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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

Is $0$ a natural number? - Mathematics Stack Exchange
Inclusion of $0$ in the natural numbers is a definition for them that first occurred in the 19th century. The Peano Axioms for natural numbers take $0$ to be one though, so if you are working with these axioms (and a lot of natural number theory does) then you take $0$ to be a natural number.

What is 0 raised to 0 - Mathematics Stack Exchange
Thus, if we define $0^0$ to either $0$ or $1$ then we get problems with these functions not being continous (without jumps if you plot them) where they are defined, which is why we keep $0^0$ undefined in most cases.

Is it true that $0.999999999\\ldots=1$? - Mathematics Stack Exchange
The expression " 0.9-repeated" is defined to be the least real-number upper bound of the sequence 0.9. 0.99, 0.999,..... , which is 1. The rationals (and the reals) can also be extended to an arithmetic system (an ordered field) in which there are positive values which are less than every positive rational.

trigonometry - Why $\sin(n\pi) = 0$ and $\cos(n\pi)=(-1)^n ...
$\cdots \sin(-\pi), \sin(0), \sin(\pi), \sin(2\pi), \sin(3\pi),\cdots$ Which is exactly where the sine function has its roots, so it is always equal to $0$. For the cosine case, use the identity $\cos(x) = \cos(x + 2\pi) $ (period of the cosine function is $2\pi$) and plug $\cos(0)$ and $\cos(\pi)$ to verify this.

What is $\gcd (0,a)$, where $a$ is a positive integer?
Thus it makes no sense to define $\rm\ gcd(0,8)\ $ to be $\,0\,$ or $\,1\,$ since $\,0\,$ is not a common divisor of $\,0,8\,$ and $\,1\,$ is not the greatest common divisor. The $\iff$ gcd definition is universal - it may be employed in any domain or cancellative monoid, with the convention that the gcd is defined only up to a unit factor.

I have learned that 1/0 is infinity, why isn't it minus infinity?
1 x 0 = 0. Applying the above logic, 0 / 0 = 1. However, 2 x 0 = 0, so 0 / 0 must also be 2. In fact, it looks as though 0 / 0 could be any number! This obviously makes no sense - we say that 0 / 0 is "undefined" because there isn't really an answer. Likewise, 1 / 0 is not really infinity. Infinity isn't actually a number, it's more of a concept.

What does $QAQ^{-1}$ actually mean? - Mathematics Stack Exchange
0. Identifying a "rotated shear" matrix. 0. From x-X, y-Y and z-Z axes angles to Euler angles. 0. What ...

If a series converges, then the sequence of terms converges to $0$.
I understand the notion behind using the cauchy criterion condition, but I'm confused as to how this shows that the sequence converges to 0 $\endgroup$ – User3213651 Commented Mar 4, 2020 at 21:03

real analysis - If $x≥0$ and $0≤x<ϵ$, for all $ϵ>0$, then $x=0 ...
$\begingroup$ The statement says that the conclusion follows if the inequality is true for all $\epsilon > 0$. $. Fixing $\epsilon$ at a particular value is not meaningful, especially if that value is possibly outside of the range of $\epsilon$ that you are allowed to consider. $\end

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