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 incentre 
and 
. Let lines 
and 
intersect the circumcircle of at 
and 
, respectively. Consider points 
and 
such that 
and 
are parallelograms (with 
, and 
). Let 
be the point of intersection of lines 
and 
. Prove that points 
, and are concyclic.
positive real numbers such that 
. Prove that :
+1 w
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 
?
round any of its endpoints. Is it possible that after several rotations the needle returns to initial position with the endpoints interchanged?
, where 
are linearly independent over 
, this is a unique representation.
is invariant for both endpoints. They start at different parity. So they can't interchange positions.
![${\mathbb Z}[\omega]$](http://latex.artofproblemsolving.com/d/1/2/d12fd2e831d0da64de29e065253c03b71f18bad6.png)
where 
. Draw the needle as a directed segment from to in the plane.
. Rotations around the other endpoint correspond to ![\[ z \mapsto z + \omega^k - \omega^{k-1} \]](http://latex.artofproblemsolving.com/2/e/0/2e04e71d6294a14fa622b3807200a6ed85095f58.png)
for some choice of 
.. To prove this we only need show the following claim, which proves the relevant invariant.
in .![${\mathbb Z}[\omega]/(\omega-1)$](http://latex.artofproblemsolving.com/3/a/0/3a0dc4b76ee55649dcdacc1f56d8a38433a9862d.png)
is not trivial. Write ![${\mathbb Z}[\omega] = {\mathbb Z}[T] / (T^4+1)$](http://latex.artofproblemsolving.com/5/5/2/5528f519ce9305f3ab167cca0e09a578155aae11.png)
, then ![\[ {\mathbb Z}[\omega]/(\omega-1) \cong {\mathbb Z}[T] / \left( T^4+1, T-1 \right) \cong {\mathbb Z} / 2 = {\mathbb F}_2 \]](http://latex.artofproblemsolving.com/8/9/1/891f618efc341f8fd237341a0972a89a0df9eabf.png)
as desired. 
and the other is at 
, and WLOG rotate about the left endpoint first. Suppose a sequence of rotations works. Right before we switch endpoints to rotate around, draw the needle's current position on the plane. Also draw the line joining and . Then the drawn segments clearly form a cycle/polygon. Furthermore, because the endpoints must be interchanged, there must be an odd number of vertices on this graph. On the other hand, every drawn segment has length and is either horizontal, vertical or has slope 
. Because and 
are linearly independent over 
, it then follows that to end at the same position we started, we need an even number of horizontal and an even number of vertical edges. By rotating the argument 
the same is true for edges with slope and edges with slope 
, so there are an even number of edges and thus vertices: contradiction. 
.. A move consists of taking a complex number and adding 
where is some integer from to . (Effectively what this does is that we rotate the needle by some amount, and then jump to the other endpoint.) We now show that the number of moves must be even, if we return to . More formally, if![\[ \sum_{k = 0}^{7} a_k\omega^k = 0 \]](http://latex.artofproblemsolving.com/a/b/c/abcfbd631cd0597cdacf245fbbdf59dc6e9ead09.png)
for some integers 
through 
, we must have that![\[ \sum_{k = 0}^{7} a_k \equiv 0 \pmod{2}.\]](http://latex.artofproblemsolving.com/4/1/c/41c7d30468c67c6fe0d6a84cc77bf2b8e2ac730f.png)
through are zero, by subtracting pairs of 
while preserving the parity.
. As such 
. Likewise, 
. This then forces 
, so the statement is true.
be a primitive eighth root of unity. Suppose otherwise, and consider the locus of all positions of the segment. Modulo the first and last positions, all such segments 
form an equilateral polygon with an odd number of sides.'s that sum to ; equivalently, there exists a polynomial ![$f \in \mathbb Z[X]$](http://latex.artofproblemsolving.com/3/e/3/3e3d314244e5e514fdf63e59038fc713b6a48ed1.png)
such that 
is odd and 
.
, and hence 
, contradiction!
