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factorial  Why does 0! = 1?  Mathematics Stack Exchange
The theorem that (n k) = n! k!(n−k)! (n k) = n! k! (n − k)! already assumes 0! 0! is defined to be 1 1. Otherwise this would be restricted to 0
Is $0$ a natural number?  Mathematics Stack Exchange
the 1 1. st number is 1 1. ; in making limits, 0 0. plays a role which is symmetric to ∞ ∞. , and the latter is not a natural number. Pros of considering 0 a natural number: the starting point for set theory is the emptyset, which can be used to represent 0 0. in the construction of natural numbers; the number n n.
algebra precalculus  Zero to the zero power – is $0^0=1 ...
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.
Is $0^\infty$ indeterminate?  Mathematics Stack Exchange
If we type those expressions into Mathematica, however, it tells us that 0^infinity is 0 and 1^infinity is indeterminate. – user3680. Commented Oct 9, 2013 at 23:42. All the answers here assume 0∞ 0 ∞ is 0 + ∞ 0 + ∞. But Wikipedia doesn't assume that. It claims that 0∞ 0 ∞ is an indeterminate form because 0 + ∞ 0 + ∞ has the ...
Why does 0.00 have zero significant figures and why throw out the ...
The measurement 0.00010 has 2 sigfigs (or in SN, 1.0e4). A measurement with the same apparatus that reports 0.00000 should seemingly also have at least 2, but we cant determine that from the string. When ambiguous perhaps it must be written as 0.0e4 or 0.00e3 depending on the edge case convention?
Prove that $ AA^T=0\\implies A = 0$  Mathematics Stack Exchange
Supose A ≠ 0. Then, for any X, A + X ≠ X. Multiplying by AT at the right, we get: (A + X)AT ≠ XAT. AAT + XAT ≠ XAT. XAT ≠ XAT. Arriving in a contradiction, we get A = 0. I realise it may not be totally correct to multiply by AT without making some proper assumptions, but I know I can pick any X.
A thorough explanation on why division by zero is undefined?
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.
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.
limit when zero divided by infinity  Mathematics Stack Exchange
33. 0∞ 0 ∞ is not an indeterminate form. On the contrary, those limits tell you that the limit of the entire quotient is 0 0. This may be easier to see if you rewrite to. limx→∞ f(x) 1 h(x) lim x → ∞ f (x) 1 h (x) where limx→∞ f(x) = 0 lim x → ∞ f (x) = 0 and limx→∞ 1 h(x) = 0 lim x → ∞ 1 h (x) = 0, and the product ...
logarithms  Why I can't calculate $0*log(0)$ but can $log(0^0 ...
7. If you need to calculate 0 log 0 0 log 0, you're probably either: Doing something wrong. Implementing an algorithm that explicitly states that 0 log 0 0 log. . 0 is a fib that doesn't mean "compute zero times the logarithm of zero", but instead something else (e.g. "zero") If log00 log 0 0 worked in your programming language, it's ...
