AoPS Academy
, 
,
, 
, why 
or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.. It is usually regarded that 
, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.? The issue here is that 
isn't even rigorously defined in this expression. What exactly do we mean by ? Unless the example in question is put in context in a formal manner, then we say that is meaningless.? Suppose that 
. Then we would have 
, absurd. A more rigorous treatment of the idea is that 
does not exist in the first place, although you will see why in a calculus course. So the point is that is undefined.
? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that is an indeterminate form.
, 
or , obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](http://latex.artofproblemsolving.com/2/6/0/260f97a1cf1ea8722ff97a29b83bac7bfe83e577.png)
etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function 
given by the mapping 
is undefined for 
. On the other hand, 
is undefined because dividing by is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context. is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](http://latex.artofproblemsolving.com/9/9/3/993c78b62ec2c84b5d77daa8e9a8cd6f70fcf904.png)
So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let 
for each 
which means that via this order topology each subset has an infimum and supremum and 
is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In it is perfectly OK to say that,![\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}](http://latex.artofproblemsolving.com/6/d/8/6d8fc1eca5c791d430a7168cd3b37a02104fa318.png)
So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,![\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]](http://latex.artofproblemsolving.com/a/0/b/a0b67076efcc014c78b2023be1151c84b57c1503.png)
So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define 
as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](http://latex.artofproblemsolving.com/8/4/e/84ed3d6ab237df970333b87beaef5311caa46185.png)
which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, 
means complex infinity, since we are in the complex plane now. Here's the catch: division by is allowed here! In fact, we have![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and are left undefined. We also have
Furthermore, we actually have some nice properties with multiplication that we didn't have before. In 
it holds that![\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]](http://latex.artofproblemsolving.com/8/e/0/8e035b4841d20c5ad671c46355cc667ca01e533b.png)
but 
and 
are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with , by defining them as![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by also being allowed with the same indeterminate forms (in addition to some other ones).
Prove that
Proposed by Phan Ngoc Chau
be a triangle, and let 
be a point on side 
. A line through intersects side 
at 
and ray 
at 
. The circumcircle of triangle 
intersects the circumcircle 
of triangle again at point 
distinct from point 
. The lines 
and 
intersect again at 
and 
respectively.
grid has had the central unit square removed. For which positive integers 
is it possible to cut the remaining area into 
and 
rectangles?
positive integers have no single-digit prime factors?
(meaning 
takes pairs of positive real numbers to positive real numbers) such that for any real numbers 
,
holds; and
then 
.
(meaning takes pairs of positive real numbers to positive real numbers) such that for any real numbers ,
holds; and then .
and 
. It's easy to see that these work; we now show that these are the only ones.
denote the given assertion. We first show that is "injective." Suppose that 
for some 
; then by 
and 
, we have
which implies that 
.
and the injectivity we just proved, we have
, we see that 
and in particular 
Now 
gives 
, we have 
Thus, the function 
is multiplicative; making the transformation 
produces an additive 
. But we know that 
for all 
. Choosing ![$a\in [1,e]$](http://latex.artofproblemsolving.com/a/4/4/a444827f623722efa689aa2e89dceafb0782f16f.png)
and thus ![$\log(a)\in [0,1]$](http://latex.artofproblemsolving.com/1/f/d/1fd2452befec220985f36ce322cd6cec60f19b31.png)
yields 
so 
is bounded below on a nontrivial interval. Since is also additive, we conclude that 
for some 
. Thus 
and 
Substituting this back into gives 
so or .
and normal division 
. They clearly works. Let denote the first given condition.
gives 
thus replacing 
gives 
. Therefore 
.
gives 
. Thus 
gives 
. In particular, this implies 
and 
(by pluggin 
and 
respectively).
. Thus 
is multiplicative. Moreover 
(by 
) thus is an involution. Moreover, the second condition gives 
for 
. It turns out that these are sufficient conditions to determine 
.
