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Thread: Binary Operation

  1. #1
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    Binary Operation

    A binary operation $\displaystyle *$ is defined on the set $\displaystyle S=\{a,b,c\}$ by $\displaystyle x*y=x$ for all $\displaystyle x,y\inS$. Determine which of the properties (Associative, Identity, Inverse, Commutative) are satisfied by $\displaystyle (S,*).$

    Solution:
    Let $\displaystyle x,y$, and $\displaystyle z$ be any three elements of $\displaystyle S$ (distinct of not). Then $\displaystyle x*(y*z) = x*y=x$, while $\displaystyle (x*y)*z=x*z=x$. Thus (S,*) is associative. Now (S,*) has no identity since for every element $\displaystyle e \in S$, it follows that $\displaystyle e*a=e*b=e$ and so it is impossible for $\displaystyle e*a=a$ and $\displaystyle e*b=b$. Since (S,*) has no identity, the question of inverses does not apply here. Certainly, $\displaystyle (S,*) $ is not commutative since $\displaystyle a*b=a$ while $\displaystyle b*a=b$.

    Question:
    I suppose $\displaystyle e$ is an identity element. Since (S,*) has no identity why it contradict itself by saying "(S,*) has no identity since for every element "$\displaystyle e\in S$"?

    $\displaystyle e*a=a$ and $\displaystyle e*b=b$ make sense, but what is $\displaystyle e*a=e*b=e$?
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  2. #2
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    Quote Originally Posted by novice View Post
    A binary operation $\displaystyle *$ is defined on the set $\displaystyle S=\{a,b,c\}$ by $\displaystyle x*y=x$ for all $\displaystyle x,y\inS$. Determine which of the properties (Associative, Identity, Inverse, Commutative) are satisfied by $\displaystyle (S,*).$

    Solution:
    Let $\displaystyle x,y$, and $\displaystyle z$ be any three elements of $\displaystyle S$ (distinct of not). Then $\displaystyle x*(y*z) = x*y=x$, while $\displaystyle (x*y)*z=x*z=x$. Thus (S,*) is associative. Now (S,*) has no identity since for every element $\displaystyle e \in S$, it follows that $\displaystyle e*a=e*b=e$ and so it is impossible for $\displaystyle e*a=a$ and $\displaystyle e*b=b$. Since (S,*) has no identity, the question of inverses does not apply here. Certainly, $\displaystyle (S,*) $ is not commutative since $\displaystyle a*b=a$ while $\displaystyle b*a=b$.

    Question:
    I suppose $\displaystyle e$ is an identity element. Since (S,*) has no identity why it contradict itself by saying "(S,*) has no identity since for every element "$\displaystyle e\in S$"?
    $\displaystyle e$ is not an identity element, it is just a letter. You can think of it as "candidate for identity element." The text says "for every element $\displaystyle e \in S$", not "for an identity element $\displaystyle e \in S$". The text does not contradict itself.

    Quote Originally Posted by novice View Post
    $\displaystyle e*a=a$ and $\displaystyle e*b=b$ make sense, but what is $\displaystyle e*a=e*b=e$?
    Look at the definition of identity element. The logic here is for left identity element in particular. The point is that if $\displaystyle e$ is an identity element then $\displaystyle e*a = a$ and $\displaystyle e*b = b$, therefore $\displaystyle e*a\ne e*b$. This produces a contradiction since we know $\displaystyle e*a=e*b=e$. Therefore, no matter what choice of $\displaystyle e$ we make, $\displaystyle e$ is not an identity element.
    Last edited by undefined; Aug 4th 2010 at 11:08 AM. Reason: typo
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    Quote Originally Posted by undefined View Post
    $\displaystyle e$ is not an identity element, it is just a letter. You can think of it as "candidate for identity element
    If $\displaystyle e$ is only a possible candidate, the what operational property is there that can produce $\displaystyle e*a=e*b=e$, or is this just non-sense created for making the point.
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  4. #4
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    Quote Originally Posted by novice View Post
    If $\displaystyle e$ is only a possible candidate, the what operational property is there that can produce $\displaystyle e*a=e*b=e$, or is this just non-sense created for making the point.
    $\displaystyle e$ is an element of $\displaystyle S$. Direct application of the definition of $\displaystyle *$ gives $\displaystyle e*a=e*b=e$, regardless of what $\displaystyle e$ is. It is not nonsense.
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  5. #5
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    Thank you sir. All is clear now.
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