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).
be positive real numbers satisfying![\[ xy + yz + zx = 3xyz. \]](http://latex.artofproblemsolving.com/9/2/0/9200874fd6bf9a32ff0c4e82382a450be68cc444.png)
Prove that![\[
\sqrt{\frac{x}{3y^2z^2 + xyz}} + \sqrt{\frac{y}{3x^2z^2 + xyz}} + \sqrt{\frac{z}{3x^2y^2 + xyz}} \le \frac{3}{2}.
\]](http://latex.artofproblemsolving.com/f/5/f/f5fb85d81e93935601ceb5f30c25aea7aa3f23b2.png)
be an acute triangle. Let 
be any point on the side 
. On line 
, choose any point 
. Let 
and 
be the tangents from to the circumcircle of 
, and let 
and 
be the tangents from to the circumcircle of 
. Find the locus of the intersection of 
and 
.
, 
is bisector of 
and 
. and are feet of perpendicular from 
to and 
,respectively.Prove that orthocenter of triangle 
is on 
.
memorized but I'm sure there are people out there with more!
indicates my favorites.
is defined over the positive integers as follows: 
, 
for 
prime, and for all relatively prime positive integers 
and 
, 
. If 
is the smallest positive integer such that 
, find the units digit of .
If convex quadrilateral satisfies 
, 
, 
, 
, and 
, what is the value of 
? Express your answer in simplest radical form.
and 
be the greatest odd divisor of . Let 
for even 
denote the product of every odd positive integer less than . If 
for positive integers 
and 
where 
is minimized, find the number of divisors of 
. There exists exactly one positive real number such that the graph of the equation 
consists of a line and a point not on the line. The distance from the point to the line can be expressed as 
, where and are positive integers and is not divisible by any square greater than . Find 
.
. If 
are positive real numbers such that 
, 
, and 
, find . Express your answer as a common fraction. Let 
be a convex pentagon satisfying 
, 
, 
. Let be the intersection of lines 
and . If 
has a perimeter of 
and an area of 
, find the area of .
Let 
be an acute scalene triangle with longest side . Let 
be the circumcenter of . Points and are chosen on such that 
and 
. If 
, 
, and 
, the area of the circumcircle of can be expressed as 
. Find .
such that 
is divisible by .
Alex has a 
by grid of squares. Let be the number of ways that Alex can fill out each square with one of the letters , 
, 
, or such that in every row and column, the number of 's and 's are the same, and the number of 's and 's are the same. (For example, a row with squares labeled 
or 
is valid, while a row with squares labeled 
or 
is not valid.) Find the remainder when is divided by 
.
by grid of squares along the gridlines into two or more pieces such that if three pieces meet at a point , then there are actually four pieces with a vertex at ? An example is shown below.
be real numbers satisfying the system of equations
What is the value of 
? In , the incircle is tangent to 
at , and 
is the reflection of across the midpoint of . Suppose that the inradii of 
and 
are and respectively, and the distance between their incenters is 
. What is the inradius of ? Express your answer as a common fraction. be a positive integer and let 
be the set of all -tuples of 's and 's. Two elements of are said to be neighboring if and only if they differ in only one coordinate. Bob colors the elements of red and blue such that each blue -tuple is neighboring to exactly two red -tuples and no two red -tuples neighbor each other. If 
, find the least possible value of .
be the remainder when![\[100x^{100} + 99x^{99} + 98x^{98} + \cdots + 3x^3 + 2x^2 + x\]](http://latex.artofproblemsolving.com/d/5/4/d5401c220a8682335a1306ae790a80947a38d3f6.png)
is divided by 
Solve for 
![\[
\frac{4000^2+4050^2}{(2000+2025)^2+(2000-2025)^2}
\]](http://latex.artofproblemsolving.com/b/9/9/b991bda876c059439578dd1eeb54b1f8010df755.png)
in the given equation gives 
Call this statement 
. This immediately gives 
is injective.
and 
gives 
Here is some constant. Also, comparing 
and 
and using injectivity, we have 
Here 
is another constant. Now this gives 
holds for all 
, which forces 
. So in fact 
Now 
gives 
Call this new statement 
.
gives 
Comparing 
and 
and mutilplying y 
gives 
Using 
in each of these terms, we have 
, and replacing 
, we get 
Now we use the statements 
to simplify that into 
Setting 
, there, we get 
, and since 
because of injectivity, we get 
, which in turn implies 
can be written as 
, which becomes Cauchy's equation after setting 
. Since the codomain of is bounded from below, we must have 
, and only 
fits.