. Therefore 
and 
for all 
. If is non-linear, then it's a well known theorem that the graph of any nonlinear Cauchy function is dense in a plane. However, it's clear that is bounded below in ![$[2019,10^{10}]$](http://latex.artofproblemsolving.com/d/a/9/da96aef6e76e046e1b578efb2a16023272e8cdba.png)
thus it's contradiction.
for some constant 
which implies 
. From 
, we get 
. This gives the two solutions described at the beginning.
be the assertion 
. Note that, if for a fixed 
, 
, then comparing and 
we get 
. Thus, for any fixed 
, 
is injective. Next, implies that, 
, and thus, 
, using the injectivity. Now, 
implies 
. As done above, set 
. This expression yields . yields 
, that is, 
. In particular, inputting 
here, and using 
, we get that 
. In particular, it suffices to determine 
to recover the operation 
. Now, using this repeatedly, 
, which is equal to 
, that is, 
for every 
. Note, plugging here, we get 
. Let 
, this yields 
. Set 
, this yields 
. Finally, let 
, this brings us 
. (I believe using second condition), we get 
for some 
constant, which yields, 
, and thus, 
for some constant. Plugging this for 
, we get 
, yielding 
or 
, as claimed.
and 
. It is easy to check that these work. We now show these are the only solutions.
, then 
.
whence . 
, then .
whence and .
and 
.
, so by Claim 1, , as desired. For the second part, check that 
.
.
as requested.
, so that 
. Our given functional equation rewrites to 
However, by Claim 1, is injective, and by Claim 4, is an involution, so plugging in 
as gives 
. Hence, 
satisfies 
. The second condition implies that 
for all 
, so is bounded, and thus for some 
., but substituting yields 
. Hence, 
or 
, as desired. 
be denoted by 
.
takes all positive real values as we vary 
keeping 
fixed. 
, so 
is constant for fixed .
over the positive reals s.t. 
implies that is surjective, and gives 
.
.
gives 
. As is surjective, we have that is multiplicative.
over the real numbers by 
. multiplicative gives 
additive.
for all 
. So is bounded below in 
, so 
for some real constant . This gives 
. As 
, 
or 
, so 
or 
. It's easy to see that both of them work, and we're done.
to 
? for some power 
, i.e. can this be stated as well known? Or would one need to reduce it to the lemma used in the official solution (pg 2) in order to get points?
to ? for some power , i.e. can this be stated as well known?
was a solution way before I found that 
was a solution...
was a solution way before I found that was a solution...
and has already been found to be a multiplicative involution on 
with 
and satisfying another condition which becomes
if 
and 
if 
.
or 
(the identity function or the reciprocal function).
denote the fixed points of . Consider the function 
defined by 
.
.
and is the identity.
such that 
. We now use the other condition to show that this implies is the reciprocal function.
be arbitrary. WLOG, we can take 
, 
and 
(replacing and 
by their reciprocals and powers if necessary).
. Hence 
for all 
and is the reciprocal function.
denote the subset on which is the reciprocal. Then
(This is actually a decomposition of abelian groups where the factors are the image and kernel of which is a homomorphism.) without the inequality condition it would be equivalent to all ways of finding two multiplicative subsets whose intersection is 
and together generate under multiplication.
and 
. It is easy to check they work. We now show these are the only solutions.. Define 
be defined by 
(this is a definition that will help us later on).
.![\[(1\diamondsuit 1) \cdot (1\diamondsuit 1) = 1\diamondsuit (1\diamondsuit (1\diamondsuit 1)) = 1\diamondsuit ((1\diamondsuit 1)\cdot 1) = 1\diamondsuit (1\diamondsuit 1) = (1\diamondsuit 1) \cdot 1 = 1\diamondsuit 1.\]](http://latex.artofproblemsolving.com/e/b/9/eb914eaccd361cf941e38eaa40498ef1c901be86.png)
As 
, this implies the desired. 
