Plus Minus Times Divide Zero Through Six

@William Park – "The debate over the definition of 00 has been going on at least since the early 19th century. At that time, most mathematicians agreed that 00 = 1, until in 1821 Cauchy[20] listed 00 along with expressions like 0 / 0 in a table of indeterminate forms. In the 1830s Libri[21][22] published an unconvincing argument for 00 = 1, and Möbius[23] sided with him, erroneously claiming that \scriptstyle \lim {t \to 0^+} f(t)^{g(t)} \;=\; 1 whenever \scriptstyle \lim {t \to 0^+} f(t) \;=\; \lim_{t \to 0^+} g(t) \;=\; 0. A commentator who signed his name simply as "S" provided the counterexample of \scriptstyle (e^{-1/t})^t, and this quieted the debate for some time. More historical details can be found in Knuth (1992).[24]

More recent authors interpret the situation above in different ways: Some argue that the best value for 00 depends on context, and hence that defining it once and for all is problematic.[25] According to Benson (1999), "The choice whether to define 00 is based on convenience, not on correctness."[26] Others argue that 00 should be defined as 1. Knuth (1992) contends strongly that 00 "has to be 1", drawing a distinction between the value 00, which should equal 1 as advocated by Libri, and the limiting form 00 (an abbreviation for a limit of \scriptstyle f(x)^{g(x)} where \scriptstyle f(x), g(x) \to 0), which is necessarily an indeterminate form as listed by Cauchy: "Both Cauchy and Libri were right, but Libri and his defenders did not understand why truth was on their side."[24]

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@William Park – But I read in a lot of books that 0 to power of 0 is undefined and every number except zero to the power of 0 is as 1 .

@William Park – Think about it. Anything to the zeroth power is 1, but 0 to any power is zero. The reason it is undefined is because it combines those things and could be 0 or 1.

@William Park – Anything to the zeroth power is one, but 0 to any power is zero. A question is indeterminable if they combine the two.

@William Park – Of course. High school classes always say that it is undefined.

@Simone Peroni – A counterargument to that was already provided by someone else. Consider applying the same reasoning for $0^2$ . You can also write it as $0^2=\frac{0^3}{0^1}=\frac{0}{0}$ But that means that $0^2$ can also be anything, which is false since we know it's specifically $0$ .

@Simone Peroni – $a=0$ and $m>n>0\Rightarrow { a }^{ m-n }\neq \frac { { a }^{ m } }{ { a }^{ n } }$ .

$a=0$ and $m\ge n>0\Rightarrow { a }^{ m-n }={ 0 }^{ m-n }$ and $m-n\ge 0$ .

$a=0$ and $m\ge n>0\Rightarrow \frac { { a }^{ m } }{ { a }^{ n } }$ is undefined.

What you've really proved was that $\frac { { a }^{ m } }{ { a }^{ n } }$ is undefined when $a=0$ and $m\ge n>0$ . There's no question to that. However, ${ a }^{ m-n }\neq \frac { { a }^{ m } }{ { a }^{ n } }$ when $a=0$ and $m>n>0$ .

@Simone Peroni – Yep, I agree with you

@Simone Peroni – I was pointing out that your demonstration was illogical. Idk if you saw that.

@Osama Kawish – $0^0 = \frac{0^1}{0^1} = \frac{0}{0}$

@Lew Sterling Jr – Ah I was referring to that Euler advocated the convention $0^0 = 1.$

@Hari Krishnan – Zero to the 50th power (or even to an infinite power) is still zero, so zero divided by zero is not infinity.

@André Winston – $\frac { { x }^{ m } }{ { x }^{ n } } ={ x }^{ m-n }$ is true only $\forall x\neq 0$ .

@André Winston – True, but based on limits, graphing, and notations,they still are saying that it 1 and not indeterminate.

@André Winston – If we will use that line of reasoning, though, ${ 0 }^{ 1 }$ would also not be determined, because you could consider it as ${ 0 }^{ 2-1 }$ or $\frac { { 0 }^{ 2 } }{ { 0 }^{ 1 } }$ , where you'll also get a division by zero.

Don't you think ${ 0 }^{ 2-1 }={ 0 }^{ 1 }\neq \frac { { 0 }^{ 2 } }{ { 0 }^{ 1 } }$ ?

@Davin Gery – I see your point.