[SOLVED] Binary Operations


by Arden1528
Tags: binary, operations, solved
Arden1528
#1
Sep11-03, 07:54 PM
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I believe I understand the critera to see of something is a "group"
1.It must be binary
2.It must be associative
3.Must contain a Identity
4.Must have a inverse
My question is the how to read a problem.
For example:
All real numbers minus zero under the operation a*b = abs(ab), does this make a group. Not looking for a answer, as this is homework, but I am looking for a little help. Thanks
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mathman
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#2
Sep11-03, 10:38 PM
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There is a problem in trying to define an identity. I could tell you more, but try to work it out yourself. You have to examine what happens with positive numbers and negative numbers.
HallsofIvy
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#3
Sep12-03, 04:46 AM
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Thanks
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Actually, I have a point about your grammar!!

When you say "It must be binary", what is "it"?

The only noun you have mentioned to that point is "group" so grammatically this would be "the group must be binary" which makes no sense. A group consists of a set of objects together with a binary operation on the set. It is the operation that is "binary" and it is the operation that must be associative. On the other hand it is the sat that "contains an identitity. It also would be better to say that "every member of the group has an inverse" rather that just "Must have an inverse".

It is easy to show that the identity in a group is unique.
If "a" and "b" are both identities, then, since a is an identity,
a*b= b. Since b is an identity, a*b= a. Thus, a= a*b= b.

There is, as mathman pointed out (and he should know!) a definite problem with a*b= |ab|. What is the identity?

MathematicalPhysicist
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#4
Sep12-03, 06:39 AM
P: 3,169

[SOLVED] Binary Operations


Originally posted by Arden1528
I believe I understand the critera to see of something is a "group"
1.It must be binary
2.It must be associative
3.Must contain a Identity
4.Must have a inverse
My question is the how to read a problem.
For example:
All real numbers minus zero under the operation a*b = abs(ab), does this make a group. Not looking for a answer, as this is homework, but I am looking for a little help. Thanks
when you say binary do you mean to the closure axiom?
mathman
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#5
Sep12-03, 07:40 PM
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A binary operation is a well defined term in mathematics. Specifically, it is a function combining two variables into a one variable result.
Arden1528
#6
Sep12-03, 07:53 PM
P: n/a
Well this is where I become a little hazy. I missed the first day of this class, so I missed the binary operation(*) lecture. My book says that
" a binary operation * on S is a function that associates to each ordered pair (x1,x2) of elements of S, which we denote by x1*x2." Could someone please translate this for me. A classmate tried to help me by showing me a example. If you are given all Reals with addition
(R,*) with a*b=a+b. You should be able to take any two real numbers and add them to get a real number, this would make it a binary operation. Mr. Math could you help me out please...
StephenPrivitera
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#7
Sep13-03, 04:39 PM
P: 364
Maybe it would help to know what a function is. A function from the set S to P means that for each element of S there is associated one and only one element of P. For example, consider the function F={(x,y) : y=sqrt(x)} from the natural numbers to the reals. By this we mean that for each x within the set of natural numbers there is associated one and only number to this x from the set reals. The function may look something like this: F={(1,1),(2,1.4142...),(3,1.73...),(4,2),(5,sqrt(5))...}.

A cross product SxS = {(x,y) : x in S, y in S}. A cross product is not a function because for x there are many associated y. For example, {1,2} x {1,2}= {(1,2),(1,1),(2,1),(2,2)}. For 1, there are associated both 1 and 2.

A binary operation is a function from SxS to S. The domain of the function is SxS. The range is S. For each element in the domain, there is one and only one element associated to it from the range. Really, you don't have to understand the first two paragraphs to understand a binary operation, but it helps. You could say the binary operation + on the reals is defined as a+b=c where a is real, b is real and c is real. This definition doesn't mention anything about a function or cross product. But the operation could be written + ={((a,b),c) : a,b,c are reals}. For each (a,b) there is associated only one element from the range (namely c). By this definition, 1+2=4pi is true so long as 1+2 does not equal any other number.

I haven't done your problem. But you're looking for an element in the set of reals such that a*I=I*a=|aI|=|Ia|=a. If there's no identity, there can't be an inverse. So check for the identity first.
MathematicalPhysicist
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#8
Sep19-03, 07:30 AM
P: 3,169
Originally posted by mathman
A binary operation is a well defined term in mathematics. Specifically, it is a function combining two variables into a one variable result.
in my text about group theory there is no mention about the term "binary" but to the the term "closure" so i think they are the same thing.
mathman
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#9
Sep19-03, 07:42 PM
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Closure and binary are two different things. A binary operation takes two arguments into one result. Something is closed under an operation (binary or otherwise) if the result of the operation is in the same set.
MathematicalPhysicist
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#10
Sep20-03, 03:00 AM
P: 3,169
if you can do me a favour and check the text perhaps it has a mistake in it.
the text:
http://members.tripod.com/~dogschool/groups.html

thanks mathman.
mathman
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#11
Sep20-03, 10:22 PM
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I didn't read it thoroughly, but as for closure, he is correct. Among the axioms for a group is that it is closed under the binary operation, i.e. the statement c=a.b (where . symbolizes the operation of taking a and b into c) imples that c is also in the group.

However, closure is a very general term and can be used in other contexts. Rational numbers are not closed under limiting operation, while real numbers are (assuming the operation is looking at a convergent sequence). In other words a convergent sequence of rational numbers can have an irrational number as a limit, but a convergent sequence of real numbers is a real number.
MathematicalPhysicist
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#12
Sep21-03, 06:55 AM
P: 3,169
so there five axioms for a group.
the ones mentioned above by Arden1528 and closure.
phoenixthoth
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#13
Oct9-03, 06:00 AM
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Originally posted by HallsofIvy
Actually, I have a point about your grammar!!

When you say "It must be binary", what is "it"?

The only noun you have mentioned to that point is "group" so grammatically this would be "the group must be binary" which makes no sense. A group consists of a set of objects together with a binary operation on the set. It is the operation that is "binary" and it is the operation that must be associative. On the other hand it is the sat that "contains an identitity. It also would be better to say that "every member of the group has an inverse" rather that just "Must have an inverse".

It is easy to show that the identity in a group is unique.
If "a" and "b" are both identities, then, since a is an identity,
a*b= b. Since b is an identity, a*b= a. Thus, a= a*b= b.

There is, as mathman pointed out (and he should know!) a definite problem with a*b= |ab|. What is the identity?
the title of the thread is "Binary Operations," so i think one could infer the first "it" from that.


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