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Old 2012-01-14, 16:57   #408
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the only binary operators I can think of that seem to fit either commutative or associative seem to fit both:

a*b=b*a a*(b*c)=(a*b)*c
a+b=b+a a+(b+c) = (a+b)+c
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Old 2012-01-14, 17:08   #409
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Quote:
Originally Posted by science_man_88 View Post
the only binary operators I can think of that seem to fit either commutative or associative seem to fit both:

a*b=b*a a*(b*c)=(a*b)*c
a+b=b+a a+(b+c) = (a+b)+c
Consider the inverse of each of these. For instance, is subtraction commutative and / or associative? Is division?

Another point: you ought to qualify what values your a, b and c may take. I'm being generous and assuming that you mean they are real numbers. I shouldn't have to make that assumption, not least because I may be making a false assumption.


Paul
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Old 2012-01-14, 17:51   #410
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Quote:
Originally Posted by xilman View Post
Consider the inverse of each of these. For instance, is subtraction commutative and / or associative? Is division?

Another point: you ought to qualify what values your a, b and c may take. I'm being generous and assuming that you mean they are real numbers. I shouldn't have to make that assumption, not least because I may be making a false assumption.

Paul
a-(b-c) = (a-b)+c as far as I know
and a-b!=b-a as far as I know

division a/b and b/a are inverse
a/(b/c) = a*(c/b)
(a/b)/c = a/(b*c)

yeah I'm not always thinking which is a bad habit, I was thinking naturals so you have a point ( especially since it makes no sense since a/b may be non-integer).

Last fiddled with by science_man_88 on 2012-01-14 at 17:55
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Old 2012-01-15, 01:16   #411
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\text{  }A\rightarrow \text {Cartesian product}\leftarrow B
\text{  }\uparrow \text {                                      }\uparrow
\text{   }a\leftarrow \text { binary relation } \rightarrow  b
\text {                   }\uparrow
\text {              function}\rightarrow \text {only b for every a}
\text {                   }\uparrow
\text {         binary operator}
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