Show that zero is the identity for this operation and (6 – a)...?
. Define a binary operation ∗ on the set {0, 1, 2, 3, 4, 5} as
a * b = a + b if a + b < 6
= a + b – 6 if a + b ≥ 6
Show that zero is the identity for this operation and each element a of the set is
Invertible with (6 – a) being the inverse of a.
Doubt : Should not 6 also be a n element of the set ? if yes or no - want explanation
Update:This is one question from NCERT math book for class 12 as stated above.( Miscellaneous Exercise on Chapter 1 Q. No. 14 ) The proof is simple but the inverse operator must work on all the elements of the given set. Hence it should work on 0 as well which implies 0*(6-a) = 0*(6-0) = 0*6 = 0 + 0+6 -6 = 0 - fine. But this implies 6 has to be an element of the set. Time extended to get more answers.
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Sure!. Yes!
To show that 0 is the identity, let a be any one of the integers in the set. Then:
0 * a = 0 + a = a = a + 0 = a * 0.
Since 0 * a will NEVER be greater than 6 (as 0 + a is at most 5), 0 is clearly the identity element. Let us show that 6 - a is the inverse for the * operation.
Suppose that for a is an element of the set. Then 6 - a is also in the set. Since 6 - a + a = 6 for each element a in the set, then:
a * (6 - a) = a + (6 - a) - 6 = (a + (-a)) + (6 + (-6)) = 0.
and
(6 - a) * a = (6 - a) + a - 6 = 0
This completes your proof.
Essentially, you are defining " * " as an operation in Modulo 6 This means you have six valid values (0,1,2,3,4,5) in the set. So, no, "6" is not part of the set.
Will leave you with the proof - it's the "additive unity" for the operation (a * 0 = 0 * a = a)
Yes! .