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- 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
- Why is $1^{\\infty}$ considered to be an indeterminate form
The indeterminate forms are often abbreviated with stuff like "$1^\infty$" but that's not what they mean This "$1^\infty$" (in regards to indeterminate forms) actually means: when there is an expression that approaches 1 and then it is raised to the power of an expression that approaches infinity we can't determine what happens in that form
- Is it true that $0. 999999999\\ldots=1$? - Mathematics Stack Exchange
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- Proof that $(AA^{-1}=I) \\Rightarrow (AA^{-1} = A^{-1}A)$
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- Why is $1 i$ equal to $-i$? - Mathematics Stack Exchange
There are multiple ways of writing out a given complex number, or a number in general Usually we reduce things to the "simplest" terms for display -- saying $0$ is a lot cleaner than saying $1-1$ for example
- 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
- If $A A^{-1} = I$, does that automatically imply $A^{-1} A = I$?
A-1 A means that first we apply A transformation then we apply A-1 transformation When we apply A transformation we reach some plane having some different basis vectors but after apply A-1 we again reach to the plane have basis i ^ (0,1) and j ^ (1,0) It means that after applying A-1 we reach to the transformation which does nothing
- 有问题，就会有答案 - 知乎
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