Lajos Lóczi:
On commutative and associative powers
Addition and multiplication both obey the law of commutativity
a+b=b+a and
a^{.}b=b^{.}a,
and associativity
(a+b)+c=a+(b+c)
and
(a^{.}b)^{.}c=a^{.}(b^{.}c),
for any real numbers a,b and c. However,
powers fail to be commutative and associative, as we know in general
a^{b}b^{a} and
a^{(bc)}(a^{b})^{c}.
In spite of this, equality may hold in special cases. Our aim is to
determine all positive real pairs (x,y) and triples
(x,y,z) satisfying
and
respectively, then, with their help study the integer and rational
solutions as well.
1 The commutative case
First, we want to find the positive real solutions of (1). (The
reason we confine our attention to the positive case only is that
powers in general are not defined for negative bases in the set of
real numbers, furthermore, the case when one of the variables is equal
to zero does not seem too compelling either.) To start with, let us
try to guess some solutions. We soon find x=2 and y=4
(or vice versa, because of the symmetry). Since (1) does not
seem to admit any other positive integer solutions, let us try some
radicals. If we are lucky, we discover and
\(\displaystyle y=3\sqrt{3}\), since
\(\displaystyle (3\sqrt{3})^{\sqrt{3}}=(\sqrt{3}^{\,3})^{\sqrt{3}}={\sqrt{3}}^{\,3\sqrt{3}}.\)
The point here is that \(\displaystyle 3\sqrt{3}\) can also be written as . On further investigation we hit on the pair
\(\displaystyle x=\root3\of{4}\) and \(\displaystyle y=4\cdot\root3\of{4}\), since
\(\displaystyle (4\cdot\root3\of{4})^{\root3\of{4}}=(\root3\of{4}^{\,4})^{\root3\of{4}}={\root3\of{4}}^{\,4\cdot\root3\of{4}}.\)
This time the equality \(\displaystyle 4\cdot\root3\of{4}=\root3\of{4}^{\,4}\) played
a key role. We notice that essentially in all examples
y=vx with vx=x^{v} is true,
which will be the right track to follow.
1.1 Introducing a new variable
We look for y in the form y=vx with v
being a real parameter, that is we set . Then (1) becomes
(vx)^{x}=y^{x}=x^{y}=x^{vx}=(x^{v})^{x}.
Since every expression here is positive, raising the leftmost and
rightmost expressions to the power \(\displaystyle \frac{1}{x}\), we get the crucial relation
vx=x^{v}. Now multiplying by we
have v=x^{v1}. If v1, that is if
x\(\displaystyle ne\)y,
raising to the power \(\displaystyle \frac{1}{v1}\) we derive . The corresponding y reads
\(\displaystyle y=vx=v\cdot
v^{\frac{1}{v1}}=v^{\frac{1}{v1}+1}=v^{\frac{v}{v1}}=x^v.\)
If v=1, then y=x.
The expression obtained for x will appear several times
later on, so we set \(\displaystyle h(v)=v^{\frac{1}{v1}}\). The domain of
definition of the function h is the positive reals except 1,
and its range is apparently a subset of the positive reals.
We now have the possible solutions, which are actually
the solutions to (1): if v=1 and x is an
arbitrary positive real number, then y=x is clearly a
solution, which we will call a trivial solution. If
v\(\displaystyle ne\)1, then
x=h(v) and
y=v^{.}h(v) are nontrivial
solutions, since, as we have just seen,
y=vx=x^{v} and
y^{x}=(x^{v})^{x}=x^{vx}=x^{y}.
Our first goal has been achieved by establishing that all positive
real solutions (x,y) to
x^{y}=y^{x} are of the
form (x,x) and
(h(v),v^{.}h(v))
(x>0,v>0,v\(\displaystyle ne\)1).
The function h has interesting properties. If v>0
and v\(\displaystyle ne\)1,
then h(v) is the value which multiplied by v or
raised to the power v yields the same:
v^{.}h(v)=h(v)^{v},
as we have proved it. Another functional equation h satisfies
is \(\displaystyle v\cdot
h(v)=h(\frac{1}{v})\), or equivalently , as one can easily check.
This implies for example, that nontrivial solutions can be written
as \(\displaystyle (h(v),h(\frac{1}{v}))\) too, consequently a solution
(x,y) is transformed into the solution
(y,x) by the substitution .
The function h is not defined for v=1, nevertheless
it can be shown that it is strictly decreasing and (with e=2.71828... being the base of
the natural logarithm).