and the other at . We will let 
and 
and after 
moves, we let 
denote the new endpoint of the needle after a move. Note that if then we can set up the following recurrence relation for 
![\[ z_{n} = \omega^{e_{n}}(z_{n-2}-z_{n-1})-z_{n-2} = (1- \omega^{e_{n}})z_{n-1}+\omega^{e_{n}}z_{n-2}\]](http://latex.artofproblemsolving.com/4/1/c/41c8e53e652c5e96cd471fdbacb2f92e05d4a376.png)
for an arbitrary sequence of integers 
. This recurrence works because we translate to a new point 
on the unit circle, rotate it by 
and undo the translation, which is exactly the operation we desire. is odd and 
are integers, then ![\[ \omega^{a_1}+\omega^{a_2}+ \cdots + \omega^{a_k} \not = 0 \]](http://latex.artofproblemsolving.com/c/8/3/c8365b7b5921c9a82fe7334cff111fa28799e0f8.png)
then 
is even. Indeed, note that ![\[ \Re(t_1\omega^1+t_2\omega^2 \cdots +t_8\omega^8) = t_1\cdot \Re(\omega^1)+t_2 \cdot \Re(\omega^2) \cdots +t_8 \cdot \Re(\omega^8)=0 \]](http://latex.artofproblemsolving.com/4/9/4/494a80516578c37d743fd357ba5482b01bbee943.png)
![\[ \implies (t_1+t_3-t_5-t_7) \frac{\sqrt{2}}{2} + t_2+t_4 = 0 \implies t_1+t_3-t_5-t_7=0 \]](http://latex.artofproblemsolving.com/f/9/f/f9fbeef15255e342809c0b6d58893034221c02e0.png)
So, 
is even. We also note that ![\[\omega \cdot (t_1\omega^1+t_2\omega^2 \cdots +t_8\omega^8) = 0 \implies t_8\omega^1+t_1\omega^2 \cdots +t_7\omega^8\]](http://latex.artofproblemsolving.com/a/f/0/af0ed3863ec1e2d86c7f2d7597e97424acae65a2.png)
So, 
is even by the same logic. Thus, 
is even as desired. 
denote the number of terms in the expansion of , we can set up the following recursion ![\[ z_{n} = (1- \omega^{e_{n}})z_{n-1}+\omega^{e_{n}}z_{n-2} = z_{n-1} + \omega^{e_n+4} \cdot z_{n-1} + \omega^{e_n}z_{n-2}\]](http://latex.artofproblemsolving.com/c/6/3/c63c4e287786626a944e450a5e7935f11337c8c1.png)
![\[ \implies A_n = 2A_{n-1} + A_{n-2} \]](http://latex.artofproblemsolving.com/2/8/7/28717d4f0ecbab658c885103bee6adbe595c82fc.png)
Where 
and 
. Note that this recursion implies that for all odd , we have 
. Thus, 
has an odd number of terms in it's expansion. Thus, 
for all odd . So, the two ends of the needle cannot swap.
![$\mathbb{Z}\left[\frac{\sqrt{2}}{2}\right]^2$](http://latex.artofproblemsolving.com/9/c/0/9c00968aacadfe8470d61c5f8f7eecaf71153545.png)
. WLOG the needle has length . We will only focus on one endpoint, and prove that it cannot lie on the starting point of the other endpoint.
around another results in both 
and 
being invariant modulo 
. However, since the two endpoints of the needle have distance , the two endpoints have different parities, so they cannot swap.
are vectors with the same magnitude that sum to and the argument of each vector is a multiple of 
degrees, then is even.. The idea here is that since is irrational, both the integer and parts of both 
and 
coordinates must be zero. Thus 
appears the same number of times as 
, and 
appears the same number of times as 
. Furthermore, 
in total occur the same number of times as 
in total due to the coordinate. Thus, the total number of vectors is even. degrees. However, in order for it to get back to its original position with the orientation swapped, the number of such vectors must be odd, contradiction.
. Induct gives 
parity unchanged.
Assume this exists. If we trace out the path as we move along the sequence, it forms a polygon(not necessarily degenerate) with odd number of vertices.![[asy]
size(4cm);
pair Exp(real r){
return (cos(r), sin(r));
}
draw((1, 0) -- (0, 0) -- Exp(pi/4) -- 2 * Exp(pi/4) -- (2 * Exp(pi/4) + Exp(0)) -- (2 * Exp(pi/4) + 2 * Exp(0)));
draw(2 * Exp(pi/4) + (2,0) -- (1, 0), Dotted);
dot((0, 0), black);
dot((1,0), red);
dot(Exp(pi/4), red);
dot(2 * Exp(pi/4), black);
dot(2 * Exp(pi/4) + (1,0), red);
dot(2 * Exp(pi/4) + (2,0), black);
[/asy]](http://latex.artofproblemsolving.com/6/1/2/6124521a2bc1408652ed0d9e0271b22939d9bec1.png)
such that ![\[\sum_{k = 0}^{2\ell} \omega^{a_k} = 0.\]](http://latex.artofproblemsolving.com/4/5/4/45407360bba2600886aab65352ba7608d4c85ee0.png)
Substitute 
and we get that the polynomial ![\[P(x) = \sum_{k = 0}^{2\ell} x^{a_k}\]](http://latex.artofproblemsolving.com/c/2/5/c25c7865d9701a3252501b29d78dfb322c9d06c5.png)
must have a root at 
. Furthermore, ![$P(x)\in\mathbb Z[x].$](http://latex.artofproblemsolving.com/9/6/a/96ab7f9852da14f353985515279469b1abe435e9.png)
Since 
it follows the minimal polynomial of (equivalently 
) must divide 
Thus, 
However, this implies 
However, this is false since 
is the sum of an odd number of 
:o
so 
and since 
is just 
therefore 
so 