It is clear that we can replace 
by 
and get:
Suppose that 
for some 
Then 
gives us 
contradiction. So 
But this means that
Combining the two boxed results gives us 
such that
for all positive real numbers and 
.
for all 
. Put 
thus 
for all 
. Observe that as 
we see 
has no upper bound. Thus, for any 
it is possible to pick with 
; hence 
is a subset of , consequently is surjective over 
. If 
then plugging 
and 
subsequently, we conclude 
or is injective. Thus, is a bijection on positive reals.
yielding 
for all 
. Swapping 
fixes the LHS, hence 
for some constant and all 
. Hence 
for all . Playing the Devil's trick again, we put 
in the original equation; so 
and swap 
; injectivity of then yields that 
is a constant function. Thus, 
and so 
for all 
. Now plug 
in 
to conclude that 
for all 
. Now we immediately get 
for all and so 
(putting 
); hence is additive too. Now is strictly increasing so if 
then 
and vice-versa. Thus, 
for all 
and is the identity function. It also works. 
to note that is injective and 
. Now take with 
gives 
. Therefore is bijective.
Replace 
by 
in the above relation and consider symmetric equation with 
swapped we deduce
Fix and vary to get using bijective that 
so 
, i.e. 
.
Replace by to get
Now use injectivity to get
. Taking a similar consideration as before with 
we see that 
. Therefore 
, so 
becomes Cauchy FE. Extend by 
for all 
and note that extended satisfies Cauchy on the reals, and 
for all 
, so is linear, from which we conclude that 
, as desired.
such thatfor all positive real numbers and .
is the only solution to the given equation. Note that it indeed works.
denote the given assertion.
Therefore, let 
. Then we have 
and some 
.
is injective . Now 
imply that . So 
and hence is bijective .
.
. Then, 
.
. (By induction on this, we get 
and observe that:
Now using 
and 
.
. So , iterating on both sides we get 
. is an involution. Now comparing and 
we see that
Plugging this back in we get 
and 
.
.
, we get unbounded values of , and for a fixed value of , we get that is surjective over 
. So, letting 
tend towards infinity gives surjectivity.
, 
gives 
, so is bijective.
. As varies for fixed , we get that the image of 
is 
. Thus, if 
, 
, so is increasing. As it is both increasing and bijective, our function must be continuous.
. This gives that 
, so it is fixed as varies. Note that we must have 
to make sure the argument is positive, and as approaches this lower bound, the LHS gets arbitrarily close to 
. The RHS cannot be less than , since it will otherwise be exceeded as approaches 
. Likewise, it can't be more than . So, we must have 
. As 
, LHS approaches , so we can get by similar logic that it must always be . Thus, 
.
, which gives 
. As , LHS approaches 
while RHS is always larger, but approaches 
by above. Using a similar argument, if 
, then we eventualy have RHS>LHS, and if 
, we have the opposite. Hence, 
. is both an involution and increasing. If it is nonconstant, though, we can find 
such that 
. Thus, is the only solution.
such that
for all positive real numbers and . is injective function.
the given assertion.
: 
so 
, so is injective.
. is injective we have that 
.
so 
.
so 
.
it's simple to show that 
so 
. we have that 
so 
.
, thus 
. and 
.
etc.
denote the given assertion, by comparing ![\[P\left(\frac{1}{x}, \frac{1}{f(1)}\right) \implies \frac{1+xf\left(\frac{1}{f(1)}\right)}{f\left(\frac{1}{f(1)}\right)}=f(f(1)+f(x))\]](http://latex.artofproblemsolving.com/8/1/1/811b4efc48217026e70946539a8d3ea55150d4b8.png)
and ![\[P\left(1,\frac{1}{f(x)}\right)\implies \frac{1+f\left(\frac{1}{f(x)}\right)}{f\left(\frac{1}{f(x)}\right)}=f(f(x)+f(1))\]](http://latex.artofproblemsolving.com/1/9/8/198edb0bf4d14a9b3080c10c46ecc4ef6a437e0a.png)
we will get 
where 
From here, it's also clear that is injective. Also, by comparing ![\[P(f(x),1) \implies \frac{f(x)+f(1)}{f(x)f(1)}=f\left(1+f\left(\frac{1}{f(x)}\right)\right)\]](http://latex.artofproblemsolving.com/b/c/6/bc6babd50b111a2e007f9b1a01d9e36524efd067.png)