Figure 1: \(\displaystyle h(v)=v^\frac{1}{v1}\)
1.2 A little calculus
We now would like to have a picture of the graph of the solutions
of (1). The trivial solutions (y=x for
x,y>0) are the bisector of the first quadrant in the
xy plane. The nontrivial solutions however are not of
the usual form y=f(x), since y is not
directly expressed in terms of x, but both x and
y are functions of the parameter v. Nonetheless, it is
seen that for any x from the range of h there exists
exactly one y different from x such that
x^{y}=y^{x}, which implies
the existence of the function \(\displaystyle f\colon x\mapsto y\). Using this function,
nontrivial solutions can be written as (x,f(x)),
namely, if h^{[1]} denotes the inverse function of
h (which exists because of its strict monotony), then
x=h(v) implies
v=h^{[1]}(x), hence
(h(v),v^{.}h(v))=(x,h^{[1]}(x)^{.}x)
is what we required (with
f(x)=x^{.}h^{[1]}(x)). However,
this manipulation will not give any further information, since
expressing v from x=h(v)that is
determining the function h^{[1]}does not seem to be
possible via elementary functions.
Now we investigate the behaviour of the functions
h(v) and v^{.}h(v) in the
parametric representation, from which we will able to sketch the graph
of the nontrivial solutions. Plotting this graph together with the
graph of the trivial solutions in the same system of coordinates
yields the complete positive real solutions to (1).
We will need the concept of limit and derivative of a function
(together with some wellknown limits), so these proofs will be
omitted, or only sketched.
To simplify things a bitand to see some other nice relationswe
again introduce a new variable: we reparametrize our coordinate
functions. Let u denote the exponent in h(v),
that is let \(\displaystyle u=\frac{1}{v1}\). This yields \(\displaystyle v=1+\frac{1}{u}\). In terms of this new
variable the two functions become
\(\displaystyle h(v)=\left(1+\frac{1}{u}\right)^u\) and \(\displaystyle v\cdot
h(v)=\left(1+\frac{1}{u}\right)^{u+1}\).
Call these new functions of u g_{1}(u)
and g_{2}(u), respectively. It is easy to see
that graphs of g_{1} and g_{2} are
reflections of each other about the axis
in the xy plane, since the image of a point u of
the xaxis under this reflection is (u1), and the
substitution u\(\displaystyle mapsto\)(u1) transforms g_{1} into
g_{2}, because
g_{1}(u1)=g_{2}(u) and
g_{2}(u1)=g_{1}(u).
Therefore it is sufficient to examine the function
g_{1}, properties of its counterpart
g_{2} will be easily deducible thereafter. Since
v runs through the positive reals except 1, it is easy to see
that after the substitution u will run through the set , hence this will be the domain of
definition of g_{1}. The behaviour of
g_{1} at the endpoints of its domain is established by
using the following wellknown limits. , from below. , from above, since the substitution
reduces it to \(\displaystyle \lim_{w\to+\infty}\root{w}\of{w+1}=1\). Further, , since the base tends to 0 from
above, while the exponent tends to (1). Finally, from above, shown by the
substitution w=u. The function g_{1} is
continuous on its domain. It can be proved that it is strictly
increasing on (\(\displaystyle infty\),1) and on (0,+\(\displaystyle infty\)). Its graph is shown on Figure
2.
Figure 2: \(\displaystyle g_1(u)=\left(1+\frac{1}{u}\right)^u\)
Now we are in a position to draw the nontrivial solutions of the
equation x^{y}=y^{x} from
the parametrization
(g_{1}(u),g_{2}(u)), , see Figure 3. (Naturally,
this is equivalent to the original parametrization
(h(v),v^{.}h(v)),
v>0, v\(\displaystyle ne\)1.) As u increases from  to 1, the
points (g_{1}(u),g_{2}(u))
define the lower right part of the curve on the figure, since the
first coordinate grows from e to +\(\displaystyle infty\), while the second coordinate decreases
strictly from e to 1. Conversely, as u increases from 0
to +\(\displaystyle infty\), the
points (g_{1}(u),g_{2}(u))
describe the upper left part of the curve, because the first
coordinate grows from 1 to e, and the second coordinate
strictly decreases from +\(\displaystyle infty\) to e. The straight line with
slope of 45^{o}, as we already know, comes from the trivial
solutions. Thus Figure 3 contains all the positive pairs
(x,y) for which the corresponding powers
commute. Symmetry of the solutions manifests in the reflection
symmetry of the graph with respect to the axis
y=x. Trivial and nontrivial solutions meet at the point
(e,e).
