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OpenStudy (anonymous):

Let * be a binary operation on the set {0, 1, 2, 3, 4, 5}, defined as a *b={(a+b, if a+b <6 & a+b-6,if a+b≥6) Show that zero is the identity element for this operation and each element a ≠ 0 of the set is invertible with 6 – a being the inverse of a.

OpenStudy (anonymous):

Just show that a*0=0*a=a for all a in the set {0,1,2,3,4,5} for the first part. For the second part, show that (6-a)*a=a*(6-a)=0.

OpenStudy (anonymous):

Also note that since you cannot assume, a priori, that this operation is commutative, you must show both a*0=0 and 0*a=0 (for instance).

OpenStudy (anonymous):

Ok tell me if Im doing it right

OpenStudy (anonymous):

a*0=a+0=a 0*a=0+a=a a*0=0*a=a

OpenStudy (anonymous):

0 is the identity element

OpenStudy (anonymous):

Looks good so far

OpenStudy (anonymous):

(proof of commutativity is straightforward since a [+] b is either a+b=b+a or else a [+] b is a+b-6=b+a-6.) let's see a+0 is always less than 6 so a[+]0 = 0[+]a = (a+0) = a, yep

OpenStudy (anonymous):

You may want to note that in each of those examples, clearly a+0 is less than 6

OpenStudy (anonymous):

Yes ,what about the second part ?

OpenStudy (anonymous):

@Broken: I know, but when people are first learning things like relation theory and algebra, I find it best when they don't dive in automatically assuming things are as nice as they are used to! (ie. commutative, have inverses, etc.)

OpenStudy (anonymous):

Just show that (6-a)*a=a*(6-a)=0 for all a in {0,1,2,3,4,5}

OpenStudy (anonymous):

(6-a)*a=6-a+a=6 ? what did i do wrong?

OpenStudy (anonymous):

ok 6-a+a-6=0 :)

OpenStudy (anonymous):

Mhm, very good. Thus, you can conclude that 0 is the identity element for this operation (part 1) and that all elements have inverses (part 2).

OpenStudy (anonymous):

but how do i know if a+b <6 or a+b≥6

OpenStudy (anonymous):

and sorry how is a-6 inverse of a?

OpenStudy (anonymous):

Well, in part 1, you know that all of them must be less than 6, since you are adding 0 to a number less than 6. For part 2, clearly a and 6-a add up to 6 which is greater than or equal to 6. I don't understand your question, I never said a-6 is the inverse of a, since 6-a is the inverse of a.

OpenStudy (anonymous):

yes sorry i meant 6-a

OpenStudy (anonymous):

Since (6-a)*a=a*(6-a)=0, which is the identity element, then a and 6-a are considered inverses to each other. When dealing with operations other than standard addition or multiplication on the reals, inverse does not always mean either negative or reciprocal like you are used to. All an inverse of an element "a" is, is another element "b" such that a*b=0, which 0 is the identity element.

OpenStudy (anonymous):

Ok Thank You I appreciate your help , you deserve a medal ;)

OpenStudy (anonymous):

No problem, always glad to help :)

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