,![\[f(f(x))=1\diamondsuit (1\diamondsuit x) = (1\diamondsuit 1)\cdot x = 1\cdot x = x.\]](http://latex.artofproblemsolving.com/7/6/b/76b6cde549273ae9daba965eeafa438bf9bc8328.png)
Plug in 
to get![\[a\diamondsuit x = a\diamondsuit (1 \diamondsuit (1\diamondsuit x)) = a\cdot (1\diamondsuit x) = af(x).\]](http://latex.artofproblemsolving.com/7/e/d/7ed0cd7f333eefc457921bfe7d0c6427aced95be.png)
This prompts us to rewrite the original given information as![\[af(bf(c)) = af(b) \cdot c,\]](http://latex.artofproblemsolving.com/9/1/9/91988f7d0c8153106b4fd1e02916604e20982698.png)
or![\[g(bf(c))=cf(b).\]](http://latex.artofproblemsolving.com/e/3/f/e3fffa0d92ea363f87f32004b8d59bf333537956.png)
We additionally have for all 
. Note that this implies is bijective, so the original statement is equivalent to![\[f(bc) = f(b)f(c)\]](http://latex.artofproblemsolving.com/b/7/c/b7c2cc4d34724bfb97435fcb87e972cfcb8e5c90.png)
for all 
. Hence, 
satisfies the Cauchy equation. Note that by the second given piece of information, we have 
for all . In particular, this means that has the lower bound of 
on the interval ![$[\log 1,\log 2]$](http://latex.artofproblemsolving.com/2/9/b/29b12cbc56968ad0bef6f29e205f10597cd735c8.png)
. Hence, is linear and 
is an exponential function 
for some real . By the equation , we have 
, so 
. Hence, by 
, we get the claimed solutions of 
.
, so that our conditions become ![\[ f(a,f(b,c))=cf(a,b) \qquad\text{ and } \qquad f(a,a)\geq 1, \text{ when } a\geq 1\]](http://latex.artofproblemsolving.com/f/d/e/fdea209d9a3d5323ac26ff5c701ad788b70c9380.png)
and 
, which clearly work. The main idea is to show 
is a multiplicative, determining completely. As usual, let denote the assertion 
. is 'injective', that is, 
and 
both imply 
.
and 
. For the latter, observe that 
by this `injectivity'. Then comparing 
and 
gives our other form of `injectivity'.
implies ![\[ f(a,f(1,c))=ac\]](http://latex.artofproblemsolving.com/b/3/a/b3a007ab07d0c3bd7f53f5ab1ca8d8d624ca5633.png)
Then by putting 
in the above, we obtain 
. However, 
from which injectivity gives 
. Now since 
, 
. Hence letting in this gives 
. Finally, 
gives
as required. is multiplicative
gives ![\[ f(1,f(b,c)) = cf(1,b) = f(c,b) \]](http://latex.artofproblemsolving.com/f/a/7/fa76a0a2c487a10842c1f2c916f9a7182b1ca130.png)
by the previous claim. In the above, put 
so that ![\[f\left(1,f\left(\frac{1}{f(1,c)},c\right)\right)=f(c,\frac{1}{f(1,c)})=1=f(1,1)\]](http://latex.artofproblemsolving.com/e/8/5/e85223a89194a008887b7e34e927b1eef46cba69.png)
giving 
by injectivity. Recall that - setting gives, after some computation, 
. Finally, 
hence is multiplicative as claimed.
, so that 
. The second condition gives 
for all 
. Then for any 
, 
so 
. Since any non-linear additive function must be dense in the plane, it follows that is linear, from which we deduce 
for some constant .
, we find that 
, so 
or 
which correspond to the claimed solutions.
with 
and let denote the first assertion.
For fixed 
the function 
is injective.
