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We all get it $$x^0=x^{1-1}=x^1\cdot x^{-1}=x/x =1$$ as long as $x\neq 0.$ But this supposes the rule $x^{m+n}=x^mx^n$ for $m$, $n\in \mathbf{Z}.$ Unfortunately, each proof of the rule $x^{m+n}=x^mx^n$ that I have seen uses the fact that $x^0=1$ in it so to use the fact that $x^{m+n}=x^mx^n$ in a proof that $x^0=1$ would be a circular argument. So, I suppose that I am asking for one of two things: (1) A proof that $x^0=1$ for non-zero x, using only the field axioms (and maybe definitions like $x^{-1}=1/x$), or (2) A proof that $x^{m+n}=x^mx^n$ for $m$, $n\in \mathbf{Z}$ that does not employ $x^0=1$ in its proof. The latter seems more unlikely. Any insight is appreciated.

Chris Christopherson
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    That's either a definition or we have to wait until calculus, limits and stuff...BTW, Field Theory has nothing to do with this. – DonAntonio Sep 11 '22 at 23:13
  • Ha of course it does. To even begin a course like real analysis (calculus) you start walking around in a field. I get that the tag might not bring in the right crowd though so ill take it off. – Chris Christopherson Sep 11 '22 at 23:17
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    I think you have things slightly backwards. By definition, if $n$ is a nonnegative integer, $x^n=\underbrace{x\cdot x\dots\cdot x}_{\text{$n$ times}}$, and it is a firmly established mathematical convention that the product of zero numbers equals the multiplicative identity, $1$. One reason why this convention makes sense is precisely because we want the identity $x^nx^m=x^{n+m}$ to hold. – Joe Sep 11 '22 at 23:18
  • @Joe I guess then its just slightly frustrating because there is an easy proof that $$0\cdot a=0$$ using the axioms. So we are okay if the empty sum has a proof but we just decided that the empty product does not need one? – Chris Christopherson Sep 11 '22 at 23:22
  • Obtaining the general rule does not require assuming $x^0=1$, that result comes out as a special case by using definitions of "product" and "power of". – Nij Sep 11 '22 at 23:27
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    @ChrisChristopherson: Well, the fact that the empty sum is equal to $0$ is also just a definition, convention, whatever. There are reasons that this definition makes sense, yes. But ultimately there is nothing to stop me from defining the empty sum as equal to $27$. You can argue that the empty sum ought to equal $0$ because $0\cdot a = 0$, but that is a motivation for the definition of the empty sum, and not a proof. – Joe Sep 11 '22 at 23:29
  • @Nij this is great! Can you show me such a proof ? – Chris Christopherson Sep 11 '22 at 23:30
  • @Joe wait is the empty sum not defined by a number $a$ added to itself zero times? In other words, the empty sum is by definition $0a$ right? But there is only one choice for $0a$ under the field axioms: $$0.$$ I guess I wish there was a way to show that there is only one choice for $x^0$ under the field axioms. Also does this site offer reasons for downvotes? – Chris Christopherson Sep 11 '22 at 23:38
