Proof that (R, +) is not a Cyclic Group
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- Опубліковано 25 жов 2014
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Proof that (R, +) is not a Cyclic Group.
We prove that the set of real numbers under the operation of addition does not form a cyclic group. This is really an easy observation but in this video we go through the details.
I've seen a proof that Q is not finitely generated before (which, I guess already depends on knowing that Q is not cyclic, because proving that any finitely generated subgroup of Q is actually a cyclic subgroup of Q is the crux of it) - so I just figured that since (Q, +) is infinitely generated already and (Q,+) is a subgroup of (R,+) then R cannot be cyclic either.
What if n belonged to R, the set of all real numbers? Then couldn't you add or subtract, like "x + (1/2)x = (3/2)x"?
n does belong to R, since the integers Z is a subset of the real numbers R. But that's irrelevant.
We are looking for a generator of R under the group operation addition (look up the definition of generator).
And the sorcerer showed that if there was a generator of R under addition, then we would fall into an absurdity hole.
Thank you! Excellent!
👍
What are generators of Q under addition.
Since it is non cyclic so no element of Q can act as a generator.
Can you solve (Q,+)is not cyclic ?please
is it exactly the same?
Excellent!
Thank you!
Can any one show.. y the Pair of groups (R*,•)& (C*, •)is not isomorphic..?
Isn't it because C* has lots of elements of finite order and R* only has 1 and -1 ? Oh right - that's why I've seen this before mentioning that you can find an order three element in C* which doesn't exist in R*. (I guess, the order 4 element i in C* is the most obvious one)
(Q*, • ) is not cyclic how to prove it
bakwas