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### Inverses within Semigroups

```Date: 05/06/2002 at 14:49:38
From: Clayton
multiplication/division, exponentiation/logarithm

I have been studying set theory and have come across the definition
of inverses within semigroups. The book I have been reading is "Set
Theory and Logic," by Robert R. Stoll.

In his discussion of semigroups he defines inverses, "An element a of
a semigroup X, <X,*,e>, is inertible iff there exists an element a'
of X such that a * a' = a' * a = e."

He goes on to discuss the semigroups for addition in which e = 0 and
multiplication for which e = 1. He does not discuss a semigroup for
exponentiation. I would like to know the value of e for such a
semigroup if it exists. In other words what CONSTANT value e satisfies
the equation, a^b = e?

I have been exploring complex exponentiation in the hope that e for
exponentiation may have a complex solution. I have tried solving the
equation a^b = e, directly by:

log  a^b = log  e
a          a

b = log  e
a

If a is in R, however, and is allowed to vary from -oo to +oo, there
is no sensible solution for b that I can find.

Clayton
```

```
Date: 05/06/2002 at 17:04:22
From: Doctor Peterson
multiplication/division, exponentiation/logarithm

Hi, Clayton.

You've jumped over the preliminaries, which must never be ignored:
before you can talk about inverses, you have to look at the
definitions of a semigroup and of an identity (your "e"). You are
taking the definition of an inverse as if it were the definition of
the identity, skipping an important step.

But first, we have to see whether the set of real or complex numbers
forms a semigroup under exponentiation. I'll take the definition
given in Eric Weisstein's MathWorld:

http://mathworld.wolfram.com/Semigroup.html

Semigroup

A mathematical object defined for a set and a binary
operator in which the multiplication operation is associative.
No other restrictions are placed on a semigroup; thus a
semigroup need not have an identity element and its elements
need not have inverses within the semigroup.

(The fact that you have identified a semigroup as <X,*,e> suggests
that the existence of an identity is being assumed, making this
discussion refer to a monoid rather than a mere semigroup; but that
doesn't affect my discussion.)

So to define a semigroup, we first have to choose the set; since
you've mentioned complex numbers, I'll assume you intend that to be
the set we're using. You've specified the operation, exponentiation.
Now we have to show that it is associative:

a^(b^c) = (a^b)^c  for all a, b, c

But that is NOT true for exponentiation! Here is a simple
counterexample:

2^(3^4) = 2^81

(2^3)^4 = 2^(3*4) = 2^12

Clearly these are not equal, so we don't have a semigroup. (That's
probably why your book didn't mention it!)

Now let's pretend this doesn't matter, and take the next step (so you
can see what to do in other problems). What would the identity be?

The identity would be an element e for which

a^e = e^a = a  for all a

Let's see what this would mean. First, we want

a^e = a  for all a

Clearly e=1 works here.

We also want

e^a = a  for all a

Play with that a while, and you will see that there is no such e. So
even if we had a semigroup, it would not have an identity, so the
concept of an inverse would be meaningless.

As you see, when we choose an arbitrary set and operation, there is
no guarantee that it will satisfy the definition of even as broad a
concept as a semigroup. You can never assume anything here.

Incidentally, your subject line suggests that you are confusing the
concept of inverse in a semigroup with that of an inverse operation,
such as the logarithm. You are evidently looking for an explanation of
inverse functions (and the logarithm in particular) in terms of
inverse elements of a semigroup. The concepts are different, and
exponentiation does not even give a semigroup at all, as I said, but
there are some interesting relationships.

Let's first look at the case of addition.

We have seen that there is an (additive) identity, called 0, defined
by the fact that

a + 0 = 0 + a = a  for all a

We have seen that the (additive) inverse of an element a, which we
can write as "-a", is defined by:

a + -a = -a + a = 0

Now we can define the inverse operation to addition, which we will
call subtraction, by

a - b = a + -b

It is no coincidence that we use the same symbol, "-", for both
inverses; but it's important to distinguish them and realize that we
could have used different symbols, and that the properties of
subtraction have to be proved from the combined properties of addition
and the additive inverse. We are simply defining a new operation here.

