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anonymous
 5 years ago
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+b6,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.
anonymous
 5 years ago
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+b6,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.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just 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 (6a)*a=a*(6a)=0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Also 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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok tell me if Im doing it right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a*0=a+0=a 0*a=0+a=a a*0=0*a=a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.00 is the identity element

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(proof of commutativity is straightforward since a [+] b is either a+b=b+a or else a [+] b is a+b6=b+a6.) let's see a+0 is always less than 6 so a[+]0 = 0[+]a = (a+0) = a, yep

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You may want to note that in each of those examples, clearly a+0 is less than 6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes ,what about the second part ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@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.)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just show that (6a)*a=a*(6a)=0 for all a in {0,1,2,3,4,5}

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(6a)*a=6a+a=6 ? what did i do wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mhm, 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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but how do i know if a+b <6 or a+b≥6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and sorry how is a6 inverse of a?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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 6a add up to 6 which is greater than or equal to 6. I don't understand your question, I never said a6 is the inverse of a, since 6a is the inverse of a.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes sorry i meant 6a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since (6a)*a=a*(6a)=0, which is the identity element, then a and 6a 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok Thank You I appreciate your help , you deserve a medal ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No problem, always glad to help :)
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