Figure 3
1.3 Some simple consequences
The analysis carried out above implies for example, that powers can
not commute, if the base and the exponent are different numbers being
less than 1, since in the nontrivial case g_{1} and
g_{2} both have values strictly greater than
1. Similarly, if x,y>e and x y,
then x^{y}=y^{x} does not
admit any solutions. Moreover, substituting the parametric
representations into the expression x^{y} yields
\(\displaystyle h(v)^{v\cdot h(v)}=v^{\frac{v^{\frac{v}{v1}}}{v1}}.\)
It can be proved that the range of this function is the interval
(e^{e},+\(\displaystyle infty\)), which implies for example, that if
x^{y}<e^{e} and
x\(\displaystyle ne\)y, then
x^{y}\(\displaystyle ne\)y^{x}.
1.4 Integer and rational solutions
After the positive real solutions, we turn our attention to the
positive integer and positive rational solutions of
x^{y}=y^{x}. Here again we
have trivial solutionsevery integer or rational x>0 with
y=x is appropriateand the more involved nontrivial
ones. Integer solutions can be inferred using the graph on Figure 3:
the upper left branch contains only the integer grid point
(x,y)=(2,4), since the first coordinate satisfies
1<x<e, but e<3 implies x=2 (and
correspondingly y=4). Due to symmetry, the only grid point on
the lower right branch is (4,2). These are the integer solutions of
(1).
In order to acquire rational solutions, the values of parameter
v are to be determined for which both components of the pair
(h(v),v^{.}h(v)) are
positive rational numbers. If h(v) and
v^{.}h(v) are rational, then v has
to be also rational, since h(v) is positive. So we may
suppose \(\displaystyle v=\frac{p}{q}\) with positive integers p and q,
being relatively prime. Taking into account that we are looking for
nontrivial solutions, v\(\displaystyle ne\)1 must hold, that is pq. As we have
seen, the substitution \(\displaystyle v\mapsto\frac{1}{v}\) only interchanges
x and y, so it is sufficient to study the case
v>1 (or equivalently, p>q), that is to
inspect the upper left branch of the graph on Figure 3. So
p>q\(\displaystyle ge\)1. Plugging \(\displaystyle \frac{p}{q}\) into v, we get
\(\displaystyle h(v)=\left(\frac{p}{q}\right)^{\frac{q}{pq}}.\)
Let m=pq, then m\(\displaystyle ge\)1 is an integer. First we show that if
m>1, then \(\displaystyle h\left(\displaystyle\frac{p}{q}\right)=\root{m}\of{\frac{p^q}{q^q}}\)
can not be rational. Since p and q are relatively prime,
the fraction \(\displaystyle \frac{p^q}{q^q}\) is also in its lowest terms. The
m^{th} root of such a fraction can only be rational if
both its numerator and denominator are perfect m^{th}
powers.
Indeed, let \(\displaystyle \root{m}\of{\frac{r}{s}}=\frac{a}{b}\), with r,s
and a,b being relatively prime numbers,
respectively. (r,s,a,b>0 may be
assumed.) Then \(\displaystyle \frac{r}{s}=\frac{a^m}{b^m}\) holds, further,
a^{m} and b^{m} are
relatively prime. Since the reduced form of a rational number is
unique (the numerator and the denominator are positive), we conclude
that both r and s are perfect m^{th}
powers.
Hencesupposing to the contrary that is
rationalp^{q}=a^{m} and
q^{q}=b^{m} hold. Now
q and m are relatively prime, since
p=q+m and the greatest common divisor (gcd) of
p and q is 1, from which we verify that p and
q are also perfect m^{th} powers. To this end,
pick an arbitrary prime from the prime factorization of p. If
its exponent is denoted by k, then its exponent in the
factorization of p^{q} is
k^{.}q, now divisible by m. Because the
gcd of m and q is 1, we see that m divides
k, therefore p is indeed a perfect m^{th}
power. A similar argument indicates that q is also a perfect
m^{th} power.
Now the equality m=pq can not hold, since the
difference of two different m^{th} powers is greater
than m. Indeed, let
t_{1}>t_{2}>0 be integers, then
t_{1}^{m}t_{2}^{m}=(t_{1}t_{2})(t_{1}^{m1}+t_{1}^{m2}t_{2}+...+t_{1}t_{2}^{m2}+t_{2}^{m1}),
and the right hand side is strictly greater than
(t_{1}t_{2})^{.}m^{.}t_{2}^{m1},
which is at least m.
Thus we have proved that if m=pq>1, then
\(\displaystyle h\left(\displaystyle\frac{p}{q}\right)\) is irrational.
Finally, in the case m=1 (i.e. p=q+1),
\(\displaystyle h(v)=\left(\frac{q+1}{q}\right)^q\) is clearly rational. Then
\(\displaystyle v\cdot
h(v)=\left(\frac{q+1}{q}\right)^{q+1}\). Using the more usual n
instead of q, we have the rational solutions as
\(\displaystyle x=\left(1+\frac{1}{n}\right)^n\) and ,
or vice versa, with integer n\(\displaystyle ge\)1. (If n=1, the formula gives
the integer solution x=2 and y=4 we already have.)
