Proving a^0=1: Step-by-Step Guide

[PeroK]

There is no logical issue - its done all the time. But in this case what you are doing by these definitions is extending, in a reasonable way, via the property you would like, namely x^(a+b) = x^a*x^b, what a^x is. Note - you can only go as far as the rationals via this. It just screams something more elegant should exist - that's the feeling I got from the final sentence of the original post. And it does - and it even exists for the reals - not just rationals. IMHO that more elegant way is better. But to be fair I don't think most students really care - only a few like the OP see surely there is something better than just defining things - and of those that do even less want to pursue it - even though if they did they will learn a lot about some more advanced math - namely calculus which will be to their credit.

Thanks
Bill

Well, it's all a matter of taste, I guess, and your definition of elegance. If teaching this to A-Level students, say, my preferred options would be:

1) Define $a^0 = 1$, with the necessary justification.

2) Take as an axiom that $\forall \ n \in \mathbb{Z}: a^na^m = a^{n+m}$, which implies that $a^0 = 1$.

3) Say that once you've done a course in real analysis and rigorously defined $\log x = \int_1^x \frac{1}{t}dt$ and defined $\exp(x)$ as the inverse of the log function and defined $a^x = \exp(x \log(a))$, then you can prove that $a^0 = 1$.

My guess, from this thread, is that many students would prefer 2). As a 16-year-old I would have been very unhappy with 3). Not to say baffled by it!

 

[bhobba]

Well, it's all a matter of taste, I guess, and your definition of elegance.

Well said

My guess, from this thread, is that many students would prefer 2). As a 16-year-old I would have been very unhappy with 3). Not to say baffled by it!

My proposal is you do not start with full blown analysis - you would have rocks in your head to try that. Only an introductory hand-wavy treatment is necessary to do what I suggest. Then you gradually move to honors calculus with integrated analysis, then an advanced rigorous treatment of linear algebra and multi-variable calculus. That Hubbard book is very good and as he says can be done at different levels - because of that some good high schools like Roxbury Latin use it in grade 12.

Yes it is a big issue the national curriculum places like here in Aus and England have. I personally am not a fan of it - here in Aus we are moving to all teachers having Masters and surely with teachers that qualified you just need to set some minimum standards and leave how to get there and how far you go up to the teacher and the students they have. At least one school here in Aus does that - the way it gets around the national curriculum is there is an out if you are on an individual learning plan. The first year at that school is spent doing that plan and there is a 50 minute meeting each day with your 'home' teacher ensuring you remain on track and is updated if required.

Thanks
Bill

 

[nuuskur]

Uh, I remember I started real analysis I on my first semester and it was considered to be hand-wavy (with many results revised and given proofs for in analysis III). Even the hand-wavy course put rocks in my head Nowadays, undergraduate levels start with Calculus I and II - that is like analysis I-lite abridged edition.

 

[Rada Demorn]

You better leave f(x) undetermined and admit that:

Nothing to add really. I am correcting the mistyping error in #64 post:

$f(x+x+x) = f((x+x)+x)= \frac{f(x+x)}{f(x)} = \frac{f(x)/f(x)}{f(x)}= \frac {1} {f(x)} = \frac{1}{f(x)}$

 

[bhobba]

Uh, I remember I started real analysis I on my first semester and it was considered to be hand-wavy (with many results revised and given proofs for in analysis III). Even the hand-wavy course put rocks in my head

The hand-wavy I am talking about is less rigorous than even what is done in HS in Aus where they introduce an intuitive idea of limit. Calculus Made Easy doesn't even do that - dy/dx is literally that - dx is a small quantity and dy = y(x+dx) - y(x). The chain rule is trivial dy/dx = (dy/dg)*(dg/dx) - dg simply cancels. Its all quite simple - intuitively. You would do it in grade 10, combined with a bit of pre-calc. Then more rigorous grade 11 and rather challenging grade 12. Just my thought - probably would crash in practice.

Still for those interested in starting to learn calculus, Calculus Made Easy is a good choice.

Thanks
Bill

 

[Stephen Tashi]

It is shown that all elementary functions are continuous in their domain, therefore due to continuity of $x\mapsto a^x$ we have $a^{q_n} \to a^0$.

How do we show the elementary function $f(x) = a^x$ is continuous at $x= 0$ without having a definition of $f(0)$?

 

[Stephen Tashi]

Still for those interested in starting to learn calculus, Calculus Made Easy is a good choice.

That may (or may not) be true, but it is not a good suggestion vis-a-vis the original topic of this thread, which concerned proof.

The thread is evolving in the direction of mathematical humor - challenge: Prove $x^0 = 1$ without assuming something that relies on having defined $x^0$. Similar challenges are: Prove $x + 0 = x$ without defining this to be a property of $0$, Prove $(x)(1) = x$ without defining this to be property of $1$, etc.

