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seen Nov 3 at 18:48

Aug
18
comment How to transform/expand a simple sum to prove equality of two sets?
Thanks for these hints. I will try now to check for equality on my own and i will try to understand the manufacturing of this explicit formula :-)
Aug
18
comment How to transform/expand a simple sum to prove equality of two sets?
Prooving this one should be not that difficult however it is not clear to me how to develop this term out of the sum of the powers of $-1$. That is the point - i would like to know how you have seen this!
Aug
18
revised How to transform/expand a simple sum to prove equality of two sets?
added 24 characters in body; edited title
Aug
18
comment How to transform/expand a simple sum to prove equality of two sets?
To be honest i do not see why you need to use $(-1)^{i+1}$ instead of $(-1)^i$ - the pattern seems clear to me, but the only reason to use $i+1$ seems to be to "heal" the problem to get to this pattern, isn't it that?
Aug
18
comment How to transform/expand a simple sum to prove equality of two sets?
@André Nicolas: It is true that $(1-(-1)^n)/2$ has this same property, but what is the way to get to this assumption?
Aug
18
revised How to transform/expand a simple sum to prove equality of two sets?
added Mathematica output
Aug
18
asked How to transform/expand a simple sum to prove equality of two sets?
Aug
15
comment Proof that $\mathbb{Z}$ has no zero divisors
Sorry for the confusion, but i have now understood this one. Can you explain me more detailed why do you use the term ring and not just the set of integers?
Aug
15
accepted Proof that $\mathbb{Z}$ has no zero divisors
Aug
15
revised Proof that $\mathbb{Z}$ has no zero divisors
added 8 characters in body
Aug
15
comment Proof that $\mathbb{Z}$ has no zero divisors
@Carl Mummert: I've added the rules i should base my proof on.
Aug
15
revised Proof that $\mathbb{Z}$ has no zero divisors
edited the allowed rules
Aug
15
comment Proof that $\mathbb{Z}$ has no zero divisors
@Arturo Magidin: $\mathbb{Z}$ are normal integers $\{\cdots,-2,-1,0,1,2,\cdots\}$ with the "normal" multiplication like every knows it.
Aug
15
comment Proof that $\mathbb{Z}$ has no zero divisors
@Carl Mummert: I often look back :-) What other ideas do you have?
Aug
15
awarded  Editor
Aug
15
revised Proof that $\mathbb{Z}$ has no zero divisors
added 359 characters in body
Aug
15
asked Proof that $\mathbb{Z}$ has no zero divisors
Jul
16
accepted Proof that a linear transformation is isomorphic
Jul
16
awarded  Commentator
Jul
16
comment Computation of characteristic polynomial fails for me
@Theo Buehler: Thanks for the help - und danke sehr ;)