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    @ChrisChristopherson: My point is that you can, in principle, define the empty sum however you like. The same goes for any definition in mathematics. I can define $a+b$ to mean the product of $a$ and $b$, even though that would be facetious. If you decide to define the empty sum as $0a$, then yes, according to the field axioms, it must be equal to $0$. And if you decide to define $x^0$ in a way that is consistent with the usual rule $x^m\cdot x^n=x^{m+n}$, then $x^0$ must be equal to $1$. Does that help? – Joe Sep 11 '22 at 23:48
  • See this answer. – 2'5 9'2 Sep 12 '22 at 00:26
  • @ChrisChristopherson how much calculus do you know? – CyclotomicField Sep 12 '22 at 00:42
  • @CyclotomicField I'd say a good amount. Not nearly enough though and I will never get to know enough. I earned an M.A. in Math. – Chris Christopherson Sep 12 '22 at 01:04
  • Take as a definition: $x^n= x \cdot x \cdot x \cdots$, with $n$ copies of $x$. If we divide this equation by $x$, the RHS loses one copy of $x$, and due to that, using the same definition, the exponent $n$ on the LHS becomes $n-1$. Hence $x^n/x = x^{n-1}$. Now, setting $n=1$, we have $x^1/x = x^0$. But clearly by our definition, $x^1=x$, and by standard axioms $x/x=1$. Therefore $x^0 = x^1/x = x/x = 1$. (This claims nothing about addition of exponents, and doesn't even suppose that division decreases the exponent. I guess it assumed we've agreed on what division is.) – Eric Snyder Sep 12 '22 at 05:41
  • I think my abstract algebra teacher just defined it as 1 because it would validate the identity. – TC159 Dec 20 '22 at 02:44
  • You can't prove a definition. Equations such as $x^0=1$ and $x^1=x$ as not field axioms or derived from field axioms. They are definitions. – Somos Dec 20 '22 at 03:09
  • You tagged this as algebra-precalculus but I just wanted to point out that you can perform the integral (a Laplace transformation) on $L({t^0}) = \int_0^\infty t^ne^{-s*t}dt$ and you'll end up obtaining $\frac{0!}{s^{(0+1)}} = \frac{1}{s}$ once you evaluate that integral. The inverse laplace transformation of $L^{-1}(\frac{1}{s})=1$. I could do this with the integrals and work it out properly for you but maybe there is some intuition to be gained by changing your domain? – jake mckenzie Dec 20 '22 at 18:41
  • You want a proof from the field axioms. There is no mention of exponentiation in the field axioms, just addition and multiplication, so at the very least you are going to have to accept a definition of exponentiation as a supplement to the field axioms. The standard way to define $x^n$ for $x$ in the field and $n$ a positive integer is inductive; define $x^1$ to be $x$, and $x^n$ for $n>1$ to be $x^{n-1}x$. Then, using induction again, $x^{m+n}=x^{m+n-1}x=x^mx^{n-1}x=x^mx^n$. So, from the field axioms, you're only one definition and two inductions away from $x^{m+n}=x^mx^n$. – Gerry Myerson Dec 22 '22 at 21:01

