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abstract algebra - Prove that 1+1=2 - Mathematics Stack Exchange
Jan 15, 2013 · Possible Duplicate: How do I convince someone that $1+1=2$ may not necessarily be true? I once read that some mathematicians provided a very length proof of $1+1=2$. Can you think …
Word，插入多级列表，但是改了1.1，第二章的2.1也变成1.1，随着改变 …
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 …
Why is $1/i$ equal to $-i$? - Mathematics Stack Exchange
May 11, 2015 · While 1/i = i−1 1 / i = i 1 is true (pretty much by definition), if we have a value c c such that c∗i = 1 c ∗ i = 1 then c= i−1 c = i 1. This is because we know that inverses in the complex …
limx→0, (1+x)^1/x=e 为什么？ - 知乎
Jun 26, 2020 · 对于 (1+1/n)^n ＜ 3的证明如下图 (图片来自 崔尚斌数学分析教程)
If $A A^{-1} = I$, does that automatically imply $A^{-1} A = I$?
Mar 30, 2020 · 1 Short Answer Yes AA -1 = A -1 A = I when the Det (A) ≠ ≠ 0 and A is a square matrix. Long Answer A matrix is basically a linear transformation applied to some space. For the sake of …
How can one find the factorization $a^4 + 2 a^3 + 3 a^2 + 2 a + 1
If you find a way of proving $\sqrt {a^4 + 2a^3 + 3a^2 + 2a + 1} \in \mathbb {N}$ without calculating the concrete factorization, it's fine.
Formal proof for $ (-1) \times (-1) = 1$ - Mathematics Stack Exchange
Jun 13, 2020 · Is there a formal proof for $(-1) \\times (-1) = 1$? 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?
Proof that $(AA^{-1}=I) \\Rightarrow (AA^{-1} = A^{-1}A)$
I'm trying to prove a pretty simple problem - commutativity of multiplication of matrix and its inverse. But I'm not sure, if my proof is correct, because I'm not very experienced. Could you, plea...
Formula for $1^2+2^2+3^2+...+n^2$ - Mathematics Stack Exchange
$ (n+1)^3 - n^3 = 3n^2+3n+1$ - so it is clear that the $n^2$ terms can be added (with some lower-order terms attached) by adding the differences of cubes, giving a leading term in $n^3$. The factor 1/3 …