless it really is as simple as I think), any help would be appreciated, full proofs get lifesaver. Assume that * is an operation on S with the identity element e and that x*(y*z)=(x*z)*y...........................1 for all x,y,z in S. Prove that * is commutative and associative. LET X,Y,Z BE ANY 3 ELEMENTS AS GIVEN COMMUTATIVE PROPERTY TST Y*Z=Z*Y... SINCE EQN. 1 IS TRUE FOR ALL X,Y,Z IN THE SET AND E IS THE IDENTITY ELEMENT OF THE SET LET US PUT X=E...WE GET E*[Y*Z]=[E*Z]*Y....SINCE E IS IDENTITY ELEMENT E*Z=Z.....E*[Y*Z]=Y*Z...SO WE GET Y*Z=Z*Y.........PROVED......COMMUTATIVE...SIMILARLY BY PUTTING Y=E OR Z=E WE CAN SHOW XY=YX...XZ=ZX....OK SO MULTIPLICATION IS COMMUTATIVE ...PROVED... ASSOCIATIVITY X*[Y*Z]=[X*Z]*Y........................1 BUT WE PROVED ABOVE THAT MULTIPLICATION IS COMMUTATIVE SO X*[Y*Z] = [Y*Z]*X....AND... [X*Z]*Y = Y*[X*Z] THAT IS WE HAVE FROM EQN.1 X*[Y*Z]=[Y*Z]*X=[X*Z]*Y=Y*[X*Z] THAT IS [Y*Z]*X=Y*[X*Z].........................................2 BUT AGAIN DUE TO COMMUTATIVE PROPERTY X*Z=Z*X...SO PUTTING IN EQN.2 WE GET [Y*Z]*X=Y*[Z*X]....THAT IS MULTIPLICATION IS ASSOCIATIVE ...PROVED...

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