,
,
,
,
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 a convex pentagon with 
and 
For each point 
in the plane, define 
The least possible value of 
can be expressed as 
where 
and 
are positive integers and 
is not divisible by the square of any prime. Find 
be the set of positive integers. A function 
satisfies the equation ![\[\underbrace{f(f(\ldots f}_{f(n)\text{ times}}(n)\ldots))=\frac{n^2}{f(f(n))}\]](http://latex.artofproblemsolving.com/2/e/7/2e73ff2f75b103e9d6a4004a42e4ed7f4ee64a67.png)
for all positive integers 
.
is cut along the planes 
,
.
,
is a nonnegative integer. Call a number gyatt if it has 
or 
factors. What is the sum of all skibidi numbers less than 
that are also gyatt numbers?
unit
,
and 
are integers. What is the value of 
?
and 
for all 
in the domain of 
,
?
Y by pifinity, AbdulelahAltaf, Adventure10, Mango247, and 1 other user
We define a chessboard polygon to be a polygon whose sides are situated along lines of the form 
or 
, where 
and 
are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping 
rectangles. Finally, a tasteful tiling is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a 
rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner.
![[asy]size(300); pathpen = linewidth(2.5);
void chessboard(int a, int b, pair P){
for(int i = 0; i < a; ++i) for(int j = 0; j < b; ++j)
if((i+j) % 2 == 1) fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6));
D(P--P+(a,0)--P+(a,b)--P+(0,b)--cycle);
}
chessboard(2,2,(2.5,0));fill(unitsquare,rgb(0.6,0.6,0.6));fill(shift(1,1)*unitsquare,rgb(0.6,0.6,0.6)); chessboard(4,3,(6,0)); chessboard(4,3,(11,0)); MP("\mathrm{Distasteful\ tilings}",(2.25,3),fontsize(12));
/* draw lines */
D((0,0)--(2,0)--(2,2)--(0,2)--cycle); D((1,0)--(1,2)); D((2.5,1)--(4.5,1)); D((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1)); D((8,2)--(8,3)); D((12,0)--(12,2)--(11,2)--(13,2)); D((13,1)--(15,1)--(14,1)--(14,3)); D((13,0)--(13,3));[/asy]](//latex.artofproblemsolving.com/e/7/1/e712b33951fe4d8c30f4664d0857d871b13fbfb6.png)
a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully.
b) Prove that such a tasteful tiling is unique.
or 
, where 
and 
are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping 
rectangles. Finally, a tasteful tiling is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a 
rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner.![[asy]size(300); pathpen = linewidth(2.5);
void chessboard(int a, int b, pair P){
for(int i = 0; i < a; ++i) for(int j = 0; j < b; ++j)
if((i+j) % 2 == 1) fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6));
D(P--P+(a,0)--P+(a,b)--P+(0,b)--cycle);
}
chessboard(2,2,(2.5,0));fill(unitsquare,rgb(0.6,0.6,0.6));fill(shift(1,1)*unitsquare,rgb(0.6,0.6,0.6)); chessboard(4,3,(6,0)); chessboard(4,3,(11,0)); MP("\mathrm{Distasteful\ tilings}",(2.25,3),fontsize(12));
/* draw lines */
D((0,0)--(2,0)--(2,2)--(0,2)--cycle); D((1,0)--(1,2)); D((2.5,1)--(4.5,1)); D((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1)); D((8,2)--(8,3)); D((12,0)--(12,2)--(11,2)--(13,2)); D((13,1)--(15,1)--(14,1)--(14,3)); D((13,0)--(13,3));[/asy]](http://latex.artofproblemsolving.com/e/7/1/e712b33951fe4d8c30f4664d0857d871b13fbfb6.png)
a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully.b) Prove that such a tasteful tiling is unique.
.
of the forced squares together with all of the lower squares with color 
squares with the color other than 
.
the weight 
.
even.
for those in horizontal dominos and 
for those in vertical dominos. A 
distasteful block with lower left corner at 
,
.
.
must be in the polygon. We can't have the domino at 
in the same tiling as the domino at 
,
must connect in another direction; either (a
in the same tiling as 
.