Suppose 
Then, 
and 
give 
Since 
we must have 
as needed. 
such that 
for all 
for all From 
we have 
so we are done by injectivity. 
for all 
yields 
On the other hand, 
gives 
Therefore, 
for all This implies that 
as desired. yields 
and 
yields 
Now we have two cases:
There exists 
such that 
Furthermore, we have 
Therefore, we have 
Let be the integer such that 
Since is multiplicative, we have 
and 
Thus, 
Hence, 
for all 
in this case. 
If 
then 
is multiplicative, the condition implies that is strictly increasing. Therefore, since is an involution, it must be the identity. 
and 
which can be easily checked to work.
. The condition is![\[ P(a,b,c):\quad f(a,f(b,c))=f(a,b)\cdot c. \]](http://latex.artofproblemsolving.com/4/4/7/44704040feb2abb00ee8d620510bd26e98e45018.png)
The key is to analyze 
.
.. We have
Since is positive, this implies .
is an involution, and hence a bijection. implies 
, i.e. 
. Therefore, is an involution, which means it is also a bijection.
is multiplicative.
, we have 
, so 
. Since is bijective, this implies for all . Note that 
where we used 
in the middle. So now, implies 
, i.e. 
.
. Since is multiplicative, is additive. We want to involve the second condition, and bring in 
somehow. We have![\[ P(a,1,g(a)):\quad f(a,g(g(a))=ag(a) \implies f(a,a) = ag(a) \implies g(a)=\tfrac{1}{a}f(a,a) \ge \tfrac{1}{a} \]](http://latex.artofproblemsolving.com/a/f/2/af28348da9d157c2920bf2a5bad3461a4eb2475f.png)
for . Therefore, 
. In particular, is bounded below on some nontrivial interval. Combined with being additive, we have 
for some constant 
. Now, 
, so . From , we have 
, so now 
, so 
, so 
for some constant 
. Plugging this into gives 
. Therefore, or . Both are easy to confirm that they work.
denote the first condition. The answer is multiplication and division, which clearly work. gives 
. Left-multiplying by and applying the given condition demonstrates 
.
gives
but since 
we actually have 
, i.e. the function 
is multiplicative. Also note tells us is involutive.
transformation, so 
is Cauchy. We show 
, hence is linear. Indeed, it's equivalent to 
, which follows from
Hence 
for some constant 
and 
. But since is involutive we have 
. now gives the desired conclusion.
denote the given assertion. is "injective" (if 
, then ).
.
.
, .
, so 
.
.
.
.
.
is multiplicative.
so is additive. is bounded over the interval 
.
, then ![\[f(a)=\frac{1}{a}\cdot (a\diamondsuit a)\ge \frac{1}{a}\]](http://latex.artofproblemsolving.com/6/8/d/68dd3e17486d53745d79dcd22991ab827149bcaa.png)
, so 
. We have ![\[g(x)=\ln(f(a))\ge \ln(\frac{1}{a})>-1,\]](http://latex.artofproblemsolving.com/d/9/1/d91072b3ebcd5d00549f9470303985427299111e.png)
which proves our claim., so .
.
.
, we have 
., then we have 
., then we have 
.
and 
which clearly work. Now we show that these are the only functions.. Suppose that 
, so by comparing 
we have
thus . Thus 
is "injective". we have
by "injectivity". Next, gives
in particular giving 
, so from 
we obtain
Then gives
Letting 
, this means that 
. Further is bijective and thus surjective as 
, so letting 
we get 
for all 
. Now let , so is 
and additive. Further, for ![$a \in [1,e]$](http://latex.artofproblemsolving.com/8/0/2/802ff4f60f0719598875c1f4ba69408780595968.png)
we have
so for ![$x \in [0,1]$](http://latex.artofproblemsolving.com/d/3/4/d3423a3200d0ad7660fc4265d59d5598a53d551d.png)
, is bounded below, thus by Cauchy we have for . Substituting, this yields , so
To finish, yields
hence 
or 
. 
and 
., the function 
is injective.
such that 
. Then,
Define 
.
. The function also satisfies 
, and 
.
and
Setting gives us 
. We can then characterize the operation as
Finally,
Taking of both sides yields 
, as desired. 