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Suppose we define $x^n$ in the obvious way for positive integer $n$, but abstain from defining $x^0$. We can prove by induction on $n$ that $x^m\cdot x^n=x^{m+n}$ for all values of $x$, $m$, and $n$. For $n=1$, this is the assertion that $x^{m+1}=x^m\cdot x$, which is true by definition. Assuming the statement is true for a given value of $n$, it follows that $$x^m\cdot x^{n+1}=x^m\cdot (x^n\cdot x)=(x^m\cdot x^n)\cdot x=x^{m+n}\cdot x=x^{m+n+1} \, .$$ Consider it as an exercise to justify why each of these equalities must hold.

Then, if we want to preserve the rule $x^m\cdot x^n=x^{m+n}$ for when $m$ and $n$ could be any integers, then at least in the case $x\neq0$, we must define $x^0$ as $1$, and $x^n$ as $1/x^{-n}$ for negative integers $n$. From these definitions, the proof of $x^{m+n}=x^m\cdot x^n$ is straightforward: just break it into different cases.

This reasoning is valid in any field, provided we understand the term "integer" to mean an element of the following set (which is possibly finite): $$ \{0,1,-1,1+1,-1-1,1+1+1,-1-1-1,\dots\} \, . $$

Joe
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  • This is certainly helpful thank you. Perhaps I am stubborn but I will need proof to convince me that $$x^{m+n} = x^mx^n$$ for $m$, $n\in \mathbf{Z}\backslash {0 }.$ I am working on a proof right now and will post what I come up with (even though this goes against what I was asking for in the title). I will also attempt to prove $x^{-n}=1/x^n$ for natural $n$. – Chris Christopherson Sep 12 '22 at 00:55
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    @ChrisChristopherson: It is not does make sense to prove that $x^{-n}=1/x^n$ when $n$ is a natural number, because it's a definition, not a theorem. What does make sense is to prove the statement "if we are to define $x^{-n}$ for natural $n$ in such a way that the law $x^{m+n}=x^m\cdot x^m$ holds, then we must define it as $1/x^n$". More formally, for each $a>0$, consider the function $f_a:\mathbb N\to \mathbb N$ given by $f_a(x)=a^x$. You want to prove that there is exactly one extension of $f$ to $\mathbb Z$ for which the "indices law" still holds. – Joe Sep 13 '22 at 20:21
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The main motivation for defining $x^0=1$ is because $0$ is the additive identity, $1$ is the multiplicative identity, and they're intimately tied together by the logarithm. Why? Well, logarithms have the property that $\ln(ab)=\ln(a)+\ln(b)$ which is how the arose initially. This means $\ln (a)=\ln(a1)=\ln(a)+\ln(1)$ and so if we extract $\ln(a)=\ln(a)+\ln(1)$ and subtract $\ln(a)$ from both sides we see that $\ln (1)= 0$. People used tables of logarithms to turn large multiplication problems into easier to computer addition problems. You can see the same idea here because it's essentially $x^ax^b=x^{a+b}$.

However we don't have to define the logarithm to be inverse of the exponential function. With a little calculus we can define the logarithm to be $\ln t = \int_1^t \frac{1}{x} \, dx$ and then prove it has the properties we're looking for like being the inverse of $e^x$ and the aforementioned $\ln(ab)=\ln(a)+\ln(b)$. If you don't know any calculus this equation basically represents the area between the graph and the $x$-axis of the function $f(x)=1/x$ between $1$ and $t$. However if $t=1$ then the starting and end point are both $1$, so the area has to be $0$. Proving things this way is a little challenging but it can be done without appealing the usual definitions using $x^ax^b=b^{a+b}$.

CyclotomicField
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Maybe this is too much machinery for a simple thing but one could do:

  1. Define $exp(x) = \lim_{N \rightarrow \infty} 1+ \sum^{N}_{n=1} \frac{x^n}{n!}$ by showing it is well defined for each $x$
  2. Show it is monotically increasing and thus there is an inverse, which can be defined as $\log$
  3. Define $a^x =e^{\log(a)x}, \forall a>0$
  4. Show that one recovers the usual $a^x$ for $a,x \in \mathbb{N}$
  5. Show that $a^{x+y}=a^xa^y,\forall x,y \in \mathbb{R}$

Now you get as a theorem, $a^0=1, \forall a>0$

Rohan Didmishe
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  • OP asks for a proof from the field axioms. I don't think this qualifies. – Gerry Myerson Dec 22 '22 at 20:50
  • But none of the steps assume anything other than real field axioms and definition of limits right?(of course inverses and stuff) – Rohan Didmishe Dec 22 '22 at 20:51
  • A proof of $x^0=1$ should hold in all fields, not just in the reals. There are fields that don't have convergent series, don't have a concept of "monotone increasing", but still have $x^0=1$. I'd say you're not using "too much machinery" but you are carrying too much excess baggage. – Gerry Myerson Dec 22 '22 at 21:07
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[Most of this is in the comments, mine or those by other users, but I think it's worth an answer.]

You want a proof from the field axioms. There is no mention of exponentiation in the field axioms, just addition and multiplication, so at the very least you are going to have to accept a definition of exponentiation as a supplement to the field axioms. The standard way to define $x^n$ for $x$ in the field and $n$ a positive integer is inductive; define $x^1$ to be $x$, and $x^n$ for $n>1$ to be $x^{n-1}x$. Then, using induction again, $x^{m+n}=x^{m+n-1}x=x^mx^{n-1}x=x^mx^n$. So, from the field axioms, you're only one definition and two inductions away from $x^{m+n}=x^mx^n$. Now $x^m=x^{m+0}=x^mx^0$, so $x^0=1$.