,
before 
is the only possibility: 
,
must be in at least one tiling. If it's in both, the pair of fixed dominoes we have forces another fixed domino at 
,
,
,
and reverse colors to get case (a'): that 
leads to a contradiction, as a ladder of fixed horizontal dominoes leads to failure.
to 
because of the domino at 
in the same tiling. If we connect 
,
:
is 
in the loop because it is distasteful with 
.
because that's case (a'). That leaves only 
-
?
square, it takes 
moves to go from the worst arrangement (no tasteful 
board.
would make the argument simpler, though. In that case, we can't keep going down in case (b).
![[asy]size(100); pathpen = linewidth(2.5);
void chessboard(int a, int b, pair P){
for(int i = 0; i < a; ++i) for(int j = 0; j < b; ++j)
if((i+j) % 2 == 1) fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6));
D(P--P+(a,0)--P+(a,b)--P+(0,b)--cycle);
}
chessboard(2,2,(2.5,0));fill(unitsquare,rgb(0.6,0.6,0.6));fill(shift(1,1)*unitsquare,rgb(0.6,0.6,0.6)); chessboard(4,3,(6,0)); chessboard(4,3,(11,0)); MP("\mathrm{Distasteful\ tilings}",(2.25,3),fontsize(12));
/* draw lines */
D((0,0)--(2,0)--(2,2)--(0,2)--cycle); D((1,0)--(1,2)); D((2.5,1)--(4.5,1)); D((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1)); D((8,2)--(8,3)); D((12,0)--(12,2)--(11,2)--(13,2)); D((13,1)--(15,1)--(14,1)--(14,3)); D((13,0)--(13,3));[/asy]](http://latex.artofproblemsolving.com/d/9/6/d96b48633a5257fd25edaca85aa4cd582d8aad02.png)
a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully.
(with 
nonempty tasteful tilings).
(WLOG lower left with only upper and right neighbors and facing outside the interior of the polygon) bordered above and to the right by 
respectively, so either 
or 
is a domino; also let 
denote the square to the upper right of 
cannot be tiled (with dominoes), then every tiling of 
must have at least two tasteful tilings (by removing 
(with an even number of squares and including the boundary of the actual polygon) composed of two disjoint tilings ("1 mod 2 shifts" of each other) and an interior 
composed of fixed dominoes (i.e. the same for the two tilings).
and the corner 
faces outside the polygon, we need 
in order to reach 
in this order along the (boundary) path. We can show that 
cannot be tiled (this is of course also true if 
(in regards to the distasteful configurations) that 
(which is a vertical white-black domino, viewed from bottom to top) to the tasteful tiling of 
itself (since the only distasteful configuration 
,
are not in 
such that 
for the center 
of some square in 
until it "leaves" (it loops eventually, so it must) for the first time at 
on the left and 
on the right. Clearly 
,
and 
,
,
,
,
(i.e. for the corner at 
)
.
,
is not a fixed interior domino, or else 
are all dominoes in the same tiling (one of the two tasteful ones mentioned in (*)), whence tiling instead by 
,
yields that 
can be tiled, contradicting (**) (as before, for the corner at 
is not a fixed interior domino.
lies in 
(by symmetry about 
is a fixed interior domino for 
and 
is in the interior (for 
)
is a domino, which forces 
(if 
;
,
must belong to a domino (since it's in the interior). Yet 
cannot be a domino since 
is already a fixed (interior) domino, so 
(as long as 
;
,
(in this order) along the boundary, 
,
must appear on the boundary in this order. Thus 
and 
are dominoes in the same tasteful tiling of 
)
lie on a diagonal and therefore all have the same color.
from one vertex to the next, the color of the square on the left remains the same).
rectangle can be tiled by dominoes if and only if 
is even or 
is even.
are both even
.
.
or 
.
denote the chessboard polygon which can be tiled by dominoes.
of the polygon, and WLOG is it black (other case similar). Then we have two cases:
square with two vertical dominoes to arrive in the first case. So this additional domino will not cause any distasteful tilings.
,
be any such cycle and let 
enclosed by 
.
,
,
,
,
is covered by a vertical domino. Then 
must be covered by a horizontal domino, to avoid a distasteful tiling. Then if 
is in 
lie inside 
so the line segment 
joining 
has slope 
instead).
squares are all counterclockwise.