, our findings translate to the following
for .
, notice that 
as well. Given that 
is linear and bounded below at for , it is well known that 
for some constant , and thus 
. Of course, the only that satisfy condition 2 is 
, so 
or 
.
and 
, which gives us the final solution of and 
, both of which can be checked to work.
with , because when I was solving the problem it was too hard to write a symbol. is right injective, or that 
means that 
. Assume this is not the case, and such 
exist. Then 
, which is impossible.
, so . 
, so 
. Thus 
is fixed as varies, so 
and 
.
. Rewriting the given equation in terms of eventually gives 
for any . In addition, , so and . Thus we can rewrite the given as . 
for 
, so 
. Since 
, 
or 
, and either equals for all 
or for all , as desired.
denote the given assertion, and because i'm lazy replace 
with .
.
.
; defining 
means that this can be rewritten as .
. Using this equation, rewrite the given condition as ![\[ag(bg(c)) = ag(b) \cdot c \implies g(bg(c)) = cg(b).\]](http://latex.artofproblemsolving.com/2/f/4/2f40a3153c348cfffb54352a2240c5903ccb1cd2.png)
Since , is bijective, so therefore we can rewrite this equation as 
. Letting 
tells us that satisfies Cauchy. The second piece of information tells us that in fact, 
for 
is bounded below by 
over ![$[\log 1,\log e] = [0, 1] \implies h$](http://latex.artofproblemsolving.com/0/4/8/048566e9af3c9ae6bca4113f6301a2fb03d5ee14.png)
linear 
for some . However, note that ![\[g(g(x)) = x \implies k = \pm 1 \implies f(a,b) = ab \,\, \text{or}\,\, a/b,\]](http://latex.artofproblemsolving.com/b/3/0/b30f773b63415de96a13109d965a435211deb9ff.png)
which both work.
and 
are both satisfactory definitions for . We will now show that they are the only ones. Denote the first assertion by 
.
. 
gives 
. Hence, is surjective. Suppose that 
. Then,![\[(1 \, \diamondsuit \, 1) \cdot a = 1 \, \diamondsuit \, (1 \, \diamondsuit \, a) = 1 \, \diamondsuit \, (1 \, \diamondsuit \, b) = (1 \, \diamondsuit \, 1) \cdot b,\]](http://latex.artofproblemsolving.com/d/c/2/dc20818d2c0416adcfbbef8e60274002e1fe2ffb.png)
so 
. is injective.
with as many s as we like, progressively gaining a foothold on the structure of . Let and be arbitrary positive real numbers.
.
.
.
.
.
, alternatively 
. We define the helper function 
for all 
. Then, 
, i.e. is a Cauchy function. The second given assertion is equivalent to 
for all 
, and this statement translates to 
for all 
. Hence, is not dense in the plane; it must be linear. Checking all such verifies that the only possibilities are 
and 
. We obtain our two solutions from there.
and and varying in the first equation. Furthermore, if we fix the left operand, every value of the right operand gives a distinct value for the expression: this can be also be seen by varying in the first equation, since 
must be distinct for each distinct .
is a bijection for any positive constant 
. and and varying , we see similarly that 
is a bijection for any positive constant . in the first equation, it follows that for any fixed constant , the function is also injective., it follows that is bijective.
gives us that 
, and plugging in 
gives us that 
. Hence, the function is linear for any positive constant . and letting 
, we can manipulate the given equation:
Plugging in gives , and plugging 
gives![\[ f(bc) = f(b)f(c). \]](http://latex.artofproblemsolving.com/a/4/a/a4a298e052f168302d7fadbd71804698c19086e9.png)
Letting 
, we have 
. The second condition in the problem bounds , hence 
, so 
for some . (if that wasn't a fakesolve my mind is blown....)
is 
for some constant . Plugging back into the first equation forces 
.
denote the given assertion and let 
. Consider the following claim: for all 
.
gives 
and 
gives 
, thus 
for all 
. 
for all 
.