 

[nuuskur]

How do we show the elementary function $f(x) = a^x$ is continuous at $x= 0$ without having a definition of $f(0)$?

Ahh..see? Already got stuck in the circle :(. Well, then the whole argument is null.

We could prove $f(x) = a^x$ is continuous at $0$ with the candidate function value of $1$, which is not very difficult, but we'd need to know what $a^q$ for rational powers means. This is a can of worms.

 

[mathwonk]

pardon me rada, i did not explain clearly what i was proving. there are two issues here, one is how to define a^x for all real x, and you are right i have not done this. the other issue is whether any definition at all, will yield a^0 = 1. Therse are called the existence and the uniqueness aspects of the exponential function. I have done only the uniqueness aspect. I.e. I have not shown how to define f(x) = a^x for all x, but rather i have proved that any definition at all, if one is possible satisfying the rule f(x+y) = f(x).f(y), (and also f(n) = a^n for positive integers n) must then obey the rule that a^0 = 1.

The reason of course for choosing this property of f is that this property is indeed satisfied by the function a^n, defined for positive integers n. Hence it is reasonable to hope that the function will still satisfy this property for all x. As I stated at the end above, the existence problem, of actually saying how to define a^x for all real x, (which is much harder), is usually done by integration theory.

I.e. one first proves, by considering the integral of 1/x, that there is a function g defined for all positive reals, with g(1) = 0, and whose derivative is 1/x; it follows that this function satisfies g(xy) = g(x)+g(y) for all positive reals, and is invertible. Moreover the function takes all real values. Then the inverse f of this function is defined for all reals, satisfies f(0) = 1, and f(x+y) = f(x).f(y). It follows that if we set a = f(1), then this function f(x) is a good candidate for a^x. At least it agrees with a^x when x is a positive integer, and it has the right additive and multiplicative property.

This subject is a bit difficult to explain thoroughly to students in my experience. Some people prefer to do it by approximation, but that is even more diffiocult technically.

 

[Rijad Hadzic]

The fundamental issue is that when you use some mathematical symbols, you must define what you mean by that arrangement of symbols. Until you know what you mean by those symbols, you cannot start to do mathematics using them. In this case, for example, you might write:

$2^0$

But, what does that mean? There's no immediate way to "multiply 2 by itself 0 times". Unlike $2^1, 2^2, 2^3 \dots$, which have a simple, clear definition.

My recommended approach is to define $2^0 = 1$ before you go any further. Then you know what those symbols mean.

Now, of course, you need to be careful that a definition is consistent with other definitions, and you need to understand the implications of a certain definition.

In this case, the only other candidate might be to define $2^0 = 0$. But, when you look at the way powers work, you see that defining $2^0 =1$ is logical and consistent.

Thanks for this reply. I found this reply the most satisfying and beneficial.

 

[Stephen Tashi]

note also that then f(1/n)...f(1/n), (n times), = f(1/n+...+1/n) = f(1) = a, so f(1/n) = nth root of a = a^(1/n). Thus also f(m/n) = f(1/n)...f(1/n), (m times), = a^(1/n)...a^(1/n), (m times).

Thus f is determined on all rational numbers, and hence by continuity also on all real numbers.

There is the technicality of whether $f(m/n)$ is well defined. e.g. $f(2/3) = f(4/6)$ ? $f(x) = (-2)^x$

then this function f(x) is a good candidate for a^x.

I think you have in mind proving the existence of $e^x$.

At least it agrees with a^x when x is a positive integer, and it has the right additive and multiplicative property.

Ok, we can define $a^x$ in terms of $e^x$ for $a > 0$. Then we have the problem of the case $a \le 0$.

For the benefit of @Rijad Hadzic, we should point out that valid attempts to avoid directly defining $b^0 =1$ for all $b \ne 0$ involve defining other things and then showing these other definitions imply $b^0 = 1$. One cannot avoid using definitions as the foundation.

 

[mathwonk]

as far as i know there is no good way to define a function a^x for all real x when a is negative. hence the base for an exponential function is always chosen to be positive. (the difficulty of course is that a^(1/2) should be a square root of a, which will not exist in the real case for a < 0.)

 

[Rada Demorn]

pardon me rada, i did not explain clearly what i was proving. there are two issues here, one is how to define a^x for all real x, and you are right i have not done this. the other issue is whether any definition at all, will yield a^0 = 1. Therse are called the existence and the uniqueness aspects of the exponential function. I have done only the uniqueness aspect. I.e. I have not shown how to define f(x) = a^x for all x, but rather i have proved that any definition at all, if one is possible satisfying the rule f(x+y) = f(x).f(y), must then obey the rule that a^0 = 1.

Now sir, I think that it is actually a mistake to consider the Functional Equation: f(x+y) = f(x).f(y) over the Reals. If we take this over the Natural numbers it has solution $a^x$ and I think this is obvious.