Therefore, these are the points of the curve on Figure 3 with both
coordinates being rational.
The sequences \(\displaystyle \left(1+\frac{1}{n}\right)^{n}\) and play an important role in real
analysis, because they tend to the number e (as n+): this
constant is usually defined as the limit of these very sequences. We
have established another interesting property of them, namely,
corresponding terms of these sequences are the only (positive and
different) rational numbers for which powers
x^{y} and y^{x} commute.
1.5 A different approach
In finishing the commutative case, we mention a different approach
to obtain some information on the real solutions to the equation
x^{y}=y^{x}. Namely, by
raising both sides to the power \(\displaystyle \frac{1}{xy}\) (x,y>0), we get
\(\displaystyle x^{\frac{1}{x}}=y^{\frac{1}{y}}\), which requires the equality of
two values of the same function (not necessarily for different
arguments): the original equation is equivalent to
f(x)=f(y) with .
We trivially get a solution, if x=y. The question is
whether equality can hold true for x\(\displaystyle ne\)y. Some further analysis shows (using
limits and monotony) that the function f is onetoone on the
interval (0,1], and maps the intervals (1,e) and (e,) onto . (It is strictly increasing on (1,e)
and strictly decreasing on (e,\(\displaystyle infty\)).) Using the continuity of f
implies that the original equation has nontrivial solutions: for every
1<x<e there is a unique y
(e<y<+\(\displaystyle infty\)) such that
f(x)=f(y), and conversely, for every
e<x<+\(\displaystyle infty\) there exists a unique y
(1<y<e) such that
f(x)=f(y), see Figure 4. For
x\(\displaystyle in\)(0,1] or
x=e, only y=x ensures
f(x)=f(y).
Summarizing, for x\(\displaystyle in\)(0,1] or x=e there exists a unique
y, while for x\(\displaystyle in\)(1,e) or x\(\displaystyle in\)(e,+) there are two
y's such that
x^{y}=y^{x}. This approach
easily yields the number of solutions, but not the solutions
themselves.
Figure 4: \(\displaystyle f(t)=t^\frac{1}{t}\)
2 The associative case
Our task now is to determine all positive real triples
(x,y,z) for which
(x^{y})^{z}=x^{(yz)}.
The left hand side is clearly equal to
x^{yz}. If x\(\displaystyle ne\)1, then we have
yz=y^{z}. But this is the very equation
we have encountered in the commutative case. If z=1, then every
positive y is a solution, otherwise
y=h(z).
Therefore we have all positive real triples for which powers are
associative (see Figure 5):
(1,y,z), where y,z>0,
(x,y,1), where x,y>0,
x\(\displaystyle ne\)1,
(x,h(z),z), where x,z>0,
x\(\displaystyle ne\)1,
z\(\displaystyle ne\)1.
Figure 5
2.1 Integer and rational solutions
Thanks to the investigation carried out in the commutative case,
rational solutions are already in our hands. Starting from the real
solutions, we arrive at the conclusion that positive rational powers
are associative if the triple (x,y,z) falls into
one of the categories below:
(1,y,z), with rational y,z>0,
(x,y,1), with rational x,y>0,
x\(\displaystyle ne\)1,
\(\displaystyle \left(x,\left(1+\frac{1}{n}\right)^n,\frac{n+1}{n}\right)\), with
rational 0<x1 and positive integral n,
\(\displaystyle \left(x,\left(1+\frac{1}{n}\right)^{n+1},\frac{n}{n+1}\right)\),
with rational 0<x1 and positive integral n.
The first two cases are trivial. The last two follow from the
formula (x,h(z),z), since in the
commutative case we have collected all rational numbers v for
which h(v) is also rational. (Write z instead of
v now.) There we have found that if
with integers p>q\(\displaystyle ge\)1, then h(v) is rational
if and only if p=q+1, which implies the third case
(using n instead of q, and allowing q=1
also). Finally, if \(\displaystyle v=\frac{p}{q}\) again, but this time with q>p1, then the proof
in the commutative case remains true by exchanging p and
q, and we get the only possibility q=p+1. This is
described in the fourth row. (The case p=q has already
been treated above, since here z\(\displaystyle ne\)1.)
As for the integer solutions, the third row above yields integer
values only if n=1, but the last row never does, so the
positive integer solutions to (2) are as follows:
(1,y,z), with integer y,z>0,
(x,y,1), with integer x,y>0,
x\(\displaystyle ne\)1,
(x,2,2), with integer x>1.
