Why is $1 i$ equal to $-i$? - Mathematics Stack Exchange 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
What is the value of $1^i$? - Mathematics Stack Exchange 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
abstract algebra - Prove that 1+1=2 - Mathematics Stack Exchange 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
factorial - Why does 0! = 1? - Mathematics Stack Exchange 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
If $A A^{-1} = I$, does that automatically imply $A^{-1} A = I$? 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)