2013年3月15日 · 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.

$\begingroup$ 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 $0 <k < n$. 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.

2017年1月22日 · 1) x^a × x^b = x^a+b; for x = 0 and a = 0, you would get 0^0 × 0^b = 0^b = 0, so we can't tell anything -- except confirm that 0^0 = 1 still works here! 2) x^{-a}=1/{x^a} -- so when a = 0 , x^{-0} = 1/x^0 = x^0 , which again does work for 0^0 = 1 ; 3) {x^a}^b = x^{a×b} , thus x^(1/n) is the n-th root -- and 1/n = 0 for no value of n , so ...

Whereas exponentiation by a real or complex number is a messier concept, inspired by limits and continuity. So $0^0$ with a real 0 in the exponent is indeteriminate, because you get different results by taking the limit in different ways.

$\begingroup$ This definition of the "0-norm" isn't very useful because (1) it doesn't satisfy the properties of a norm and (2) $0^{0}$ is conventionally defined to be 1. $\endgroup$ – Brian Borchers

2015年5月29日 · Is a constant raised to the power of infinity indeterminate? I am just curious. Say, for instance, is $0^\infty$ indeterminate?

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.

2019年4月23日 · In computer languages where x/0 returns an object for which multiplication is defined, you do not have that (x\0)*0 == x. So we can can a class of objects in which we call one of the objects "zero", and have a class method such that "division" by "zero" is defined, but that class will not act exactly like the real numbers do.

2015年9月11日 · On the contrary, those limits tell you that the limit of the entire quotient is $0$. This may be easier to see if you rewrite to $$ \lim_{x\to\infty} f(x)\frac1{h(x)} $$ where $\lim_{x\to\infty} f(x) = 0 $ and $\lim_{x\to\infty} \frac1{h(x)}=0 $, and the product of two functions that both have limit $0$ surely also has limit $0$.

The reason $0/0$ is undefined is that it is impossible to define it to be equal to any real number while obeying the familiar algebraic properties of the reals. It is perfectly reasonable to contemplate particular vales for $0/0$ and obtain a contradiction. This is how we know it is impossible to define it in any reasonable way.