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- algebra precalculus - Zero to the zero power – is $0^0=1 . . .
@Arturo: I heartily disagree with your first sentence Here's why: There's the binomial theorem (which you find too weak), and there's power series and polynomials (see also Gadi's answer) For all this, $0^0=1$ is extremely convenient, and I wouldn't know how to do without it In my lectures, I always tell my students that whatever their teachers said in school about $0^0$ being undefined, we
- 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
- I have learned that 1 0 is infinity, why isnt it minus infinity?
@Swivel But 0 does equal -0 Even under IEEE-754 The only reason IEEE-754 makes a distinction between +0 and -0 at all is because of underflow, and for + - ∞, overflow The intention is if you have a number whose magnitude is so small it underflows the exponent, you have no choice but to call the magnitude zero, but you can still salvage the
- algebra precalculus - Prove $0! = 1$ from first principles . . .
You can also prove it by moving the space: "0! = 1" $\Leftrightarrow$ "0 != 1", which is computer notation for "0 $\neq$ 1" :-) Then it depends on what you count as "first principles" If we're dealing with the natural numbers, this follows from the Peano axiom that the successor of a natural number is not 0 (1 being defined as the successor
- Seeking elegant proof why 0 divided by 0 does not equal 1
I began by assuming that $\dfrac00$ does equal $1$ and then was eventually able to deduce that, based upon my assumption (which as we know was false) $0=1$ As this is clearly false and if all the steps in my proof were logically valid, the conclusion then is that my only assumption (that $\dfrac00=1$) must be false
- Justifying why 0 0 is indeterminate and 1 0 is undefined
In the context of limits, $0 0$ is an indeterminate form (limit could be anything) while $1 0$ is not (limit either doesn't exist or is $\pm\infty$) This is a pretty reasonable way to think about why it is that $0 0$ is indeterminate and $1 0$ is not However, as algebraic expressions, neither is defined Division requires multiplying by a multiplicative inverse, and $0$ doesn't have one
- Is $0^\infty$ indeterminate? - Mathematics Stack Exchange
Is a constant raised to the power of infinity indeterminate? I am just curious Say, for instance, is $0^\\infty$ indeterminate? Or is it only 1 raised to the infinity that is?
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