|
- 有问题,就会有答案 - 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、时
- 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
- 知乎 - 有问题,就会有答案
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
- 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
- 甲醛含量多少范围时,正常可以入住? - 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
- 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
- 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
- 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)
|
|
|