and ![\[P(f(1),x) \implies \frac{f(1)+f(x)}{f(1)f(x)}=f\left(\frac{1}{x}+f\left(\frac{1}{f(1)}\right)\right)\]](http://latex.artofproblemsolving.com/e/7/d/e7dd89e6fa05585f0b6a8978c7df2c4c89855805.png)
we will get 
where 
Therefore, we have 
or equivalently, 
Since this holds for all positive real , it must be the case where 
which implies 
Now notice that ![\[P\left(\frac{1}{x}, \frac{1}{f(x)}\right) \implies f(2f(x))=2x\]](http://latex.artofproblemsolving.com/7/2/f/72f36b565fb2f8b0259131176877377cbd21bd04.png)
and so 
Then, ![\[P\left(\frac{1}{x}, \frac{1}{f(3x)}\right)\implies f(f(3x)+f(x))=4x=f(2f(2x)) \implies f(3x)+f(x)=2f(2x)\]](http://latex.artofproblemsolving.com/2/4/b/24b751b1a41faa66d24c744f27e7432a880a5a01.png)
and ![\[P\left(\frac{1}{2x}, \frac{1}{f(4x)}\right)\implies f(4f(x)+f(2x))=f(f(4x)+f(2x))=6x=f(2f(3x)) \implies 4f(x)+f(2x)=2f(3x)\]](http://latex.artofproblemsolving.com/f/0/6/f0653891e21e5989dcf4f515163269a8c0da583d.png)
give us ![\[4f(2x)-2f(x)=2f(3x)=4f(x)+f(2x) \implies f(2x)=2f(x) \ \ \forall x\in \mathbb{R}^{+}.\]](http://latex.artofproblemsolving.com/1/7/2/1724321d59be8c5e7e5cd64552f1d69de27b4675.png)
Hence, 
and 
Lastly, ![\[P\left(f\left(\frac{1}{x}\right), \frac{1}{y}\right)\implies f(x+y)=\frac{f\left(\frac{1}{x}\right)+f\left(\frac{1}{y}\right)}{f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)}=f(x)+f(y)\]](http://latex.artofproblemsolving.com/7/2/f/72fae9afa4e92a0b01b59a72782717e96881952c.png)
implies is additive over the positive reals, thus is linear and letting 
where 
are constants, we see that since 
, 
is the only solution. 
Now put instead of in equation. Let 
.
then 
implies that is injective.
Let 
be .
which holds for any 
which implies that is linear so 
.
which since are constant but 
are not implies that 
so 
so 
which clearly works.
, which also implies injective as a result
, we get 
for some constant over all values of 
, so 
implies 
, we have 
gives us 
, Since is surjective, varying the value of , we have that is a strictly increasing function
where 
Then, note 
contradiction. The same holds for when 
for all values of 
such thatfor all positive real numbers and . be the assertion of the given FE, which is 
Assume that , then 
and 
give or is injective.
and 
, we easily get 
Similarly, comparing 
and 
, we get 
for all . Let , we get 
, so 
and gives 
gives 
Plugging by 
into the above FE and changing the role of , by the injectivity, we get 
If 
, 
and the injectivity give 
, absurb. Thus 
, so 
gives 
immediately implies is additive, so after checking, we get for all is a solution.
. is bijective.
. For surjectivity, observe we can hit any value above 
, since we can make as small as we want we are done.
, we have 
as is a positive real. Then we have 
, so letting approach zero gives 
. Now substitute 
, so we have 
for 
. Now observe each pair 
results in exactly one value of , which is the same as the value of given by 
, but also this value is uniquely determined by so is constant, thus 
, but since the left hand side gets as small as we want we must have and 
.
, then we have 
, so 
giving 
. is the identity. Assume 
but 
. Now we have 
, giving 
. Clearly, cannot be , so 
for 
. Likewise, assume 
, which gives 
, so iff 
. This fact carries the rest of the solution, we can now proceed with a standard rational extension. over rationals. Let 
be the assertion 
. Now 
gives 
, then we can use the assertion 
, being 
to get 
. Now we can always induct, do 
to get 
and 
to get 
. Now to get all rationals we induct on the denominator, we can use 
to get all rationals with denominator , then use 
and 
to get all rationals with denominator 
, and so one and so forth we win.
for all reals. First take 
. Assume 
. Then there exists some rational 
with 
, contradiction. Symmetrical argument proves for . Now consider 
, this gives 
, giving 
by injectivity.
