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What does $QAQ^{-1}$ actually mean? - Mathematics Stack Exchange 1 $\begingroup$ When one thinks of matrix products like that, it's helpful to remember that matrices, unlike vectors, have two sets of bases: one for the domain and one for the range Thinking of applying a vector on to the right, we get that the transformation "unrotates" the vector, applies the original transformation in the original basis
what is 1 - 1 2 + 1 3 - 1 4 + 1 5 - 1 6 + 1 7 - 1 8 +1 9 Stack Exchange Network Stack Exchange network consists of 183 Q A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers
abstract algebra - Prove that 1+1=2 - Mathematics Stack Exchange The main reason that it takes so long to get to $1+1=2$ is that Principia Mathematica starts from almost nothing, and works its way up in very tiny, incremental steps The work of G Peano shows that it's not hard to produce a useful set of axioms that can prove 1+1=2 much more easily than Whitehead and Russell do
Mathematics Stack Exchange Stack Exchange Network Stack Exchange network consists of 183 Q A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers
factorial - Why does 0! = 1? - Mathematics Stack Exchange $\begingroup$ 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
How can 1+1=3 be possible? - Mathematics Stack Exchange Stack Exchange Network Stack Exchange network consists of 183 Q A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers
What is the integral of 1 x? - Mathematics Stack Exchange $\begingroup$ Well, calling Ln(-1) = i*Pi is already expanding the standard definition of Ln In the context described (where A -A are both real, and the log used is standard), Ln(-1) is undefined But I'd say the intelligent thing is to break it into two undefined improper integrals with one of the limits at 0 (where it is unbounded)