,
,
,
,
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 
km long. From any point on her path she can see exactly 
has how many digits?
and 
intersect at the point 
.
?
,
.
are chosen from 0 to 1. What is the probability that their positive difference is more than 
?
?
children respectively. If a random child from any of the families is chosen, what is the probability that the child has 
siblings? Express your answer as a common fraction.[/quote]
,
people, exactly 
such that 
.
,
.
.
has 
,
,
.
.
and the distance between 
and 
is 
Point 
such that 
What is 
and 
is the circumcenter of 
If 
what is 
,![\[ \textbf{(A)}\ 24 \qquad
\textbf{(B)}\ 30 \qquad
\textbf{(C)}\ 46 \qquad
\textbf{(D)}\ 66 \qquad
\textbf{(E)}\ 74
\]](http://latex.artofproblemsolving.com/8/2/f/82ffc15072ca99f6cd1abc312781eb4827817356.png)
Y by Davi-8191, OlympusHero, samrocksnature, HWenslawski, megarnie, Adventure10, Mango247
Let 
be a triangle with 
. Points 
and 
lie on sides 
and 
, respectively, such that 
and 
. Segments 
and 
meet at 
. Determine whether or not it is possible for segments , , 
, 
, 
, 
to all have integer lengths.
be a triangle with 
. Points 
and 
lie on sides 
and 
, respectively, such that 
and 
. Segments 
and 
meet at 
. Determine whether or not it is possible for segments , , 
, 
, 
, 
to all have integer lengths.
.
.
,
.
.
,
.
and 
.
is rational, so 
is rational. We are given that 
is an integer, so 
is rational, so 
is rational. Therefore 
and 
are rational. By similar reasoning using 
,
is rational. However, if we use the sine addition/subtraction formula, we get that 
must be rational, which implies 
is rational, which is clearly false. We have a contradiction, so at least one of those lengths must be non-integral.
unless you're assuming that 
are relatively prime. Every right triangle has legs that can be expressed as 
and 
and a hypotenuse of 
for some 
:![\[BI^2+CI^2-2\cdot BI\cdot CI\cos{135^\circ} = BC^2 = AB^2 + AC^2,\]](http://latex.artofproblemsolving.com/4/0/9/409c4d82b56eaf7521fc709dec78ba6240f98962.png)
and we're done.
and 
are rational using more similar triangles.
,
,
,
are rational.
is rational, contradiction.
is a rational thing that I forgot times 
.
is rational; hence BC = a must be. From there, triangles ABD and ACE produce 
and 
as right triangles, with 
.
and 
as squares of rational numbers.
or 
.
and 
cannot both be rational.
.
.
now gives 
a contradiction.
.
,
.
.
.
bisects 
,
.
,![\[AI^2+AB^2-2AB\cdot AI\cos{45}=BI^2.\]](http://latex.artofproblemsolving.com/f/5/0/f509c93c449d42af647b6e98e8cd1fe9f2d5e413.png)
,
is irrational so 
where 
(
)
and 
are as usual length of the sides 
and thus if both 
has to be rational and so 
has to be rational and so 
in the form 
where 
are coprime numbers with different parity .As , 
has to be a perfect square dividing the thing by 
will give us another perfect squarewriting in terms of 
we get 
is perfect square and so 
is a perfect square but as we have 
must be a non-integer. 
.![\[\angle BIC = \angle BAC+ \angle ACE + \angle ABD = 135^\circ.\]](http://latex.artofproblemsolving.com/b/b/1/bb1b240a3523caf4a2b2e892c8b9e60ceefe49c7.png)
![\[AB^2+AC^2=BC^2 = BI^2+CI^2+BI \cdot CI \cdot \sqrt{2},\]](http://latex.artofproblemsolving.com/2/c/e/2cea88d419014061133fd214cfa18ef80a1761b5.png)
![[asy]
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dot((0,8),dotstyle);
label("$C$", (0.05455029479432724,8.132494004361794), NE * labelscalefactor);
dot((0,3),linewidth(4pt) + dotstyle);
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dot((2.6666666666666665,0),linewidth(4pt) + dotstyle);
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dot((2,2),linewidth(4pt) + dotstyle);
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[/asy]](http://latex.artofproblemsolving.com/9/a/7/9a7261659c8db0bb4186e960bdfe738b2d76dc55.png)
Law of Cosines on 
gives 
If all of these segments have integer length, the left-hand side would be irrational, while the right-hand side is rational. Therefore, it is impossible for all of these segments to have integer length.
,
.
.
and 
by AA similarity. We are then able to derive the following equalities:
Thus, 
,
is rational. Analogously, 
,
,
,
is also rational. To finish, note that ![$[ABC] = \frac{bc}2$](http://latex.artofproblemsolving.com/5/d/8/5d81af12c311a4126a5272d944df341a443345e6.png)
,
we have the inradius 
is rational. However, 
,
= 
= 
= 
= 
- 
= 
,
It is well-known that 
Therefore by the Sine Law and Half Angle formula 
Therefore if 
we have the square root of an integer minus an irrational, which clearly cannot be an integer.
are integers. Let 
Then for positive integers 
with 
WLOG 
Then 
The first one yields 
but the second one gives 
a contradiction. Hence the answer is no.
and 
and 
are both integers, so 
is also an integer.
so it follows that 
from which we get 
Therefore, by the Law of Cosines, we have
From before, we know that 
is an integer, and we also know that 
and 
must be integers, so it follows that the right-hand side of this equation is a rational number. However, the left-hand side is an irrational number, contradiction.
and 
to all have integer lengths. 
