Abstract Mathematical Systems: Groups and Fields
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Closure | Associative | Identity | Inverse | Commutative | Distributive
In this section we will be examining different systems and checking for the following lists of properties and characteristics. Depending on our results, we will classify systems by the following three classifications:
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| Abelian Group |
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| Field |
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Closure: This property insures that whatever two items are operated on, the result will still be an element in the system. Below is a table of two systems. The first system has closure and the second one does not: The second example below has two results (D and E) that are not defined as elements of the orginal system. Therefore, the second system below is not closed, or we say that it does not have closure.
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Associative Property: This is the same associative property that you have known from the number system you have grown up with. Given three different elements grouped in two different ways, the results will be the same. Below is a table of two systems. The first system has the associative property and the second one does not. Notice that with the system on the left, you should check for the property at least two times. You can see in the second system below that the first combination we tried worked, but the second failed. Therefore the second system below is not associative.
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C*(A*B)=(C*A)*B C*A=B*B B=B B*(A*C)=(B*A)*C |
C*(A*B)=(C*A)*B C*A=B*B B=B B*(A*B)=(B*A)*B |
Identity Element: What is
the number in normal addition that when added results in no change in the
addition? Of course, it's zero. Zero is the identity element for normal
addition. In the system at the left below, what is the number that results in no
change for any number? It appears to be "B." Notice that whenever B is
starred by any other symbol that the result is always unchanged. Does this also
work in the reversed mode? Is there a column which also "duplicates"
the label column at the left? The answer is again "B." If you check
this in both directions, you are assured that in this case B is the
identity element for the system. In the system at the right below,
notice that there is no row which "duplicates" the row at the top.
Neither is there any column which "duplicates" the column at the left.
Therefore the second system below does not have an identity element.
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Inverse: Here we need to
check if there is an inverse for each element of the system. We see that B is
the identity element of the system below at the left. Now, we look for an
identity element (B) in each row and each column. We see that this works. This
tells us that the system at left below does pass the inverse test. On the system
at the right below, we see that here too, B is the identity element. Therefore
we will check for an identity element (B) in each row and column. We see that
there is no identity element in the A row, and that there is no identity element
in the C column. Therefore this system fails the inverse test.
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Commutative Property:
This one is easy to check if you have the system table. Draw a top-left to
bottom-right 45 degree angled straight line through the table. If you have the
commutative property the table will have duplicate entries on either side of the
diagonal. Let's examine the two systems below for the commutative property. Look
at the system at the left. With the diagonal drawn in you can see that all of
the elements match on either side of the diagonal. (colors have been added to
show the matches). This system has the commutative property. The system on the
right does not have all elements matching. You can see this by looking at the
two that are highlighted that do not match. This system does not have the
commutative property.
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Distributive Property: The final property to check is the distributive property. To even have this property requires two operations. If you have only one operation given, don't even bother checking this one. To check for the distributive property, pick three elements at random and in random order. Let's pick B, C, A. In checking for * being distributive over % on the left side below, we tried three different random selections and they all worked. So it appears that * is distributive over %. But when we checked to see if % is distributive over * (right side below), we found that it was not.
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B*(C%A)=(B*C)%(B*A) B*C=C%A C=C A*(B%B)=(A*B)%(A*B) B*(C%B)=(B*C)%(B*B) |
B%(C*A)=(B%C)*(B%A) B%B=C*B B  C |
B