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 an acute triangle with circumcenter 
and orthocenter 
, satisfying 
. The tangent line at 
to the circumcicle of intersects 
in 
. Let 
be the midpoint of 
. Prove that 
.
and 
be positive integers greater than . In each unit square of an 
grid lies a coin with its tail side up. A move consists of the following steps.
square in the grid;
for which it is possible that every coin shows head-side up after a finite number of moves.
feet and area ft^2. What is the arithmetic mean of all possible values of ?[/quote]
, in square units, is![\[ \textbf{(A)}\ 24 \qquad
\textbf{(B)}\ 30 \qquad
\textbf{(C)}\ 46 \qquad
\textbf{(D)}\ 66 \qquad
\textbf{(E)}\ 74
\]](http://latex.artofproblemsolving.com/8/2/f/82ffc15072ca99f6cd1abc312781eb4827817356.png)
is rotated 
degrees about one side. The space through which the hexagon travels during the rotation forms a solid. Find the volume of this solid. is rotated degrees about one side. The space through which the octagon travels through during the rotation forms a solid. Find the volume of this solid.
has parallel bases 
and 
, 
is the midpoint of side 
and 
. If 
cm and 
cm, what is 
? Express your answer in simplest radical form. be an cyclic quadrilateral with 
,
,
and 
Find:
we see that all prime factors of 
is 
. It is well known that there are infinitely many primes 
such that there exist satisfying 
if 
is non-constant. For any such , we can assume 
but it follows that 
. So for all such one has that 
where 
is the constant coefficient. Then 
and we can just consider 
which satisfies the same condition as . From this we get that 
for some integer 
and and it follows that 
or 
from checking 
and that it sufficies to have 
by checking 
.
. Now suppose the polynomial is non-constant. Now we prove a lemma:, we have 
.. The base couple cases are trivial by exhausting the few possibilities. Now it suffices to check all 
. Now take some 
. Then take 
where 
. After that it is clear that 
so we are done.
. Dividing through by we receive a new function satisfying the same requirements as , implying that our only solutions are of the form 
for some non-negative 
. 
for 
. To see that these work, note that 
for all positive integer 
.
is constant it must be 
, so assume otherwise. I claim only prime factors of may divide . Note that 
for integer polynomials, so let 
be equivalent to modulo some prime such that 
and thus 
. But 
so it is not divisible by , so is not divisible by any prime that does not divide and the claim is proven.
for some 
, but as is nonconstant, we have 
for infinitely many primes . As obviously divides all the nonconstant terms of 
, we then know that the constant term of is divisible by infinitely many primes, and thus 's constant term is zero. But then we can factor an out of and our arguments still hold, so by induction we get that is and we are done.
.This means that 
.Then let 
be the set of prime divisors of .If 
then 
.Write 
.Using similar arguments , we will get that 
.Thus , we can reduce the degree and ultimately get 
Since 
, we have that 
.Also 
where 
and 
was constant then it would be or by plugging 
. So now assume that its not constant, hence by Schur we have that we can set a larger enough prime such that for some we have 
and now we can set that 
and 
to get 
which would mean that 
hence 
but aince we said we can let as larger as we want we have that 
hence 
which means that the replace 
makes sense, now from here its very easy that we can repeat the same process until we get 
in which here 
has to be or hence we are done 
**
(where 
). Obviously they work. Now we will prove the converse. Assume is nonconstant (otherwise only solution is 
obviously). Define ![\[f(X):=a_dX^d + \dots + a_0\]](http://latex.artofproblemsolving.com/7/1/2/712194cdef863cd406820890051ed0f651cebb11.png)
Now see that by Schur's Theorem there exists infinitely many primes such that for some . Consider the minimal such .
or else 
contradicting minimality. But also 
, and hence and we get ![\[p \mid f(p) \iff p \mid f(0) \text{ for infinitely many primes $p$ } \iff f(0)=0 \iff a_0=0\]](http://latex.artofproblemsolving.com/a/4/e/a4ec9d539738837a41659c6987984ca928db153f.png)
and hence now we can define ![\[g(X):=\frac{f(X)}{X}\]](http://latex.artofproblemsolving.com/6/a/0/6a0cb7dc7e07ce290a88e51b4a7882a38d37d0c6.png)
and repeat the process untill we get 
.
and 
, we get the above solutions, as desired.
, determine all polynomials with integer coefficients such that divides 
for every positive integer .
be a prime factor of , with 
fixed 
(
is the maximum that satisfies)
for 
. are a subset of the prime factors of (without multiplicity).
. For 
then all prime factors of are 
, and for each we have 
, so 
as desired. is a positive or negative power of with nonnegative exponent 
for all . Thus there exist some 
for which 
for infinitely many . Both sides are polynomial in , so since they are equal infinitely many times they are the same polynomial, as desired.
where 
It follows that 
If 
then 
which is a contradiction since 
where the last value has no factor of 
then 
Thus, 
Assume 
It follows that 
which forces 
There are infinitely many such that either 
and then subtracting/adding 1 implies our result. such that 
for some 
This implies the result.
and 
Prove that
and 
.![\[P=\frac{1}{5-2a}+\frac{1}{5-2b}+\frac{1}{5-2c}.\]](http://latex.artofproblemsolving.com/c/3/7/c3708768a84c2bbb1e4fae502a9bb1e92cb13d02.png)
and height 
Volume of cone frustum with base radii 
,
because 
.
,
so we have contradiction.
be sum of coefients with odd deg (e.g 
)
similarly.
we have 
so 
or 
so putting 
.
so 
so puting 
.
where 
and 
goes from 
