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).
positive real numbers such that 
. Prove that :
and 
, there cannot exist infinitely many 
such that![\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]](http://latex.artofproblemsolving.com/5/4/d/54d20546de5736770a9929ac86163e3b06553b7d.png)
![\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/6/6/8/668d53723d307556536736fda0e4a19be848834f.png)
![\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]](http://latex.artofproblemsolving.com/1/9/f/19f2627a432354a757d3c9e8ba42e96f83f46726.png)
So:![\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]](http://latex.artofproblemsolving.com/0/7/3/073602693604bbf3cb1f5c1c2e5051e8994a5e78.png)
![\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/1/1/4/11441fab3ba768fee7a0eef4dbf68124d6637aec.png)
![\[
(n + a)(n + b) = k^2.
\]](http://latex.artofproblemsolving.com/0/e/c/0ec5e7bac2aebe9ec507727ed21f7ca028655222.png)
Assume 
. Then we have:![\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]](http://latex.artofproblemsolving.com/7/5/f/75fc52b6662df3a4c0cb2f40f49c4b3351c90a4d.png)
or it could also be that 
.![\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]](http://latex.artofproblemsolving.com/1/f/5/1f5936241f7bebcf15c68fed5267bbfad2d40c2b.png)
![\[
k_1 k_2 = \frac{n + b}{n + a}.
\]](http://latex.artofproblemsolving.com/d/4/c/d4c1958e3554b58c3def282070f849842cc10912.png)
, we have:![\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]](http://latex.artofproblemsolving.com/3/0/6/306a6a00c44d4e4db1b36c7e588f684f28fdbf44.png)
, 
must be an integer, which is not possible..
be a sequence of positive integers with 
, 
and![\[a_n = a_{n-1}^{a_{n-1}a_{n-2}}-1\]](http://latex.artofproblemsolving.com/1/b/f/1bf1a88979e7d2b33530de979d66a4e81032953d.png)
for all 
. Show that if 
is a prime less than 
for some positive integer 
, then there exists 
such that 
.
[hide=P#]a1b2c3[/hide] [hide=S#]a1b2c3[/hide]
feet and area ft^2. What is the arithmetic mean of all possible values of ?[/quote]
such that![\[
\sqrt{x - 2011} + \sqrt{2011 - x} + 10
\]](http://latex.artofproblemsolving.com/4/8/b/48b82d13b733947420a0d763c6af74636770835a.png)
is an integer.
to 
have the digit 
and are divisible by 
?
. Consider the sets 
of 
points 
for 
, which partition the vertices of 
-gon (indices are taken modulo ). If there exists a set such that all elements are vertices of the -gon, then both 
and 
are clearly sides of the -gon ("clearly" because the convex hull of the -gon will be a subset of the convex hull of the -gon). This is a contradiction since 
. So from each set a maximum of vertices can be vertices of the -gon. This gives an upper bound for of 
. This is achievable by considering the 
-gon formed by the segments 
and 
.
a maximum of vertices can be vertices of the -gon. This gives an upper bound for of 
., not 
. Luckily, you lose only one mark...
-gon such that 
.First consider all of its vertices.Let be the number of sides that have a side that is paralel to them.At first 
.Now we take 
edges that share a vertice and we see that there is a pair of paralel sides which also contains a common vertice.We remove that vertice and consider our 
-gon.We see that these edges can no longer have any paralel sides and we see that the new edge won't have paralel sides neither.Also number of edges is now less for edge which is a total of edges that don't have paralel sides so now 
.By now taking the remaining edges with the same trait we can repeat this.This way we will remove ![$[\frac{n}{4}]$](http://latex.artofproblemsolving.com/8/9/2/89210620a0e728259d9979e90a62914ff06d41fe.png)
vertices.The proof that this will get remove minimum nubmer of vertices: Suposse there exists a lesser number of vertices that we can remove.Start removing the vertices of the original one to obtain this one.We see that in every step can't go down by more than so it is impossible to remove less vertices.
we easyily obtain the answer is 
one-arcs. We now need to cover 
arcs, and each side must cover at least two arcs, so we have at most 
sides. This is easily attainable; have two-arcs covering the upper semicircle and have one-arcs covering the bottom semicircle.
-gon is inscribed in a circle. Find the maximal such that we can choose vertices from given and construct a convex -gon without parallel sides.
We claim that 
First assume 
to be the number of chords with strictly 
points in between the endpoints on the circle. Now the non-parallel condition implies 
Note that the total number of points 
To see this, for a chord having 
point in between, consider the left 
points out of the total 
( in between, and end points). Then each vertex on the circle is counted exactly once.
a contradiction.
the construction is easy; if the points are 
then draw the chords 
. For a construction, label the vertices 
in clockwise order, and take vertices 
. Edges 
and 
are parallel iff 
. Note that all edges where 
have 
unique, as with all edges where 
, which covers all edges. Furthermore, since 
is odd in the first case and even in the second, there are no overlaps "between" groups either, as desired., where 
is clockwise along the polygon from 
, to be 
, and define the pseudoperimeter of a polygon formed from vertices of our -gon in the obvious way. Suppose that we have 
edges of odd pseudolength and 
edges of even pseudolength, so we wish to maximize 
. Obviously, the pseudoperimeter of the polygon formed is always , so 
. Furthermore, the pseudolength of an edge has the same parity as , so every odd-pseudolength edge must occupy an odd residue class modulo . Since there are 
of these, we have 
. Hence
as desired. 
gon. We claim that the maximum number of vertices we can pick is 
. Label the vertices 
in the clockwise direction. It is known that edges and are parallel if and only if .
.
has elements. Now, note that the first 
sides (in the first 'half' of our gon) are non-parallel among themselves quite clearly. Similarly the last sides are also non-parallel among themselves. Also one of the first sides and one of the last sides cannot be parallel. This is because the last sides are of the form 
and the first sides are of the form 
, so![\[a+(a+1) = 2a+1 \equiv 1 \not \equiv 0 \equiv 2c +2 = c + (c+2) \pmod{2}\]](http://latex.artofproblemsolving.com/b/9/9/b99db934ed0c2db0aa045d7948e658ee33f2e903.png)
and since 
this implies that they are also distinct 
so the parallel condition is not satisfied. pair the vertices with its neighbors. For example, we pair 
, 
, 
, 
. For any pair 
let its antipodal pair 
denote the pair joining the two vertices diametrically opposite to the points in pair . Note that if a pair 
it must be a side of our polygon as there are no other vertices between the two vertices in . Thus, all four vertices in and cannot be in since 
quite clearly. Thus, out of these 4 points at most 3 can be in our polygon, which summing across all pairs implies that we can have at most vertices in and we are done.
:o
so 
and since 
is just 
therefore 
so 
