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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 $1+1=2$
Can you think of some way to We treat binomial coefficients like $\binom {5} {6}$ separately already 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.
11 there are multiple ways of writing out a given complex number, or a number in general The complex numbers are a field 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。 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? 质数就是“只能被1和它本身整除”的自然数。 然而，我们必须在此基础之上增加一条警告，宣称数字1不是质数，这简直就像马后炮一样。 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。 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
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