,
,
,
,
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.
,
isn't even rigorously defined in this expression. What exactly do we mean by 
.
,
does not exist in the first place, although you will see why in a calculus course. So the point is that 
?
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 
,
or ![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](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 
.
is undefined because dividing by ![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](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 ![\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 ![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](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 ![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and 
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 ![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by 
and 
pebbles, respectively. In the first round Barry chooses a positive integer 
and Andrew takes away 
pebbles from one of the two heaps (if 
be a triangle with 
.
lies inside the triangle in such a way that 
.
be the midpoint of 
.
.
,
.
at 
.
similarly.Prove that 
are collinear.
grid with 
rows and 
the there exist two identical rows (meaning same numbers in the same order).
have the digit 
and are divisible by 
?
and 
intersect at the point 
.
,
.
?
?
\end{table}
Y by HamstPan38825, Adventure10, Mango247
There are 
cities, 
airline companies in a country. Between any two cities, there is exactly one -way flight connecting them which is operated by one of the two companies. A female mathematician plans a travel route, so that it starts and ends at the same city, passes through at least two other cities, and each city in the route is visited once. She finds out that wherever she starts and whatever route she chooses, she must take flights of both companies. Find the maximum value of .
cities, 
airline companies in a country. Between any two cities, there is exactly one -way flight connecting them which is operated by one of the two companies. A female mathematician plans a travel route, so that it starts and ends at the same city, passes through at least two other cities, and each city in the route is visited once. She finds out that wherever she starts and whatever route she chooses, she must take flights of both companies. Find the maximum value of .
by the well-known argument, we have a monochrome triangle, so that doesn't work, and for 
a working construction is easy to find (connect vertex pairs {1,2}, {1,3}, {3,4} with red, and the rest blue). The question is, what about 
?
'
edges, so there are at most 
edges. On the other hand 
edges, so 
.
and that's it.
,
.
of vertices 
is given by the three paths 
.
,
,
can be partitioned into 
each? or else in some other way into 
,
on 
for a forest 
and 
where 
is the number of edges of 
,
.
-
.
consisting of all the blue edges, and the graph 
consisting of all the red edges. Since each graph does not have a cycle, 
edges, we have 
hence done.
edges of some color, say red. The maximal number of edges of a graph on ![\[\left\lceil \frac{\binom n2}{2} \right\rceil \leq n-1 \implies n \leq \boxed{4},\]](http://latex.artofproblemsolving.com/8/4/5/8455657f5f58ff6105675ce451b35d0bc14030f5.png)
which is easily constructable.
.
edges and thus,![\[\frac{n(n-1)}{2} \leq 2(n-1)\]](http://latex.artofproblemsolving.com/3/f/0/3f0ed5612ece892ed7db5d9ab83037660c5224ff.png)
from which we conclude that 
as desired.
.![\[\binom{n}{2} \le 2(n-1) \implies n \le 4,\]](http://latex.artofproblemsolving.com/7/3/6/73644107379b5afa70efb5187a53a84ee12064df.png)
there are three edges coming tru a single vertex in the same color, because of Dirihlet. Now the edges connecting the edges with the same color, should be colored in the other color leading to getting a monochromatic triangle - impossible. If n = 5, after bruteforcing the cases there is either a monochromatic triangle or pentagon, contradiction. For n = 4, the coloring is easy: if 
,
must contain a triangle consisted of only blue or red edges, so any value of 
fails.
.
c
.
Mark the edges 
as red and the rest as blue. This graph works, so the maximum value of 