, as needed. 
. Our given functional equation rewrites to 
. Taking gives 
, thus is an involution. Therefore is injective, so plugging 
as gives 
, so is multiplicative. Let 
, then 
and second condition becomes 
. Thus is bounded below on interval thus 
. Since implies 
or 
. Thus or 
and hence 
or 
. Both answers satisfy the condition, so we're done. 
so that 
We claim the answer is or 
, then we must have . such that . Then, the equation becomes 
so we must have .
, we have 
which means that 
here, we have 
which means that 
is injective for all . Note that 
Since is uniquely determined by 
, this shows the claim.
is also surjective. Consider the original equation 
As we fix and vary , the LHS can take on any positive real. Thus, 
is bijective for a fixed .
as well as bijectivity, for any , the corresponding second input that results in a value of 1 is always 
In particular, this is constant for all . Thus, is constant, which means that 
for some function . so that 
and 
for 
. Plugging we see that is an involution. Thus, 
gives 
The solution to this over positive reals is 
(to do this take ln and substitute 
, the constraint 
gives the necessary bound for Cauchy to work).
, we see that only work, done.
or 
. Clearly these satisfy the conditions.![\[
(1\,\diamondsuit\,1)(1\,\diamondsuit\,1)=1\,\diamondsuit\,(1\,\diamondsuit\,(1\,\diamondsuit\,1))=1\,\diamondsuit\,(1\,\diamondsuit\,1)=1\,\diamondsuit\,1
\]](http://latex.artofproblemsolving.com/2/e/9/2e9370a27c1f7414e9ba215411d66240f07bb822.png)
so 
.
then .![\[
(1\,\diamondsuit\,a)x=1\,\diamondsuit\,(a\,\diamondsuit\,x)=1\,\diamondsuit\,(a\,\diamondsuit\,y)=(1\,\diamondsuit\,a)y
\]](http://latex.artofproblemsolving.com/2/1/f/21fa224ab1f56abebd1b9f6763b402b9387567af.png)
so . 
for all .![\[
1\,\diamondsuit\,(a\,\diamondsuit\,1)=1\,\diamondsuit\,a
\]](http://latex.artofproblemsolving.com/b/6/d/b6d4d65a65944a700c66158dc48ec0aafac30b31.png)
so by Claim 1. , there exists a unique 
such that 
. Moreover, 
.![\[
a\,\diamondsuit\,\left(1\,\diamondsuit\,\frac{1}{a}\right)=(a\,\diamondsuit\,1)\cdot\frac{1}{a}=1
\]](http://latex.artofproblemsolving.com/e/b/9/eb94398174e271d462ea9820cac426f957db8f6d.png)
by Claim 2. is unique by Claim 1. 
. Then
for all 
. We also have![\[
f(f(x))=1\,\diamondsuit\,(1\,\diamondsuit\,x)=(1\,\diamondsuit\,1)\cdot x=x.
\]](http://latex.artofproblemsolving.com/5/d/d/5dd01ed9be67a359641d69c547608a93a393a6c3.png)
Let 
, which is defined because is always positive. Then for all 
. Note that![\[
a\,\diamondsuit\,\left(a\,\diamondsuit\,\frac{1}{a\,\diamondsuit\,a}\right)=(a\,\diamondsuit\,a)\cdot\frac{1}{a\,\diamondsuit\,a}=1
\]](http://latex.artofproblemsolving.com/f/8/c/f8c9026cad41d18e8ca6bd74469ca361d4e61854.png)
so 
by Claim 3. Then![\[
\frac{1\,\diamondsuit\,a}{a\,\diamondsuit\,a}=1\,\diamondsuit\,\left(a\,\diamondsuit\,\frac{1}{a\,\diamondsuit\,a}\right)=1\,\diamondsuit\,\left(1\,\diamondsuit\,\frac{1}{a}\right)=\frac{1}{a}
\]](http://latex.artofproblemsolving.com/4/8/f/48f96bfd61c0bf298951b40d7d75384921f62d20.png)
so 
for . Thus we have![\[
g(x)=\log f(e^x)\geq\log\frac{1}{e^x}=-x
\]](http://latex.artofproblemsolving.com/0/f/2/0f20ba56002120db28429a5f2cca7c01675a9dba.png)
for all . It follows that is bounded in the interval ![$[0,1434]$](http://latex.artofproblemsolving.com/6/6/b/66bb5a2c7b899d2a53d1ac710eeb5d6d05e92ec3.png)
so for some 
. Thus . Since , we must have so 
or 
. If then![\[
a\,\diamondsuit\,b=1\,\diamondsuit\,(a\,\diamondsuit\,b)=(1\,\diamondsuit\,a)\cdot b=a\cdot b.
