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Q.Construct the composition table for the operation a∗b=a+b(mod5)a * b = a+b \pmod 5 on the set A={0,1,2,3,4}A = \{0,1,2,3,4\}. Also answer the following :

(i) Is ∗* a binary operation on AA?
(ii) Find the identity element for ∗*.
(iii) Write the inverse elements of 33 and 44.
Odisha ChseOdisha CHSE +2 Science Board Exam 2024Subjective· 6mImportance★★★★★
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The composition table for a∗b=(a+b) mod 5a*b=(a+b)\bmod5 on {0,1,2,3,4}\{0,1,2,3,4\} shows closure (a valid binary operation), identity element 00, and inverses 3−1=23^{-1}=2, 4−1=14^{-1}=1.

Composition table for a∗b=(a+b) mod 5a*b = (a+b)\bmod5 on A={0,1,2,3,4}A=\{0,1,2,3,4\}:

∗*01234
001234
112340
223401
334012
440123

(i) Is ∗* a binary operation on AA? Every entry in the table lies in AA, so for all a,b∈Aa,b\in A, a∗b∈Aa*b\in A (closure holds). Yes, ∗* is a binary operation on AA.

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