Show that zero is the identity for this operation and (6 – a)...?

Question

. 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

Anonymous · Accepted Answer

Sure!. Yes!

Mewtwo · Answer

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.

L. E. Gant · Answer

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)

? · Answer

Yes! .

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Show that zero is the identity for this operation and (6 – a)...?

4 Answers