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How do i convince someone that $1+1=2$ may not necessarily be true 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). I once read that some mathematicians provided a very length proof of $1+1=2$
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Can you think of some way to 注1：【】代表软件中的功能文字 注2：同一台电脑，只需要设置一次，以后都可以直接使用 注3：如果觉得原先设置的格式不是自己想要的，可以继续点击【多级列表】——【定义新多级列表】，找到相应的位置进行修改 11 there are multiple ways of writing out a given complex number, or a number in general
The complex numbers are a field
It's a fundamental formula not only in arithmetic but also in the whole of math Is there a proof for it or is it just assumed? 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。 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. 两边求和，我们有 ln (n+1)<1/1+1/2+1/3+1/4+……+1/n 容易的， \lim _ {n\rightarrow +\infty }\ln \left ( n+1\right) =+\infty ，所以这个和是无界的，不收敛。 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.
The theorem that $\binom {n} {k} = \frac {n!} {k 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
