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).
![\[x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0\]](http://latex.artofproblemsolving.com/f/d/0/fd0ab11635ee44ad272a54c0f20a2bc47a3596de.png)
cannot all be real if 
.
not necessarily distinct factorials of positive integer numbers. Prove that the set of compact powers of two is finite.
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 .
is not a prime number.
and 
be two points in the coordinate plane. A circle is drawn such that 
is a diameter.
. Find the equation of these lines. is inscribed in a circle with radius 
. 
. 
. 
. What is the area of the Quadrilateral ?
, 
, 
, 
, 
, 
be positive integers and let 
.
divides 
and 
. Prove that is composite.
are multiples of . Evaluating 
at we get that 
is a multiple of . is composite, since 
are all strictly less than .
is prime.
is a solution then 
is also a solution. Note that does not change. Denote 
. We can claim that 
also satisfies the conditions.
. But this is impossible, since 
are all 
. 
now so we get 
, suppose is prime then it must be less than one of but as 
so absurd and done.
be prime. Note divides 
. Substitute 
we get divides 
and
, 
and 
. However, each of 
are smaller than , a contradiction.
be a fixed prime. We say a -tuple 
of integers is good if 
and 
and 
. is good then so is 
.
-tuple of positive integers. If not, WLOG is the smallest of the six numbers. Then by repeating the claim, the tuple 
is good. But now 
and 
, contradiction.
Now, assuming that is prime, we see that![\[(X-a)(X-b)(X-c)=(X+d)(X+e)(X+f)\]](http://latex.artofproblemsolving.com/f/a/5/fa55af7a46e2a02d5006a87b46caa4c436e0c0bf.png)
over ![$\mathbb{F}_S[X]$](http://latex.artofproblemsolving.com/1/4/a/14a21d85873231e62118fb1bef37dbdd0d4c09b5.png)
, so because of unique factorization, we have WLOG that 
, 
, and 
. In particular, this means that 
. However, 
, so this isn't possible. Therefore, couldn't have been prime to start with. 
, so assume for the sake of contradiction that is prime. Let 
and 
. Multiplying out, we get that 
for all integers 
. In particular, we get that for an integer , divides 
if and only if divides 
, which, because is assumed to be a prime, means that 
is just some permutation of 
modulo . WLOG, assume that 
, 
and 
modulo . But this means that 
, 
and 
are all multiples of , which is impossible, because they are positive integers which sum up to .
is prime. Notice that 
for all 
. to be 
we have that 
in 
using that is prime, yet, adding any such values of , we have that 
which is a contradiction. 
is divisible by for all integers . Thus,![\[S\mid f(a) = -(a+d)(a+e)(a+f)\]](http://latex.artofproblemsolving.com/8/e/1/8e154252998ba701b0a59a63e0b705d60e79d7a5.png)
To finish, cannot be prime because 
, so we're done. .
is prime, and take polynomials 
and 
Note that ![\[P(x)\equiv x^3 - (a+b+c)x^2 + (ab+bc+ca)x-abc \pmod S\]](http://latex.artofproblemsolving.com/7/f/5/7f59df6e2f3974a5e80bd5b4ac3dd868230000a4.png)
Note that on the other hand, ![\[Q(x)\equiv x^3 + (d+e+f)x^2 + (de+ef+fd)x+def \pmod S\]](http://latex.artofproblemsolving.com/8/4/2/84230bcf983b2168ded438594f9521775add2c1a.png)
Now, 
Note that this is divisible by 
In particular, 
is divisible by Note that each of those factors are less than but is prime, which is a contradiction.
This means that 
for all 
as the resulting polynomials are equivalent under modulo . Letting 
, 
but 
, and thus must be composite.
is prime . Then we have the following equations.
We will construct numbers 
such that:
In fact, this is not so hard to do. It is just enough to take 
and 
. It is easy to see that they satisfy the first congruence. Note that:
We see that the second desired congruence is also satisfied by given congruences. Also, observe that:
We see that the third desired congruence is also satisfied by given congruences.
are positive integers and their sum it , then they all are less than . WLOG is the largest among 
. With each construction we increase by and decrease 
by . We can keep doing it until becomes . Suppose that we needed for that constructions, in other words 
. Consider congruence:
We see that it implies that 
. WLOG assume that 
is the one which is divisible by . Then 
, which is contradiction, since it forces 
to be . composite as desired.
is prime. We will make use of these handy identities in the solution.
The following claim is the crux of the problem.
for all 
.
. For ease of notation label 
, 
and so on. Now finally for the inductive step, we will just work backward.
By the induction hypothesis, we will cancel of 
term. It is not hard to see that 
as well.
It isn't hard to expand this out and realize its a tautology. So, the induction is complete.
. This would give![\[S = a + b + c + d + e + f \mid (a + d)(b + d)(c+d).\]](http://latex.artofproblemsolving.com/3/2/d/32db74e588d686d5a1aa6a1709ed3b9955024515.png)
Since is a prime, it must divide one of the factors. But since all the factors are positive and less than , its the desired contradiction. 
as value of .
is a prime.
, 
and 
. Summing these up gives, 
Also, 
. is a solution then, 
is also a solution with 
.. WLOG that be .
. But therefore must divide one of which isn't possible as, 
. This gives a contradiction. 
was prime. Then the polynomials 
and 
have equivalent expansions modulo , hence we must have 
. But then 
: contradiction. 
hence as 
. Now , 
assume is prime, then as we have we have at least one of 
but since 
this is absurd for 
, hence our assumption false and hence is composite .