What properties does an inverse operation have? Let's combine + and -
in various ways:

(a + b) - b = a
(a - b) + b = a
a + (b - a) = b
a - (a + b) = -b

These facts all depend on associativity for their proofs; try proving
them using the definition above. (You'll find that you also have to
prove that -(a+b) = -a + -b, and that --a = a, both of which are
easy, though the former depends on commutativity.)

Now you can do the same with multiplication; the only difference,
besides the symbols, is that the number 0 has no inverse, so that
division is not defined when the divisor is zero:

Multiplicative identity, 1:

a * 1 = 1 * a = a  for all a

Multiplicative inverse, a' (I'm avoiding exponential notation to keep
things clear):

a * a' = a' * a = 1

Inverse operation, division ("/"):

a / b = a * b'

Properties of the inverse operation:

(a * b) / b = a
(a / b) * b = a
a * (b / a) = b
a / (a * b) = b'

Now we want to find an inverse operation to exponentiation. The
trouble, we saw, is that exponentiation has none of the nice
properties of addition and multiplication that define semigroups (and
better things). It is not commutative, and not even associative;
there isn't an identity, much less an inverse. But we can still do
some things, though they have to be one-sided. For example, I showed
you that 1 can be considered the "right-identity":

a ^ 1 = a  for all a

though 1^a = 1, so it is not also a left-identity, and therefore not
a true identity. But there is no inverse, so we can't just blindly
define the inverse operation as exponentiation with the inverse, as
we defined subtraction and division. A "left-inverse" would require
that

a' ^ a = 1

and a "right-inverse" would require

a ^ a' = 1

neither of which exists. (Actually, the first a' would have to be 1
for all a, and the second a' would have to be 0, neither of which is
very useful as an inverse.)

Now look at the properties we showed for the inverse operation, and
decide what the inverse operation has to be in order to behave the
same way. As with the identity, we should not be surprised if we need
different operations in each case. I'll call the inverse
operation "v" since that looks like an inverted "^":

(a ^ b) v b = a
(a v b) ^ b = a
a ^ (b v a) = b
a v (a ^ b) = vb? (undefined - there's no inverse element)

I'll take those one at a time:

(a ^ b) v b = a

What can we do to a^b to get a back? Well, we can raise it to the 1/b
power:

(a^b)^(1/b) = a^(b * 1/b) = a^1 = a

Therefore, our "left-associative inverse operation" is

c v b = c^(1/b)

Hmmm ... this is in fact exponentiating with the (multiplicative!)
inverse, just what we would have liked to do if we had had an inverse
of our own.

Next case:

(a v b) ^ b = a

What number raised to the b power gives a?

(a^(1/b))^b = a^(1/b * b) = a^1 = a

so again,

a v b = a^(1/b)

That's nice: we've defined one inverse that works for both these
cases with "left-associativity." That inverse is in fact the bth root
of a, and my notation almost works, except that the b goes before the
"radical sign" rather than after (pure coincidence).

Will this hold up?

a ^ (b v a) = b

What exponent can we raise a to to get b? Here we need something
different, and we finally come to what you were looking for:

a ^ log_a(b) = b

so we define

b v a = log_a(b)

a v (a ^ b) = "vb"

We don't have an inverse, but let's see what happens if we use the
log again:

a v (a ^ b) = log_(a^b) (a) = log(a) / log(a^b)
= log(a) / (b log(a)) = 1/b

Wow! Again we find that if we use the multiplicative inverse, our
expectations are met. Even though 1/b is in no sense the inverse
element under exponentiation, it happens to fit in here. So we have a
nice "right-associative inverse" to exponentiation:

a v b = log_b (a)

So you see that exponentiation actually has two different inverse
operations, depending on which side we look at it from: the root, or
the logarithm. That shouldn't surprise you.

