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Mathematics 17 Online
OpenStudy (anonymous):

Is the set of positive integers an example of an abelian group?

OpenStudy (anonymous):

if not what would be an example?

OpenStudy (anonymous):

you cannot ask such a question without specifying what the operation is

OpenStudy (anonymous):

is it \(\{\mathbb{N},+\}\) ?

OpenStudy (anonymous):

or perhaps \(\{\mathbb{N},\times\}\)

OpenStudy (anonymous):

the answer is "no" in any case, but it makes a difference for the reasoning

OpenStudy (anonymous):

the first does not have an identity element so the answer is no for that reason

OpenStudy (anonymous):

okay... why is it not.... and what would be an example of one?

OpenStudy (anonymous):

ummm what about R,+

OpenStudy (anonymous):

every group must have an identity element for the given operation the set of natural numbers under addition does not, since it does not include zero

OpenStudy (anonymous):

if you include zero, it still is not a group, because there also must be inverses, and you cannot solve \(2+n=0\) for \(n\) if \(n\in \mathbb{N}\)

OpenStudy (anonymous):

if you include all integers, then it would be a group under addition

OpenStudy (anonymous):

Ok ... so {Z,+}

OpenStudy (anonymous):

that would be an example, yes there are countless examples

OpenStudy (anonymous):

okay, so in order to determine what one would be.... then you take the set... and the operation... and pick any a,b,c and make sure that it is 1. closure 2. associative 3. identity 4. inverse 5. communitive

OpenStudy (anonymous):

those are the properties for an abelian group, yes

OpenStudy (anonymous):

thank you.

OpenStudy (anonymous):

yw

OpenStudy (anonymous):

and yes, \(\{\mathbb{R},+\}\) is an abelian group, however \(\{\mathbb{R},\times \}\) is not

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