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seen Jul 9 at 21:18

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 ;)
Jul
16
comment Computation of characteristic polynomial fails for me
@Theo Buehler: I am not given any specific matrix. The exercise is to prove the correctness of the relations... might it be true that $c_{n-1}$ can be computed this way and the other parts like $c_1$ not? Therefore i could proove the correctness of this without havong a look at $c_1$ if e.g. i let $n=3$.
Jul
16
accepted Computation of characteristic polynomial fails for me
Jul
16
comment Computation of characteristic polynomial fails for me
Understood. I don't know why i haven't seen this before! But is there any other algorithm or any other solution to make this one "really" recursive? This is just what was given to us in a lecture.