AoPS Academy
, 
,
, 
, why 
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.. It is usually regarded that 
, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.? The issue here is that 
isn't even rigorously defined in this expression. What exactly do we mean by ? Unless the example in question is put in context in a formal manner, then we say that is meaningless.? Suppose that 
. Then we would have 
, absurd. A more rigorous treatment of the idea is that 
does not exist in the first place, although you will see why in a calculus course. So the point is that is undefined.
? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
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 is an indeterminate form.
, 
or , obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](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 
. On the other hand, 
is undefined because dividing by is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context. is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](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 it is perfectly OK to say that,![\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 is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](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 is allowed here! In fact, we have![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and are left undefined. We also have
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 , by defining them as![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by also being allowed with the same indeterminate forms (in addition to some other ones).
,
,
be real numbers different from 0 for which 
+ 
+ 
= 0 holds+)(+)(+)≠ 0
= + + and 
= 
+ 
+ 
. Prove that numbers and are both positive or both negative.
be a triangle. Take points 
, 
, 
on the perpendicular bisectors of 
, 
, 
respectively. Show that the lines through 
, 
, 
perpendicular to 
, 
, 
respectively are concurrent.
eggs, each with a hardness value 
, where 
is a permutation of the set 
. He is throwing the eggs from an 
-floor building.
-th egg is dropped from the 
-th floor, its new hardness becomes![\[
\left\lfloor \frac{k_i}{j} \right\rfloor.
\]](http://latex.artofproblemsolving.com/7/b/3/7b3366510115e9504dac41b46a230522d6a09e68.png)
If 
, then the egg breaks and cannot be used again. and can Arthur always determine the correct ordering of the eggs according to their initial hardness values?
-gon. A hunter picks three vertices of the polygon and he lays a trap which covers the interior and the boundary of the triangular region determined by them. Determine the minimum number of times he needs to do this to guarantee that the rabbit will be trapped.
that appears after the decimal point?
years ago! for solving it, extend 
to intersect the circle with center 
...
is the closer tangent to 
, which I think the original problem states (it doesn't matter though).
, 
, and 
be the reflection of over . Clearly 
. But 
, so it suffices to show 
(equivalent since 
are the midpoints of 
respectively), which is just a simple angle chase.
to get a geometric relation. perpendicular to intersect at point 
.
) yields 
are concyclic, which leads to 
the other intersection point of two circles,let 
be the intersection point of 
and and let 
be the circumcenter of 
.Now,let 
be the intersection of the perpendicular from to 
and the perpendicular from to 
.Now,we have that 
is similar to 
so we get 
is similar to 
.Now,from an easy angle chase and using the previos similarity we get 
is similar to 
and from this we get 
so 
and that is it.
and let 
cut 
for a second time at 
, respectively. Let 
be the projection of onto 
Note that 
is the external center of homothety that maps 
It is clear that this homothety takes 
and 
, so 
Hence, 
is a diameter of 
are collinear. Furthermore, note that 
, where the last step follows from Power of a Point. Therefore, 
are concyclic, so 
, where the angles are directed. Hence, 
are concyclic, so 
, as desired. 
be the points of intersection between circles with circumcenters 
with closer to line than . Let 
, and 
denote the reflections of 
over , respectively. Also let denote the midpoint of line . We then make the following claim:
.
, that ![\[\measuredangle A_1OC = \measuredangle BO_2C = \measuredangle AMP, \]](http://latex.artofproblemsolving.com/d/c/e/dce3a79a273352aae2d04c26e2c22d320f8328d8.png)
as 
is also cyclic. 
are collinear: are collinear.
. It then suffices to equivalently show . Indeed, we have that ![\[\measuredangle AA_1C = \measuredangle MAP, \]](http://latex.artofproblemsolving.com/a/2/f/a2f53ca1321b99a2e7c1bc9e2d025cc90c002ab3.png)
and by the above claim/lemma, we obtain the desired similarlity. are collinear, in particular 
, as desired. 
the symmetric of wrt 
; be the ex-similicenter ; the intersection of and 
;
the antipodes of 
; the intersection of 
and ; the intersection of 
and we will show that :
i.e. 
is parallelogram.
but 
hence 
are collinear.
then 
.
be . A common tangent touches the circles at 
respectively. Let the perpendicular from to the line meet at . Prove that 
.
, the problem becomes equivalent to a statement regarding the which is quite neat: In triangle with orthocentre and point on arc 
such that 
(hence 
bisects ), the line through parallel to , the line 
, and the perpendicular bisector of line concur. Let 
be points on 
such that 
is a parallelogram. Then line 
passes through the midpoint of 
, i.e, the centre of the circumcircle of 
. Also, 
by projecting all on line , hence 
or 
, hence is the perpendicular bisector of , and proving lies on it was all we needed.
where 
is intersection of 
.
,
is the 
-
after inversion.
be a triangle and 
a
to 
.
such that, 
.
be a point such that 
is rectangle. Prove, 
.
