I saw that $x^0=1$ proof is $\displaystyle x^{a-a} = \frac{x^a}{x^a} = 1.$
However I don't find this a convincing proof.
How can someone prove this without using the fraction above? or, Is that fraction the only proof of $x^0=1$?
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3A real life experiment? What do you mean? – user829347 Jan 21 '22 at 22:33
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1It is defined that way. What you wrote is one reason to define it so. – David Mitra Jan 21 '22 at 22:34
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How about: a species of bacteria reproduces every 1 hour. So if you start with 1 bacteria, after 1 hour you have 2, after 2 hours you have 4, after 3 hours you have 8, etc. So the population of the bacteria is described by $P(n) = 2^n$, where $n$ is the number of hours that have passed. But then $P(0) = 2^0$ should be the population you started with, which is $1$. – user56202 Jan 21 '22 at 22:36
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@VeryForgetfulFunctor A factory production for example. – Wolf98 Jan 21 '22 at 23:51
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@Wolf98 As Ethan Bolker points out in his answer, it would not be possible to prove that $x^0=1$ using anything other than a rigorous mathematical argument. – user829347 Jan 21 '22 at 23:58
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@user56202 So if i started with 5, it will be 5? – Wolf98 Jan 22 '22 at 00:03
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@Wolf98 If you start with 5 you will have 5, 10, 20, 40, ... so it will be $P(n) = 5 \cdot 2^n$. So $5 \cdot 2^0 = 5$, so still $2^0 = 1$. – user56202 Jan 22 '22 at 01:13
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@user56202 You said "But then P(0)=2^0 should be the population you started with, which is 1" If i started with 5 it will be 5 because no hour has passed as you said so 5^0=5, i stay with the same quantity. Also i don't see why you add the 2^0. – Wolf98 Jan 22 '22 at 01:26
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1@Wolf98 When you start with $1$, the model is $P(n) = 2^n$, so the initial population should be $P(0) = 2^0$. When you start with $5$, the model is $P(n) = 5 \cdot 2^n$ so the initial population should be $5 \cdot 2^0$, not $2^0$. You add the $2^0$ to be consistent with the formula $2^n$. – user56202 Jan 22 '22 at 02:17
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@user56202 i give you +1, now i understand it, but What if i start with 5 and every hour it increases 5 times so now i have P(0) = 5^n. If no hour has passed which is n=0, i stay with my 5 bacterias, not 1. – Wolf98 Jan 22 '22 at 02:44
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@Wolf98 If you start with $5$ and it is multiplied by $5$ every hour, then (hour, population) is $(0, 5), (1, 25), (2, 125), ...$ so the model is actually $P(n) = 5 \cdot 5^n$ and $P(0) = 5 \cdot 5^0 = 5$ so still $5^0 = 1$. – user56202 Jan 22 '22 at 02:47
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@user56202 Then the key is 5^1=5. Because i also can say 5^0=5, 5^1=25, 5^2=125 etc... So 5^0=1 is the consistency of the predetermined model of exponentiation. – Wolf98 Jan 22 '22 at 03:16
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@Wolf98 That is one way to look at it, but this way ignored the fact that the original definition of exponents has to do with repeated multiplication. You can define $5^2 = 125$ if you want, but $5^2$ is naturally defined as $5 \cdot 5$ which is $25$. – user56202 Jan 22 '22 at 03:29
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user1015917's post is completely bogus; Ethan and Arthur correctly state that $x^0 = 1$ is a freely chosen definition, not something you can prove in the absence of a definition of exponentiation. If you haven't defined exponentiation, there is nothing you can prove about it! – user21820 Jan 30 '22 at 08:33
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With compound interests, your capital increases by $10\%$ every year.
$$1 \to 1.1\\ 2 \to 1.21\\ 3 \to 1.331\\ 4 \to 1.4641\\ \cdots\\ N\to1.1^N \cdots$$
Obviously we get a consistent definition if
$$0\to1$$
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So if i have 10 apples and every year it increases 10%, Does this mean i have 1 apple at the beginning or 10? It's the same reasoning of your example – Wolf98 Jan 21 '22 at 23:48
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1@Wolf98 If you start with 10 apples increasing 10% each year, you would have $10*(1.1)^t$ apples after $t$ years. Once again, if $1.1^0=1$, we have 10 apples at year zero. – Golden_Ratio Jan 22 '22 at 08:46
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You can only prove mathematical statements with arguments that start from mathematical axioms. "Real life" does not enter the picture.
That said, the assertion that for $x>0$ $$ x^0 = 1 $$ is not a theorem, it's a definition. Mathematicians could have decided that $$ x^0 = 17, \text{ or } 0, \text{ or } \ldots $$ but chose $1$ because it turns out to be the most useful. You are entitled to an explanation for that choice.
