,
,
,
,
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 
,![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](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 nonnegative integers such that 
Prove that![\[
\sum_{i=0}^n i\binom{a_i}{2}\le\frac{1}{2}\binom{a_0+a_1+\dots+a_n}{2}.
\]](http://latex.artofproblemsolving.com/c/4/9/c49d0ccfef69402d05dc44d721172498c605cf7e.png)
Note: 
for all nonnegative integers 
.
,
Y by ihatemath123, lpieleanu, RoyalPrince
Let 
and 
be positive integers with 
. There are cupcakes of different flavors arranged around a circle and people who like cupcakes. Each person assigns a nonnegative real number score to each cupcake, depending on how much they like the cupcake. Suppose that for each person 
, it is possible to partition the circle of cupcakes into groups of consecutive cupcakes so that the sum of 's scores of the cupcakes in each group is at least 
. Prove that it is possible to distribute the cupcakes to the people so that each person receives cupcakes of total score at least with respect to .
and 
be positive integers with 
. There are cupcakes of different flavors arranged around a circle and people who like cupcakes. Each person assigns a nonnegative real number score to each cupcake, depending on how much they like the cupcake. Suppose that for each person 
, it is possible to partition the circle of cupcakes into groups of consecutive cupcakes so that the sum of 's scores of the cupcakes in each group is at least 
. Prove that it is possible to distribute the cupcakes to the people so that each person receives cupcakes of total score at least with respect to .
.
.
of people such that they match less than 
bubbles. Remove that set, and repeatedly remove any set with 
.
such that 
have a matching, and no person in 
matches any bubble in 
delicious for person 
if the sum of 
into disjoint counterclockwise-oriented arcs 
such that
is the shortest arc starting at some cupcake 
that is delicious for someone;
,
is the shortest arc starting at the first cupcake counterclockwise of 
that is delicious for someone.
now. For each arc 
to consist of all the people for which 
of the sets 
;
(indexed without loss of generality) to the sets in 
such that 
.
,
we can remove them and induct down (which is false) otherwise (this case doesn’t even exist??) there exists a bubble not matched with anyone which we can induct down on preserving score. I missed the immediate match and win case and I think my second case is a subset of my first which is skullers. Like I’m pretty sure I have a solve if in my first case instead of matching and winning I match and consider the complement which eventually reduces to something like my second case.
people who you are about to delete settled. For the people who aren't settled, you need to make sure they lose exactly 
.
unsettled people value 
,
people and 
be the set of bubbles of 
matches with 
people with a bubble and give 
such that if the set of its matches is 
,
.
and 
.
,
still. Additionally, note that no person in 
.
,
is trivial.
between the people and the arcs for Pip.
of people, who are compatible with fewer than 
of the arcs. Let's imagine deleting 
for the remaining graph. We again delete 
is possible in the remaining graph, i.e.\ there are no more bad sets (and then terminate once that occurs).
.![[asy]usepackage("amssymb");
size(8cm); dotfactor *= 1.3; real w = 3; real eps = 0.4; label("People", (-w,10)); label("Arcs of Pip", (w,10)); filldraw(box((-w-eps, 9+eps), (-w+eps,7-eps)), invisible, red+1.2); filldraw(box((-w-eps, 6+eps), (-w+eps,5-eps)), invisible, orange+1.2); filldraw(box((-w-eps, 4+eps), (-w+eps,2-eps)), invisible, brown+1.2);
filldraw(box((w-eps, 9+eps), (w+eps,8-eps)), invisible, red+1.2); filldraw(box((w-eps, 7+eps), (w+eps,7-eps)), invisible, orange+1.2); filldraw(box((w-eps, 6+eps), (w+eps,5-eps)), invisible, brown+1.2);
draw((-w+eps, 9+eps)--(w-eps, 9+eps), red+dashed); draw((-w+eps, 7-eps)--(w-eps, 8-eps), red+dashed); draw((-w+eps, 6+eps)--(w-eps, 9+eps), orange+dashed); draw((-w+eps, 5-eps)--(w-eps, 7-eps), orange+dashed); draw((-w+eps, 4+eps)--(w-eps, 9+eps), brown+dashed); draw((-w+eps, 2-eps)--(w-eps, 5-eps), brown+dashed);
label((-w-eps, 8), "Bad set $\mathcal{B}_1$", dir(180), black); label((-w-eps, 5.5), "Bad set $\mathcal{B}_2$", dir(180), black); label((-w-eps, 3), "Bad set $\mathcal{B}_3$", dir(180), black);
draw((-w,1)--(w,1), deepgreen+1.3); draw((-w,0)--(w,0), deepgreen+1.3); label((0, 0.5), "Final perfect matching $\mathfrak{M}$", deepgreen);
for (int i=0; i<10; ++i) { dot((w,i), blue); dot((-w,i), blue); } label("Pip", (-w,0), dir(180), blue);
[/asy]](http://latex.artofproblemsolving.com/5/7/a/57a351245aaaaf5518f8ececbcbb58b8dc5c2576.png)
of ![\[ (\ge 1) + (\ge 1) - (\le 1) \qquad \ge \qquad 1 \]](http://latex.artofproblemsolving.com/0/5/8/058f38c90ef6a220bd70fe2840c7e84cf768f5e9.png)
meaning the induction is OK. See below for a cartoon of the deletion, where Pip's arcs are drawn in blue while 
)![[asy]
size(13cm); usepackage("amsmath"); pair O = (0,0); picture before; picture after;
real r = 1; real s = 0.9; real t = 0.65;
draw(before, arc(O, s, -20, 80), red+1.3); draw(before, arc(O, s, 100, 200), red+1.3); draw(before, arc(O, s, 220, 320), red+1.3);
draw(before, "Pip arc to delete", arc(O, r, 40, 140), blue+1.3); draw(before, rotate(-60)*"Pip arc", arc(O, r, 160, 260), blue+1.3); draw(before, rotate( 60)*"Pip arc", arc(O, r, 280, 380), blue+1.3);
label(before, "$\boxed{0.2}$", t*dir(120), red); label(before, "$\boxed{0.3}$", t*dir( 60), red); label(before, "$\boxed{0.8}$", t*dir(180), red); label(before, "$\boxed{0.43}$", t*dir(240), red); label(before, "$\boxed{0.57}$", t*dir(300), red); label(before, "$\boxed{0.7}$", t*dir( 0), red); label(before, "$Q$'s values", O, red);
draw(after, arc(O, s, -20, 200), red+1.3); draw(after, arc(O, s, 220, 320), red+1.3);
draw(after, rotate(-60)*"Pip arc", arc(O, r, 160, 260), blue+1.3); draw(after, rotate( 60)*"Pip arc", arc(O, r, 280, 380), blue+1.3);
label(after, "$\boxed{0.8}$", t*dir(180), red); label(after, "$\boxed{0.43}$", t*dir(240), red); label(after, "$\boxed{0.57}$", t*dir(300), red); label(after, "$\boxed{0.7}$", t*dir( 0), red); label(after, "$Q$'s values", O, red);
add(before); add(shift(3.2,0)*after);
[/asy]](http://latex.artofproblemsolving.com/b/8/d/b8dbd78a98423d5c16be57fce0a886df36a1824f.png)
)
