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 $0 <k < n$ 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
Justifying why 0 0 is indeterminate and 1 0 is undefined So basically, 1 0 does not exist because if it does, then it wouldn't work with the math rules Let τ=1 0 0τ=1 x0τ=x 0τ=x τ=x 0 1 0=x 0 which doesn't work (x represents any number) That means that 1 0, the multiplicative inverse of 0 does not exist 0 multiplied by the multiplicative inverse of 0 does not make any sense and is undefined
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
Seeking elegant proof why 0 divided by 0 does not equal 1 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
I have learned that 1 0 is infinity, why isnt 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
Why Not Define $0 0$ To Be $0$? - Mathematics Stack Exchange If you define $\frac00$ to be $0$, then you either have to abolish some other basic rules of arithmetic or accept the following argument: Since $3\times 0=0$, divide both sides by $0$, thereby cancelling the $0$ factor on the left and leaving $3=\frac00=0$
What exactly does it mean that a limit is indeterminate like in 0 0? The above picture is the full background to it It does not invoke "indeterminate forms" It does not require you to write $\frac{0}{0}$ and then ponder what that might mean We don't divide by zero anywhere It is just the case where $\lim_{x\to a}g(x)=0$ is out of scope of the above theorem
complex analysis - What is $0^{i}$? - Mathematics Stack Exchange $$\lim_{n\to 0} n^{i} = \lim_{n\to 0} e^{i\log(n)} $$ I know that $0^{0}$ is generally undefined, but can equal one in the context of the empty set mapping to itself only one time I realize that in terms of the equation above, the limit does not exist, but can $0^{i}$ be interpreted in a way to assign it a value?
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
definition - Why is $x^0 = 1$ except when $x = 0$? - Mathematics Stack . . . 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