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).
, 
, 
, the inequality![\[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc}
\]](http://latex.artofproblemsolving.com/f/a/9/fa964e29de91f0e2f3cb32e335a52d4521bc38fd.png)
, , be positive real numbers such that 
. Prove that![\[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}.
\]](http://latex.artofproblemsolving.com/b/2/9/b295b0b2a765e41a639657e13b724c60323fa208.png)
be a triangle. Let 
be the feet of the altitudes from 
respectively. Let 
be the projections of onto line 
. Show that 
.
MathJuice bottles and Bob starts with 
MathJuice bottles. An unfair coin is then flipped, with probability 
of landing heads. If the coin lands heads, Alice gives Bob a bottle; otherwise, Bob gives Alice a bottle. This process repeats until someone runs out of bottles.
rest on a plane. A sphere with radius 
is tangent to all of them, but does not intersect nor lie on the plane. A sphere with radius 
lies on the plane, and is tangent to all three spheres with radius 
Compute the shortest distance between the sphere with radius and the sphere with radius 
defined on all real numbers and taking real values such that ![\[f(f(y)) + f(x - y) = f(xf(y) - x),\]](http://latex.artofproblemsolving.com/f/5/2/f528fc4047e21247663d4993913eadf346315be7.png)
for all real numbers 
defined on all real numbers and taking real values such that for all real numbers 
be the assertion 
. We get:
.
the given relation.
such that 
gives 
the initial gives: 
the last one gives 
or 
then 
for some 
and 
for all the other and 
such that 
Then the initial gives: 
to the last one we get 
so 
and 
for some and 
for all the other.
then 
But 
so 
this is a contradiction.
for all 
defined on all real numbers and taking real values such that for all real numbers be the assertion 
isn't obviously a solution.
such that 
there exists a 
such that 
for all 
such that 
, we have 
which indeed is a solution.
.
. denote the assertion 
.
.
for some 
. Consider 
.
. is even. Consider 
. Consider the two equations : 
and 
.
.
denote the assertion 
gives 
implying the desired.
so 
so 
.
that works is 
so 
so 
Let 
so 
. Hence, 
Thus, is even.
so 
with the fact that 
for all real 
. and 
which give, respectively, 
Rearranging and noting that is even gives 
in 
so 
gives . Taking 
implies 
. Letting 
in the previous equation yields 
. Putting this back into the equation it was derived from, we see or that is even. From the original FE, we have 
Swapping the sign on and noting evenness, 
. Hence for all 
, which shows the function is constant. Hence, there is only one solution: 
.
denote the assertion : 
.
: 
.
: 
.
, which means that when 
, thus 
.: 
: 
, thus 
, where the first equality comes from knowing that 
.
to be and we get that 
.
to be solution to this problem.
., we have that .
.
, then we have that 
.
, then we must have that 
, then we must have that 
into the upper relation to get that 
, then we must have that 
. to get that 
.. and then we just set 
, to get that 
..
. denote the given assertion.
.
.
. is even.
.
(Using is even.)
(Equating 
and 
.)
.
be the assertion.
.
.
, then comparing with 
gives 
, which implies that is constant since 
. So only solution is 
in this case.
, then we have 
. From 
we get 
, but plugging 
and 
gives a contradiction. Thus our only solution is 
, which clearly fits.
be the assertion..
.
in above, we get 
.
. in above, we get 
.
. in above, we get .
which clearly fits.
be the assertion.
Let 
, we get 
is the only function which works.
denote the given assertion. gives Further, 
gives 
and finally 
gives Then implies is even. Subtracting from implies 
And it's easy to verify the identity zero function works.
, which works. denote the given assertion.
.
.
, so 
.
, so .
.
.
.
.
, which implies is constant, so 
.
, which works.
is surjective, for all 
. 
, which clearly fits. as 
. