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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
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
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
Why is $1$ not a prime number? - Mathematics Stack Exchange 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
If $A A^{-1} = I$, does that automatically imply $A^{-1} A = I$? A -1 A means that first we apply A transformation then we apply A -1 transformation When we apply A transformation we reach some plane having some different basis vectors but after apply A -1 we again reach to the plane have basis i ^ (0,1) and j ^ (1,0)
Arithmetic pattern $1 + 2 = 3$, $4 + 5 + 6 = 7 + 8$, and so on The other interesting thing here is that 1,2,3, etc appear in order in the list And you have 2,3,4, etc terms on the left, 1,2,3, etc terms on the right This should let you determine a formula like the one you want Then prove it by induction
False Proof of 1=-1 - Mathematics Stack Exchange 1 Indeed what you are proving is that in the complex numbers you don't have (in general) $$\sqrt {xy}=\sqrt {x}\sqrt {y}$$ Because you find a counterexample