,
,
,
,
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 
be an acute triangle with incentre 
and 
.
and 
intersect the circumcircle of 
and 
,
and 
such that 
and 
are parallelograms (with 
,
)
be the point of intersection of lines 
and 
.
,
,
denote the foot of the altitude from 
to 
.
with diameter 
intersect altitude 
at distinct points 
and 
.
passing through point 
at 
.
meets 
.
intersects 
at 
intersects 
.
be a positive integer and 
a prime divisor of 
,
.
or 
,
is integer.
if
there is a nonzero residue 
modulo 
(all considered mod 
(not necessarily distinct) such that 
.
,
.
:o
so 
and since 
is just 
therefore 
so 
such that![\[
\sqrt{x - 2011} + \sqrt{2011 - x} + 10
\]](http://latex.artofproblemsolving.com/4/8/b/48b82d13b733947420a0d763c6af74636770835a.png)
is an integer.
have the digit 
and are divisible by 
?
Y by Adventure10, megarnie, GeoKing
Suppose 
is a connected graph with 
edges. Prove that it is possible to label the edges 
in such a way that at each vertex which belongs to two or more edges, the greatest common divisor of the integers labeling those edges is equal to 1.
Note: Graph-Definition. A graph consists of a set of points, called vertices, together with a set of edges joining certain pairs of distinct vertices. Each pair of vertices
belongs to at most one edge. The graph 
is connected if for each pair of distinct vertices 
there is some sequence of vertices 
such that each pair 
is joined by an edge of 
.
is a connected graph with 
edges. Prove that it is possible to label the edges 
in such a way that at each vertex which belongs to two or more edges, the greatest common divisor of the integers labeling those edges is equal to 1.Note: Graph-Definition. A graph consists of a set of points, called vertices, together with a set of edges joining certain pairs of distinct vertices. Each pair of vertices
belongs to at most one edge. The graph 
is connected if for each pair of distinct vertices 
there is some sequence of vertices 
such that each pair 
is joined by an edge of 
.This post has been edited 1 time. Last edited by orl, Aug 15, 2008, 1:51 PM
(mario pickavet, bronze '91)
be the vertices involved in the path. If 
,
be another vertex. We can find a path 
,
,
are not among the 
'
will be equal to 
,
Otherwise, label the edges 
Since 
we are done.
s.t. 
are connected with each other and 
.
as 
as 
and 
as 4. Look at the vertex 
,
!?
such that 
and path 
(WLOG 
)
,
in succession. Then, each vertex is labeled either with 2 consecutive integers or with 
where we add edges to 
between pairs of odd-degree vertices. Then, 
such that each walk starts and ends at odd-degree vertices and each odd-degree is the endpoint of exactly one walk.
,![$[1, k_1]$](http://latex.artofproblemsolving.com/d/1/1/d119cc6926863efd3a3bafc64a8a81c175be6c2c.png)
in order, the second one with ![$[k_1+1, k_1+k_2]$](http://latex.artofproblemsolving.com/c/e/b/ceb9dee824e705b391db60eddfc1d78a53b533d9.png)
,![$[k_1+\cdots+k_{m-1}+1, k_1+\cdots+k_m]$](http://latex.artofproblemsolving.com/3/7/c/37c0ed81a5f949ec1189960c87b28beef99e05bb.png)
.
of edge labels equal to one. 
and 
to those first and last edges and of course the may not be coprime.
be the vertices of odd degree of 
,
in order. Otherwise, we will add the edges 
Let this modified graph be 
Note that this may result in the graph not being simple anymore, but that's fine since the problem still holds. Now, observe that 
from the cycle would partition the cycle into walks from 
to 
,
to 
,
,
to 
This makes it such that none of the walks can be cycles since the 
'
?
,
.
largest integers of 
where 
has an edge not in 
and replacing the edge 
with 
.
(with base case 
,
,
.
.
for which 
,
,
,
for which 
,
,
.
with 
with 
,
with 
.
for 
differ by 
.
edges, and is obviously still connected, unless it is empty. If it is empty then 
,
,
,
so that the condition is satisfied. Now, consider these labelings in conjunction with our labelings of the other edges. We know that the labelings are fine for 
since they do not change based upon the labelings of the induced graph since they are only adjacent to other 
'
.
,
for which 
,
.
in 
in a graph). Let 
be a cycle in 
.
with 
,
be the smallest such index for which 
,
be the second-smallest such index for which 
.
.
,
by minimality of 
with 
with 
with 
,
(these vertices have degree 
edges. Thus, by the inductive hypothesis we can label the edges 
so that the condition is satisfied for the reduced graph. Now, consider the entire graph. Then, for each vertex 
,
and 
,
be the vertices in the cycle. Then, label 
with 
,
with 
.
,
to get a connected graph on 
edges, so we can label the edges 
so that it works on the reduced graph. Now, it is not hard to see that it works in the original graph -- for all vertices which are not a 
,
(so it is 
)
.
for which 
,
.
with 
with 
,
with 
.
,
,
,
edges, so by the inductive hypothesis we can label it and then this labeling works for 
'
and 
all work since their neighboring edges have labels that are adjacent numbers (and so they have gcd 
consecutively. By maximality of the walk, every vertex in the walk (including the endpoints) is now valid; now, consider the subgraph 
.
,
.
is still connected, so by the induction hypothesis, we can label the edges of 
.
,
so labeling the remaining edges in arbitrary order will be satisfying the condition. Thus we're done.
degree.
.
.
is connected to goes from degree 1 to degree 2 when you include 
edges in this tree, we have the first 