\]](http://latex.artofproblemsolving.com/4/b/d/4bdffc59a1572245f7420512a00e1c2380b492ef.png)
If then![\[
a\,\diamondsuit\,b=\frac{1}{1\,\diamondsuit\,(a\,\diamondsuit\,b)}=\frac{1}{(1\,\diamondsuit\,a)\cdot b}=\frac{a}{b}.
\]](http://latex.artofproblemsolving.com/8/d/5/8d54d6af6e8c2413e1e822948b09f2dacf7fbed7.png)
We are done. 
for all 
and 
for all . It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Let be the assertion that for positive real numbers and . We start off with some minor observations. must satisfy 
.
such that 
for some . Then, comparing 
and 
we have,![\[(a\diamondsuit a)m = a\diamondsuit (a \diamondsuit m) = a \diamondsuit (a \diamondsuit n) = (a \diamondsuit a)n\]](http://latex.artofproblemsolving.com/1/e/e/1ee08ce6acb3d52cd513e2f68e43ca0cffc2a25b.png)
which is a clear contradiction. Thus, there exist no such 
and the binary operation is 'injective in the second input'. Now note that from 
we have![\[1\diamondsuit (a \diamondsuit 1) = 1\diamondsuit a\]](http://latex.artofproblemsolving.com/6/d/f/6dff575cbd8958c0c1bb8e8699ea0e2accef8f3c.png)
which due to our above observation implies 
as desired.
such that 
for all . Now we note some properties of this function. and , we have 
. we have![\[f(f(a))=1\diamondsuit f(a) = 1\diamondsuit(1 \diamondsuit a)= (1 \diamondsuit 1)a = a\]](http://latex.artofproblemsolving.com/8/0/f/80faf91a9a48fa78fb868616daf6c56af960c28e.png)
which implies that is an involution. Now, 
yeilds,
as claimed. since 
. We further note that gives
for all pairs of positive reals and . Thus, is multiplicative.
. Define the function 
such that 
. Note that,
Hence, is Cauchy-additive. Further note that for all 
,![\[f(a) = \frac{1}{f\left(\frac{1}{a}\right)} \le \frac{1}{a}\]](http://latex.artofproblemsolving.com/3/5/3/35352c2cf83b71af5aac237f44c64e3ad632b53e.png)
as a result of the second condition. But then for all and 
,
Thus, the function is bounded above in the range 
. Since we previously noted that it is also additive, we conclude that is linear over the positive reals. Thus, we let 
for constants and for all 
. Plugging this form back into the relation,
which implies that 
. Further, since is an involution,![\[a=f(f(a))=a^{k^{2}}\]](http://latex.artofproblemsolving.com/6/7/5/6755bbeacea72191a994638bad162619d8b8e7bb.png)
Thus, 
. Thus, 
for all or 
for all 
. Since is multiplicative it follows from![\[f(a)f\left(\frac{1}{a}\right)=f(1)=1\]](http://latex.artofproblemsolving.com/2/9/5/295a6c00cab12db9dc1461962d5d8f31172589d3.png)
that for all or 
for all as well, which finishes the proof.
with 
so that LaTeX is easier to type. The answers are 
or 
, which both clearly work.
