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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+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.

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  1. anonymous
    • 5 years ago
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    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.

  2. anonymous
    • 5 years ago
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    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).

  3. anonymous
    • 5 years ago
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    Ok tell me if Im doing it right

  4. anonymous
    • 5 years ago
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    a*0=a+0=a 0*a=0+a=a a*0=0*a=a

  5. anonymous
    • 5 years ago
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    0 is the identity element

  6. anonymous
    • 5 years ago
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    Looks good so far

  7. anonymous
    • 5 years ago
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    (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

  8. anonymous
    • 5 years ago
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    You may want to note that in each of those examples, clearly a+0 is less than 6

  9. anonymous
    • 5 years ago
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    Yes ,what about the second part ?

  10. anonymous
    • 5 years ago
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    @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.)

  11. anonymous
    • 5 years ago
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    Just show that (6-a)*a=a*(6-a)=0 for all a in {0,1,2,3,4,5}

  12. anonymous
    • 5 years ago
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    (6-a)*a=6-a+a=6 ? what did i do wrong?

  13. anonymous
    • 5 years ago
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    ok 6-a+a-6=0 :)

  14. anonymous
    • 5 years ago
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    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).

  15. anonymous
    • 5 years ago
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    but how do i know if a+b <6 or a+b≥6

  16. anonymous
    • 5 years ago
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    and sorry how is a-6 inverse of a?

  17. anonymous
    • 5 years ago
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    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.

  18. anonymous
    • 5 years ago
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    yes sorry i meant 6-a

  19. anonymous
    • 5 years ago
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    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.

  20. anonymous
    • 5 years ago
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    Ok Thank You I appreciate your help , you deserve a medal ;)

  21. anonymous
    • 5 years ago
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    No problem, always glad to help :)

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