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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
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
如何查看自己电脑的 IP 地址？ - 知乎 知乎，中文互联网高质量的问答社区和创作者聚集的原创内容平台，于 2011 年 1 月正式上线，以「让人们更好的分享知识、经验和见解，找到自己的解答」为品牌使命。知乎凭借认真、专业、友善的社区氛围、独特的产品机制以及结构化和易获得的优质内容，聚集了中文互联网科技、商业、影视
Why is $1$ not a prime number? - Mathematics Stack Exchange 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
arithmetic - Formal proof for $(-1) \times (-1) = 1$ - Mathematics . . . The Law of Signs $\rm\: (-x)(-y) = xy\:$ isn't normally assumed as an axiom Rather, it is derived as a consequence of more fundamental Ring axioms $ $ [esp the distributive law $\rm\,x(y+z) = xy + xz\,$], laws which abstract the common algebraic structure shared by familiar number systems