Gerry Myerson
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  • I like this but how did you apply the rule $x^{m+n}=x^mx^n$ with $n=0$ when so far you had only proven it for positive integers $m$ and $n$ ? – Chris Christopherson Dec 23 '22 at 07:49
  • You have a point. Let me think for a while.... – Gerry Myerson Dec 23 '22 at 14:20
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    OK, I've come to the point of view that you can't prove anything about $x^0$ until you have defined it, as the field axioms don't say anything about it. And since $x^0=1$ is so closely related to $x^{m+n}=x^mx^n$, via $x^m=x^{m+0}=x^mx^0$, you can't prove $x^{m+n}=x^mx^n$ for all integer $m,n$ without first defining $x^0$. In short, I think what you're looking for, can't be done. – Gerry Myerson Dec 24 '22 at 22:11
  • I have to say that I agree. Although, it is still not really sitting well with me though so I'll have to learn to come to terms with it. – Chris Christopherson Dec 25 '22 at 06:06
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You can not prove this. It is merely a convention, or a definition, not a statement that needs to be proven.

All of the "proofs" in other answers are circular and they use this convention in an implicit way in order to "prove" it.

Yanko
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  • Right. Similar to the "proof" I gave in the question itself. – Chris Christopherson Dec 25 '22 at 06:07
  • It is actually provable, if you watch what's going on. – wendy.krieger Dec 25 '22 at 09:08
  • The trick is to prove (a) 0 is the IE of addition, and (b) $x^0$ is the IE of multiplication. It thus immediately follows that since two elements listed as IE for the same operation, is one element, then $x^0=1\ \forall x$. – wendy.krieger Dec 26 '22 at 09:40
  • @wendy.krieger ahh... well the convention that $x^0$ is "the IE of multiplication" is indeed equivalent to the convention that $x^0=1$... – Yanko Dec 29 '22 at 14:45
  • The thing here is that 0 in any guise is the IE of addition. That a.x^0 = a is not a convention, but an expression of count (1 a and 0 x), also implies x^0 for all x. It doesn't require a convention. You don't need to have elephants in the room to count 0 elements, and you don't need to have x in the term to count 0 copies of x. To suppose that 0 as a counting number is different to 0 as a real is clearly wrong, since both are the IE of addition, and x^0 is the IE if multiplication =1, and this remains true for x=0. – wendy.krieger Dec 31 '22 at 09:31
  • $x^0\times a = a$ for all $a,x$ is only true if you take it as a convention, or you assume something equivalent (e.g. $x^0=1$ for all $x$). – Yanko Dec 31 '22 at 13:01
  • 0 is a counting number. No convention needed. Zero simply means an empty column, So in aabbc, there are zero elements d, and the thing is $a^2b^2c^1d^0 = a^2b^2c$ for all d, including d=0. – wendy.krieger Jan 01 '23 at 10:29
  • If taking $x^0=1$ is a convention, then the axioms are woefull incomplete, and admits the possibility of $2=3$. It is provable by showing that it functions as the IE in every case, and therefore is the IE. – wendy.krieger Jan 01 '23 at 10:33
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The proof that $x^0=1$ lies in that $x^0 \times a = a\ \forall a$, and therefore $x^0$ is the identity of multiplication. But since it can be demonstrated that $1$ is the identity, you show that $\forall x\ x^0=1$.

You will hear stories about the limit of $0^x$ as $x -> 0$, but the two ways of moving left on this line, is by division (which can get you to 0, but invilves division by 0), or by taking roots (which does not involve division, but you can't make $x=0$, since it remains positive.

Allowing $0^0$ to be anything else but one simply means that you can write any statement, and it will be true. In essence, for integer exponents, such as $0$, $1$, etc, it is the number of times the base is involved in the product. If you suppose $3 = 3\times 0^0 = 3 \times 8 = 37$, simply because each line has 0 repetitions of 0, and this might be freely set, the lines don't necessarily have to agree arithmetically.

wendy.krieger
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I thought $x^0$ must be a contant which is not zero, for all of $x$. Say it is $a\neq 0$. But, then $a^0=a$ too. Taking $x$-th power of both sides, since $(a^0)^x=a^0$, we get $a^x=a$,$\;\forall x$. But this is possible iff $a=1$.

I pushed down my joke below, because mathematics is not personal. Social media is. My joke was: I hope there is not a gap to be attacked by Gary in this thread. It is just for fans or fun:

Bob Dobbs
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