Can you prove that \mathbf{R} and \mathbf{R}-\mathbf{Q} have same cardinality?
One way would be to say that \mathbf{R}-\mathbf{Q} is not countable and must have cardinality <= \mathbf{R} and invoke the Continuum Hypothesis to conclude that its cardinality is aleph-1 same as that of...
Thanks for the example and the suggestion...
Firs thing I tried with was dihedral groups and was unsuccessful, quaternion groups never crossed my mind...
I tried with the Klein four subgroup (dihedral group with x^2=e,y^2=e) which satisfies the condition, but is Abelian and thus obviously...
If all subgroups of a group are normal, is the group abelian?
I know that the answer is NO...
Can you give a counter-example..?
Better still can you logically deduce some property the counter-example must have, which will ease our way to finding it...
Before negating, let's look at the statement logically..
Decompose the compound statement into simple ones..
p: 0<B<A <==>(0<B) AND (B<A)
The equivalence of both the statements can be easily verified...
By De Morgan's Rule, p^{c} is (0<B)^{c} OR (B<A)^{c}..
So the negation would look...
I think you are confusing between "all" and "every"... You should use the latter instead of the former in most cases above..
There is a subtle difference...
"every" tells you that you are considering a particular case now and then will scan your argument over each 'x' and thus conclude for...
I think all your doubts will be cleared if you pick up some good book of analysis and study about the field axioms...
Its not clear what you mean by x,y and '+'...
If you do mean "real numbers" and 'addition', just keep in mind that all the questions you have asked follow from the definition...
hi friends
can you help me in collecting systematically some important phenomenon or 'effects' in astronomy and astrophysics like rossiter effect, yarkovsky effect.:cool:
even recently discovered phenomenon and related explanations are welcome.:approve: