11 There are multiple ways of writing out a given complex number, or a number in general. Usually we reduce things to the "simplest" terms for display -- saying $0$ is a lot cleaner than saying $1-1$ for example. The complex numbers are a field. This means that every non-$0$ element has a multiplicative inverse, and that inverse is unique.
The theorem that $\binom {n} {k} = \frac {n!} {k! (n-k)!}$ already assumes $0!$ is defined to be $1$. Otherwise this would be restricted to $0 <k < n$. A reason that we do define $0!$ to be $1$ is so that we can cover those edge cases with the same formula, instead of having to treat them separately. We treat binomial coefficients like $\binom {5} {6}$ separately already; the theorem assumes ...
There are infinitely many possible values for $1^i$, corresponding to different branches of the complex logarithm. The confusing point here is that the formula $1^x = 1$ is not part of the definition of complex exponentiation, although it is an immediate consequence of the definition of natural number exponentiation.
Is there a formal proof for $(-1) \\times (-1) = 1$? It's a fundamental formula not only in arithmetic but also in the whole of math. Is there a proof for it or is it just assumed?
Possible Duplicate: How do I convince someone that $1+1=2$ may not necessarily be true? I once read that some mathematicians provided a very length proof of $1+1=2$. Can you think of some way to
49 actually 1 was considered a prime number until the beginning of 20th century. Unique factorization was a driving force beneath its changing of status, since it's formulation is quickier if 1 is not considered a prime; but I think that group theory was the other force.
This is same as AA -1. It means that we first apply the A -1 transformation which will take as to some plane having different basis vectors. If we think what is the inverse of A -1 ? We are basically asking that what transformation is required to get back to the Identity transformation whose basis vectors are i ^ (1,0) and j ^ (0,1).
