Why does $x^0=1$? Also, why does this rule not apply when $x=0$? I understand that math can be difficult to explain, but this subject never seems to be thoroughly explained in simple terms.
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2The rule can be extended to $0$. That is, we can define $0^0=1$ and this makes the most sense in most places. The one thing that needs to be understood is that $x^y$ is not continuous at $(0,0)$. – robjohn Feb 12 '18 at 03:54
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2In fact, $0^0=1$ also. This is a definition made, which is saying that an empty product is $1$. Similarly, $0!=1$. – Dave Feb 12 '18 at 03:55
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$\frac{x^y}{x^z}=x^{y-z} $, right. Well $\frac{x^y}{x^y} = x^{y-y}=x^0=1$ – Ryan A Feb 12 '18 at 04:02
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"Why does $x^0$ (for $x\neq 0$) equal $1$?" Have you ever seen a formal definition of $a^b$? Are you aware that $e^x=\text{exp}(x)=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots$? And if you plug zero into this? And if you generalize this for other bases? – JMoravitz Feb 12 '18 at 04:03
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@robjohn I am intrigued by your explanation, being a high school student. It is true that this subject has never been given much importance at schools, and hence not much thought either. But, coming to the original question, can $0^0$ defined to be $1$ just like $0!=1$? $ 0!=1$ makes sense but $0^0$ seems a little bit vague to me. Any help would be appreciated :) – QFTheorist Feb 12 '18 at 16:18
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@alphasquared: There are many places where $0^0=1$ makes the most sense. For example, there are $a^b$ functions from a set of $b$ elements to a set of $a$ elements. There is $1$ function from the empty set to itself: the empty function. Another example is in the binomial theorem: $(1+0)^2=\binom{2}{0}1^2\color{#C00}{0^0}+\binom{2}{1}1^10^1+\binom{2}{2}1^00^2$ which works only if $0^0=1$. – robjohn Feb 12 '18 at 16:48
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@robjohn thanks a lot. It's all clear now. – QFTheorist Feb 12 '18 at 16:51
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This answer, this answer, and this answer also talk about $0^0$. – robjohn Feb 12 '18 at 17:00
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Say we have a value $x$ and we want to raise it to a power $n$. This means that we multiply $x$ by itself $n$ times. $$x^n = \underbrace{x\cdot x\cdot x\cdot\ldots\cdot x}_{n\text{ times.}}\tag1$$ It is confusing to most people when we say that $x^0 =1$, because how can we multiply $x$ by itself $0$ times? Here, we have to look at one of the Power Rules.
The first power rule is as follows: $$x^a\cdot x^b = x^{a + b}.$$ This is provable from $(1)$. Since $x^a$ and $x^b$ are all products of $x$, then when we multiply them together, the number of times $x$ is being multiplied by itself in total is of course $a + b$ times. Therefore, the product of $x^a$ and $x^b$ is always $x^{a+b}$.
So this means that since $n = n + 0$, we get that $x^n = x^n\cdot x^0$. Therefore, $x^0$ must be equal to $1$. $$x^0 = 1.$$ But what happens when $x = 0$? Well here, we have to look at the second power rule. $$\frac{x^a}{x^b} = x^{a-b}.$$ This is also provable using the same method we used to prove the first power rule. But this means that when $x = 0$, we introduce division by $0$ which cannot exist (for obvious reasons). Since every number $x$ to the power of $1$ is equal to itself (this is also a power rule) then we can write $0^1 = 0$. Now if we want to solve for $0^0$, well according to $(1)$, we have that $0^0 = 1$. But, $$0^0 = 0^{1-1} = \frac{0^1}{0^1} = \frac{0}{0}.$$ This is where division by $0$ is introduced, and is why $0^0\neq 1$. But since $0 - 0 = 0$ then, $$\frac{0}{0} = \frac{0^0}{0^0}.$$ Therefore, because we cannot have division by $0$, we get that $0^0\neq 0$. Now looking back at the first power rule, since $0\times 0\times 0\times\ldots\times 0 = 0$ then we get that $0^n = 0$, for all $n > 0$. Of course, if $n < 0$ then this would introduce division by $0$ as well. Furthermore, this means that $x^0 = 1$, but we need $x\neq 0$.
To conclude, we get that $0^0$ is just a special case of a power of $0$, and it is thus undefined. For all $x\neq 0$, we can define $x^0$ as simply being equal to $1$. And for all $n > 0$, we can define $0^n$ as simply equal to $0$.
Mr Pie
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$$x\ne 0$$
$$x^0 = x^{a-a}$$
$$x^0 = \frac {x^a}{x^a}$$
$$\frac {x^a}{x^a}=1$$
$$x^0=1$$
Trobeli
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