By the way, I've never studied "non-associative algebras," so the
ideas here are mostly my own, once I got past the standard concepts
of "left inverse" and so on. I've never heard of a "left-associative
inverse," but it seems to make some sense here.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 05/07/2002 at 17:45:28
From: Clayton
multiplication/division, exponentiation/logarithm

Dr. Peterson,

right on the head of the nail. It's not so much that I didn't
understand the properties of logs etc. but rather that I was curious
as to "why" the inverse operations of exponent and logarithm didn't
behave in the same way as addition and multiplication.

In other words, set theory, in many cases, adopts a more general or
abstract definition of something (say a semigroup) to attempt to
define many specific concepts from algebra or other "higher" order
maths. I was curious why exponentiation and logarithm didn't seem to
fit the general definition of inverses for semigroups, which you
cleared up for me.

I didn't know about some of the interplay between multiplicative
inverses in the exponents and the operations of logarithm and
exponentiation that you pointed out. However, I have long wondered
whether a notation such as "a v b" might be less unwieldy than the
klunky "log_b (a)" and hence lead to more intuitive understanding of
the logarithm and it's properties. Most of the time, mathematical
notation lends itself to the operation at hand, but in the case of
logarithm I think the notation falls short.

A related question I've had is this, (I don't want to belabor you,
but you seem interested in my questions): Given the natural numbers
and the operation of addition, we can construct the negative numbers
using the concept of an identity element, 0. We can then construct
the operation of multiplication in terms of iterated addition. We can
then construct fractions using another identity element, 1. We then
proceed to construct exponentiation as iterated multiplication. We
construct logarithm as the inverse of exponentiation as you did in

What then? Can we construct another operation using iterated
exponentiation? Although this doesn't mean it doesn't exist, I have
never seen such an operation defined. Maybe because it wouldn't have
any use, but then again, the fellow who discovered matrices thought
they would have no application at all!

I don't want to get too deep, or ask too many questions, but is there
a relation, that you are aware of, between this hierarchy of iterated
operations (if it exists) and fractals? Fractals are, in general,
self-similar in one way or another and are constructed using
iterations. Likewise, a heirarchy of iterated operations would be
self-similar in a sense as well. Also, there seems to be a close
relation between exponentials and fractals. For example, the length of
a Koch snowflake after n iterations is,

L = 3(4^n)(3^-n)a

where a is the length of one side of the triangle upon which the Koch
snowflake is constructed. This is an exponential function. Maybe
iterated exponentiation would have an application here. :)

Again, thank you for taking the time to answer my questions.

Clayton
```

```
Date: 05/07/2002 at 22:14:43
From: Doctor Peterson
multiplication/division, exponentiation/logarithm

Hi, Clayton.

Your mention of the awkwardness of logarithm notation brings up a
point I'd thought of mentioning. Both exponents and logs have an odd,
asymmetrical notation with one number "off the line," and that
actually matches up with their asymmetrical behavior. In working with
my temporary "a v b" notation, I found it actually awkward to use,
because it's hard to remember which number is the base. Of course, I
have gotten used to using a^b for exponents (and being very careful
about parentheses as a result), so I would probably get used to a new
logarithm notation. My first thought, partly based on the observation
that my "a v b" has the base come second, would be to write the base
as a subscript, showing its inverse relationship to exponents:

a
a  = log_b(a)  inverse of  b  = exp_b(a)
b

Of course that would be far too easily confused with other uses of
subscripts. It's fun to think about how we could do this, though.

As for iterated exponentiation, that kind of operation has indeed
been studied. I haven't found it very interesting, but it has some
use. Here is a page about it:

big numbers - Susan Stepney
http://www-users.cs.york.ac.uk/~susan/cyc/b/big.htm

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Exponents
College Logic

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