![\[
(t-a)(t-b)(t-c) \equiv (t+d)(t+e)(t+f) \pmod{S}
\]](http://latex.artofproblemsolving.com/8/8/6/8869db968e5197f8922c1fbf869591364313b8bb.png)
for all integer 
. Suppose for the sake of contradiction that was prime. Take 
. Then, divides one of , 
, 
. However, since are positive integers, that would mean that one of 
is equal to . This is a contradiction as the rest of the variables could not be positive, so must be composite.
be a prime
in 
we're done.
be an integer. Note that divides the quantity 
Now there are 
cases. is one of 
WLOG suppose it is 
Letting 
gives 
so switching the positive sign yields 
If was prime, then it would have to divide at least one of these factors, say 
(the other cases are symmetrical to this). Then 
which is a contradiction due to size issues. is one of 
WLOG suppose it is 
Letting 
gives 
and repeat the size argument from the previous case. must be composite, as desired.
for some prime 
Then we get a vieta-like relation modulo 
This implies that and 
are roots to the same polynomial with coefficients modulo 
are multiples of 
for some integer 
which violates the condition that 
or is prime. Therefore must be composite.
which satisfy the given conditions. We call such a pair a good pair and if 
then we say that it is a 
good pair. We start off by proving the following nice result about good pairs. for which there exists a good pair , the pairs 
and 
are also good.good if they are good at all, since the sum 
does not change in either instance. We show that if is good then the pair must also be good. This is a mere calculation. Note that,![\[ (a-1)(b-1)(c-1)+(d+1)(e+1)(f+1) = (abc+def) -(ab+bc+ca-de-ef-fd) + (a+b+c+d+e+f)\]](http://latex.artofproblemsolving.com/7/9/d/79d92d2b07649a2309675c9bbd40793110c64920.png)
where each of the bracketed terms are known to be multiples of . The proof of the other is entirely similar, so this finishes the proof of the claim.
good pairs for any prime . By way of contradiction, say there exists such a prime and a good pair . Then, WLOG say the minimum element of the set 
is . By way of our claim, it follows that the pair 
is also good. Note that here, 
, 
and 
are all positive integers. Then,![\[p|(0)(b-a)(c-a)+(d+a)(e+a)(f+a)=d'e'f'\]](http://latex.artofproblemsolving.com/1/4/8/1486fa5dd4ccf8439f3f3f1adb0a7413d458a2b6.png)
Since is a prime, this implies that divides one of 
, 
and 
. But, since 
where the first 3 terms are non-negative and the last two terms are strictly positive,![\[p = (b-a) + (c-a) + (d+a) + (e+a) + (f+a) \ge d'+e'+f' > d',e',f' \]](http://latex.artofproblemsolving.com/f/f/8/ff81a816a7e657c13279ad8009a283d59f17697a.png)
which is a clear contradiction. Thus, there cannot exist such a prime which proves the desired result.
.,
, then 
. However, are all less than , contradiction if is prime.
is prime. Observe that![\[
a + b + c \equiv -d - e - f \pmod{S}
\]](http://latex.artofproblemsolving.com/6/5/b/65b17328dc50401bca6c02b93af26af0764e5fed.png)
![\[
abc \equiv -def \pmod{S}
\]](http://latex.artofproblemsolving.com/a/e/f/aefdb33a5e701a2c2673a276cadafcb2caf88893.png)
![\[
ab + bc + ca \equiv de + ef + fd \pmod{S}
\]](http://latex.artofproblemsolving.com/d/0/c/d0cd03b97a83dca5632a66f383f808bf142b6186.png)
By Vieta, we have that for every positive integer 
, the following relation holds:![\[
(n + a)(n + b)(n + c) \equiv (n - d)(n - e)(n - f) \pmod{S}.
\]](http://latex.artofproblemsolving.com/8/7/5/875c367d3f930c1e5f03cd7ef8c614aa4dcf8ed1.png)
By setting 
, we obtain that divides one of the numbers 
, 
, or 
.
. In particular, we have 
, i.e., 
.![\[
a + d < a + b + c + d + e + f = S.
\]](http://latex.artofproblemsolving.com/8/7/1/871fdf7fedb6beda5cc926f7cd824bb07efd123e.png)
.
we get cannot be prime, qed.
is prime. Then, 
and 
when taken modulo 
Let 
for 
It is clear that 
and 
and 
So, 
are the solutions of some cubic modulo But, by Lagrange's Theorem, this can only have at most 
solutions, so these sets are the same, in some order. So, 
where 
and 
But, then 
so we have a contradiction.
be a fixed prime. We say a -tuple of integers is good if and and . is good then so is .-tuple of positive integers. If not, WLOG is the smallest of the six numbers. Then by repeating the claim, the tuple is good. But now and , contradiction.
are congruent modulo Thus, 
However, 
However, each of the factors are less than 
which finishes.
. We have 
. must divide one of them but all three are less than 
, a contradiction. So, is composite.
![$S|(c+f)[(b+c)(c+a)+(d+e)(c-f)]$](http://latex.artofproblemsolving.com/b/e/8/be8ea894c975ae3a08306af2d92bc70de652e1e4.png)
,
,
and 
are the same polynomial, mod 
.
.
.
and 
,
for all integers 
.
or 
or 
which means one of 
are nonpositive, contradiction.
.
.
,
and consider 
Taking mod 
.
has a root at 
but 
and so we are done.
has all of its coefficients divisible by 
.
,
,
and clearly inputting 
implies 
,
.
,
,
Letting 
we have 
which is absurd if 
in 
contradiction.
.![\[
(x+a)(x+b)(x+c) - (x-d)(x-e)(x-f).
\]](http://latex.artofproblemsolving.com/7/4/6/74644154e284b0dc05bfe757c76adecafde3cbbe.png)
If 
,