Remember that for ordinary positive integers, $$ x^m x^n = x^{m+n} $$ (true because $x^n$ is defined as the result of multiplying $x$ by itself $n$ times).
That rule for adding exponents is so useful that mathematicians decided to preserve it when extending the meaning of raising $x$ to a power. Then $$ x^0 x^n= x^{0+n} = x^n . $$ The only way that can happen is if you define $x^0$ to be $1$.
The same desire to preserve the rule for adding exponents is why $$ x^{-1} = \frac{1}{x} \text{ and } x^{1/2} = \sqrt{x}. $$
For another good reason to define the product of no numbers as $1$ see Empty set and empty sum
Ethan Bolker
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Great answer! For the sake of pedantry: preserving the rule for multiplying exponents imputes the definition of fractional exponents. – Golden_Ratio Jan 22 '22 at 08:49
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I don’t think that it is arbitrary. This follows from the rules for powers. I don’t see how you can preserve the rules for powers when $x^0\neq 1$. Your answer just states that it is arbitrary but does not explain why it is arbitrary. – MachineLearner Jan 22 '22 at 16:20
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@MachineLearner I state pretty clearly that the definition is in principle arbitrary but that the way its done is the only way that preserves the rule on adding exponents - as your comment says. – Ethan Bolker Jan 22 '22 at 17:13
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Why should you define something when it follows from other simple statements? Could you give some references that state this as a definition? – MachineLearner Jan 22 '22 at 23:08
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@MachineLearner What simple statements do you think it follows from? The "proof" in the first line of your question assumes you already know $x^{-a} = 1/x^a$, which is a consequence of the definition of $x^0$ as $1$, so that "proof" is circular reasoning. This site expands on my answer: https://www.math.toronto.edu/mathnet/questionCorner/powerof0.html – Ethan Bolker Jan 23 '22 at 01:05
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We know that $a^n/a^m = a^{n-m}$ and it is clear that $a^n/a^n = 1$, when $a\neq 0$. Hence, we obtain $a^0 = 1$ for $a\neq 0$. It makes sense to define $0^0$ in order to get consistent results. But it does not make sense to define $a^0=1$ for $a\neq 0$, when if follows from simple arithmetic reasoning. – MachineLearner Jan 23 '22 at 13:47
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@MachineLearner You only "know that" from the definition of exponentiation when $n > m$. The whole point is to make sense of it when $n=m$. That does not follow from "simple reasoning" from known things. The definition $a^0 = 1$ makes sense (as you say) but it is a definition, not a consequence. – Ethan Bolker Jan 23 '22 at 14:04
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The definition of exponentiation also implies it for $n\leq m$. – MachineLearner Jan 23 '22 at 14:09
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@Ethan Bolker Would it be sufficent to define the extension of natural exponentiation to whole numbers with the property $x^nx^m = x^{n+m}$, where $n,m \in \mathbb{Z}$? By plugging in $m=0$ it follows that $x^0 = 1$ for $x^n \neq 0$. Wouldn't it then make sense to say that $x^0=1$ is a theorem? – jan Feb 05 '22 at 23:36
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@Ciprum No, it's not a theorem because it does not follow from the fact that exponents add when they are both positive. You can't just "plug in $n=0$". The expression $x^0$ has no meaning until you define it somehow. Defining it to be $1$ (as long as $x \ne 0$) turns out to be the most useful. – Ethan Bolker Feb 06 '22 at 01:01
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But I did define $x^0$, maybe I wasnt clear enough. I defined it as the value of $f(0)$ where $f: \mathbb{Z} \to \mathbb{R}$ is the function that satisfies $f(a+b) = f(a) + f(b)$ $\forall a,b \in \mathbb{Z}$ and $f(c) = x^c$ $\forall c \in \mathbb{N}, \exists x \in \mathbb{R}$. So in the argument in my previous comment, just read $x^0$ as f(0). – jan Feb 06 '22 at 18:41
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@Ciprum We agree that you did define $x^0 = 1$. Then it's not a theorem. Your theorem is that there is a unique way to extend your function $f$ defined on the positive integers to all the integers, preserving the rule for adding exponents. Then you define $f$ on the extended domain in that unique way, – Ethan Bolker Feb 06 '22 at 19:57
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If your notion of exponentiation is ultimately based on repeated multiplication (there are other ways to approach powers, where the answer may differ), then $x^0=1$ cannot be proven. It is defined to be such. And your fraction is one of many, many reasons that $1$ is the most reasonable value to give $x^0$, by a wide margin.
Arthur
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Start with the following equation
$$x=a, $$
in which $a>0$ and apply the $n^\text{th}$-root.
$$x^{\frac 1n}=a^ {\frac 1n}$$
If you take the limit $n \to \infty$ you will obtain
$$x^0=1,$$
because the infinith root of a positive number is $1$.
MachineLearner
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