Examine whether dened on the set G is a group or not.
G = {R}\(1), a * b = a + b + ab, where a,b belongs to R\(1).
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terenzreignz (terenzreignz):
Group axioms, you know them? :)
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
i am having problem with identity
terenzreignz (terenzreignz):
There are four things that must hold for a set and its defined binary operation to be true.
A set and binary operation \(\large \left<G,\ast\right>\) is a group if... what?
terenzreignz (terenzreignz):
oh okay.
Let's see if it has an identity.
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terenzreignz (terenzreignz):
Is there an element e such that
\[\Large e\ast a = a = a\ast e\] ?
OpenStudy (anonymous):
its e=0
terenzreignz (terenzreignz):
Then... what's the problem? :D
\[\Large 0\ast a = 0 + a + 0a = a = a\ast 0\]
^_^
OpenStudy (anonymous):
when we get e(a+1)=0 either a=-1 or e=0 so when a=-1 then 0/0 comes which is not defined
terenzreignz (terenzreignz):
You want to find \(\large 0 \ast -1\) ?
Then...
\[\Large 0\ast -1 = 0 +(-1) + (-1\times 0)= -1\]
So what's the problem? :)
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terenzreignz (terenzreignz):
It just so happened that by definition, \(\large -1 \ast a= -1\) for all a in G.
It doesn't have to be a problem :)