.
. Then 
yields 
, while 
yields 
, so is forced. 
for all positive real numbers . 
yields , so is bijeccctive. 
yields 
, which we call . Furthermore, 
yields 
, so 
by injectivity. is multiplicative.
yields 
. Combining this with gives ![\[f(ac) = f(a*f(c)) = f(a)*f(c)\]](http://latex.artofproblemsolving.com/d/8/1/d81a109773cff342bfecde06a4b8e7deaa2b66f7.png)
as needed. in yields 
.
.
. So in fact, ![\[f(c)*a = f(f(f(c)*a)) = f(ac) = f(a*f(c))\]](http://latex.artofproblemsolving.com/3/b/d/3bd0d1262945c31164e4a7b8380394c6ba13bc98.png)
which finishes as is surjective. 
. The second given condition then implies that 
when 
. Now let 
for real numbers 
, so that the multiplicative condition reads 
. For all 
, the second condition reads 
, so is bounded on a nontrivial interval and it follows that 
for some constant . satisfies , so 
. Using this relation with then yields 
if and 
if 
, done!
(meaning takes pairs of positive real numbers to positive real numbers) such that for any real numbers ,
holds; and then . be the assertion 
.
.
so 
.
be defined by 
. Then 
. Now we look at the second condition of the problem statement: if then 
, which means that is bounded below on a nontrivial interval. Therefore must be of the form 
for some 
, so is of the form for some 
.
we must have 
, leading to the solutions 
and 
. Then either 
or 
, which we can see both satisfy the problem statement.
OR 
Both obviously work.
where is an involution and is multiplicative.
is injective: if 
take on both sides to get 
thus implying 
so 
is an involution.
or 
so 
or is an involution.
so 
or is cauchy
translates to 
for 
which translates to 
for is bounded over some nontrivial interval(e.g. 
for all 
). Thus, 
It is easy to see 
as is an involution.
implies 
.
multiplicative if there exists a number 
.
for all a.
gives the desired result.
for all 
for some 
gives 
.
gives 
.
.
gives the desired result for any 
for any 
and Lemma 2.
for any 
gives 
.
,
.
.
gives 
.
.
.
since 
,
.
and use Lemma 1. This means that if there exists some divisive number 
.
to get 
.
.
.
is divisive and 
.
.
be an arbitrary number. Consider 
,
,
and note that both 
and 
are divisive to get that 
.
gives 
.
.
,
.
and 
are divisive. Take the one that is greater than or equal to 
.
for any 
,
,
,
is commutative.
Thus, 
Hence, 
Recall, that 
,
Thus, this implies 
such that, 
Further for some 
,
Therefore, 
.
Now Pick some 
such that 
and take 
,
.
Since, 
Since, 
.
Contradicting, 
,
Therefore, 
,
Hence, 
And let 
then 
and 
gives 
So the operation is injective. Note that 
from 
Now 
So 
then 
is 
It is not hard to get that 
Checking yields 
So 
both work.
and 
only, which clearly work. Now, we show that they are the only solutions. Let 
for a fixed 
are bijective, since surjectivity follows from fixing 
,
i.e., 
,
,
.
for all 
,
,
such that 
.
so by injectivity 
.
,
.
so since 
and 
,
Substituting this back into the first condition yields that 
so 
Replacing 
yields 
as 
,
Now, note that by the second condition, we have that for all 
Now, let 
,
.
,
so 
is well defined. Then, we have that for all 
,
so 
is Cauchy. Furthermore, since 
for all 
This, ![$[0, 1]$](http://latex.artofproblemsolving.com/a/b/1/ab178d831a786b92cb4c9ddc2d33578223036f98.png)
,
for some constant 
,
for all 
,
for some 
,
,
.
,
for all 
,
for all 
and 
are the only solutions, as claimed. 
