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 2012-01-14, 16:57 #408 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26×131 Posts 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
2012-01-14, 17:08   #409
xilman
Bamboozled!

"πΊππ·π·π­"
May 2003
Down not across

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Quote:
 Originally Posted by science_man_88 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

2012-01-14, 17:51   #410
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by xilman 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

 2012-01-15, 01:16 #411 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26·131 Posts $\text{ }A\rightarrow \text {Cartesian product}\leftarrow B$ $\text{ }\uparrow \text { }\uparrow$ $\text{ }a\leftarrow \text { binary relation } \rightarrow b$ $\text { }\uparrow$ $\text { function}\rightarrow \text {only b for every a}$ $\text { }\uparrow$ $\text { binary operator}$

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