![\[\frac{1}{f(y)}+x=f(\frac{1}{y}+f(x))\]](http://latex.artofproblemsolving.com/d/6/1/d61b799f90102cf14a4dda741b04b6fd11dfb855.png)
Answer is which holds. Let be the assertion. is bijective., compare 
and 
to get contradiction. Fix and increase to see that takes all sufficiently large positive values. We can pick 
sufficiently small thus, is surjective..
in order to observe that![\[\frac{1}{f(\frac{1}{f(y)})}+x=f(f(x)+f(y))=\frac{1}{f(\frac{1}{f(x)})}+y\implies x-\frac{1}{f(\frac{1}{f(x)})}=y-\frac{1}{f(\frac{1}{f(y)})}\]](http://latex.artofproblemsolving.com/e/0/e/e0ec6817e4de007d6120947e82374c8453bdbf2a.png)
Since this implies 
is constant and it's smaller than any positive , it must be a nonpositive constant. Let 
or 
where 
. By symmetry and injectivity we have![\[f(\frac{1}{x}+\frac{1}{y+c})=f(\frac{1}{x}+f(\frac{1}{f(y)}))=\frac{1}{f(y)}+\frac{1}{f(x)}=f(\frac{1}{y}+f(\frac{1}{f(x)}))=f(\frac{1}{y}+\frac{1}{x+c})\]](http://latex.artofproblemsolving.com/0/2/2/022f9c23f19b6522c4a79de2defeed855d4c8212.png)
Thus, 
is constant which requires . So 
as we have claimed. is involution and 
. gives 
. We get 
and injectivity implies 
is independent of . Let 
. Since 
we observe 
. Thus, 
.![\[f(f(x))+f(\frac{1}{y})+d=f(f(x)+\frac{1}{y})=x+\frac{1}{f(y)}\]](http://latex.artofproblemsolving.com/6/6/2/662168d5b320e079781cb5a5f7d3c51f3fe62441.png)
and 
. Since 
, we must have 
which implies 
. Thus, is an involution and . is additive.
yields 
. is additive and takes values on positive reals, is Cauchy function hence which implies 
. So is the only solution as desired.
distributing the modulo operation. This simplifies into 0 + a mod b since a is always divisible by a except for when it is 0 so a diamond b is 
.
and 
in a diamond b to get
and 
in a heart b and distribute the modulo as well
Therefore, 
![\[R(0) = 100(0^{100}) + 99(0^{99}) + 98(0^{98}) + \cdots + 3(0^{3}) + 2(0^{2}) + 0\]](http://latex.artofproblemsolving.com/0/0/4/0044c0eabc8eb4ba3d6b430ab908b3e92ba242f1.png)
![\[a(0)^2 + b(0) + c = 0\]](http://latex.artofproblemsolving.com/1/2/4/124caf8e3542d5c7e7d4b13b8324ad59758e5b8a.png)
![\[c=0\]](http://latex.artofproblemsolving.com/3/e/8/3e87824860754458838fc46dd729d700589ccbdb.png)
Substitute 
![\[R(1) = 100(1^{100}) + 99(1^{99}) + 98(1^{98}) + \cdots + 3(1^{3}) + 2(1^{2}) + 1\]](http://latex.artofproblemsolving.com/9/a/1/9a1757b781ed32cb6679971426b2ab28ecade5c7.png)
![\[a+b+c = \frac{(100)(101)}{2}\]](http://latex.artofproblemsolving.com/0/6/d/06d75968cb318133118fac2eac69df3e0c956e93.png)
![\[\boxed{a+b = 5050}\]](http://latex.artofproblemsolving.com/8/7/7/877ee0cf9be48f8dcee17815c0bd35deb3d3edb5.png)
Substitute 
![\[R(-1) = 100(-1^{100}) + 99(-1^{99}) + 98(-1^{98}) + \cdots + 3(-1^{3}) + 2(-1^{2}) - 1\]](http://latex.artofproblemsolving.com/0/7/5/075d546e826059f361650e03e721c5c148f65839.png)
![\[a - b + c = 100-99+98 \cdots -3 + 2 - 1\]](http://latex.artofproblemsolving.com/c/c/a/cca420756d22a06d25afe74d37d55465258c4888.png)
![\[\boxed{a-b = 50}\]](http://latex.artofproblemsolving.com/b/2/9/b292e0f9e4c3f42e823db2d4365bb991f5d172db.png)
Hence, 
and 
so 
,
.
,
forall 
forall 
,
,
,
,
(from here, we easily get that: 
)
,
,
,
,
,
,
,
,
and let 
.
.![$P(x,\frac{1}{f(y)})\implies f(f(x)+f(y))=x+y\implies [a+b=c+d\iff f(a)+f(b)=f(c)+f(d)]\implies \exists g:\mathbb{R}\rightarrow \mathbb{R},g(a)=f(x+a)-f(x)$](http://latex.artofproblemsolving.com/7/5/8/7580abc760b6c61ee8fad23138f5e465924202db.png)
.
.
and so 
,
.
which clearly satisfied the condition.