Now we need only the restriction that f(0) is not equal to zero and let x=y=0 so that f(0+0)=f(0)=f(0).f(0) which gives immediately f(0)=1 or $a^0=1$ as required.

 

[bhobba]

Anyway, with that definition of the exponential function, let's evaluate $e^0$:

$e^0 = \Sigma_{n = 0}^{\infty} \frac{0^n}{n!} = \frac{0^0}{0!} + \frac{0^1}{1!} + \dots = \frac{0^0}{0!}$

Hmm. Maybe that doesn't resolve all your misgivings after all!

I was hoping you or someone else would pick it up but the power series expansion of e(x) is (not using the compact summation formula)

e(x) = 1 + x + x^2/2! + x^3/3! + ...

So e(0) = 1.

The compact sum equation, you are correct in pointing out, has issues with zero ie what is 0^0. Its either 0 or 1:
https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero

However as far as power series expansions go its usually taken as 1.

When that happens, to avoid any possible confusion, its probably better writing it out term by term.

Thanks
Bill

 

[Stephen Tashi]

but the power series expansion of e(x) is (not using the compact summation formula)

e(x) = 1 + x + x^2/2! + x^3/3! + ...

For the purpose of proving b^0 = 1, we shouldn't call it an "expansion" if this implies it is computed by taking derivatives and evaluating them at x = 0. That would require assuming $e^0 = 1$ in the first place.

For the purposes at hand, we can define $e^x$ by the above series and then define $b^x$ in terms of $e^x$ for $b > 0$.

That leaves open the problem of showing $(-2)^0 = 1$.

 

[jbriggs444]

Now we need only the restriction that f(0) is not equal to zero and let x=y=0 so that f(0+0)=f(0)=f(0).f(0) which gives immediately f(0)=1 or $a^0=1$ as required.

$f(0)=f(0)^2$ has two solutions. The other solution is alluded to by @PeroK in #23.

Edit: However, you have covered this possibility already. Apologies.

 

[Rada Demorn]

How do you like this for a definition?

a is a Natural Number.

a^3 is a volume, a cube.

a^2 is an area, a square.

a^1 is a length, a line.

What else a^0 could be but a point, an individual, a 1?

 

[PeroK]

How do you like this for a definition?

a is a Natural Number.

a^3 is a volume, a cube.

a^2 is an area, a square.

a^1 is a length, a line.

What else a^0 could be but a point, an individual, a 1?

If we have a cube of side 2, it's volume is 8, Or, if we have a cube of bricks, with each side 2 bricks long, we have 8 bricks.

For a square of side 2, it's area is 4. Or, a square of bricks needs 4 bricks.

For a line of length 2, it's length is 2. Or, a line of bricks needs 2 bricks.

For a point, it's size is 0. Or, if we have a zero-dimensional array of bricks, then there are no bricks.

By this intuitive reasoning, we should have $2^0 = 0$.

 

[jbriggs444]

If one chose to compute a^n by choosing one member of set A (with cardinality a) for column 1, one member of set A for column 2 on up to column n and counting the number of possible combinations, one could make a reasonable argument that there is exactly one way to choose nothing at all in no columns at all for a^0.

 

[Stephen Tashi]

Perhaps a valid proof that $b^0 = 1$ could be extended to valid proof that the "empty product" convention https://en.wikipedia.org/wiki/Empty_product is not a definition at all, but rather a theorem. (Just joking.)

 

[Rada Demorn]

I always thought that a point is nil-dimensional.

Perhaps I didn't convey the meaning properly that a power is dimensionality... ( i hope you don't take all this so seriously, eh? )

 

[Rada Demorn]

If we have a cube of side 2, it's volume is 8, Or, if we have a cube of bricks, with each side 2 bricks long, we have 8 bricks.

For a square of side 2, it's area is 4. Or, a square of bricks needs 4 bricks.

For a line of length 2, it's length is 2. Or, a line of bricks needs 2 bricks.

For a point, it's size is 0. Or, if we have a zero-dimensional array of bricks, then there are no bricks.

By this intuitive reasoning, we should have $2^0 = 0$.

Very, very wrong.

2^4 is the Figurate Number represented by the vertices (points) of a hypercube ( Dimension 4 }

2^3 is the Figurate Number represented by the vertices (points) of a cube ( Dimension 3 )

2^2 ................ of a square.

2^1 ........... by the end points of a line. ( Dimension 1 )

And 2^0 is just the point or the atom itself of Nil-Dimension.

 

[mathwonk]

rada you are quite right that we need the condition f(0) ≠ 0, and this is guaranteed by my assumption that f(x+y) = f(x).f(y) for all x,y, and f(n) = a^n for n = a positive integer and a > 0, since then f(1) = a ≠ 0, and thus f(x) ≠ 0 for all x. (if f(x) = 0, for some x, then for all y, f(y-x + x) = f(y-x).f(x) = f(y-x).0 = 0. i.e. thus either f(x) = 0 for all x or f(x) never equals zero.)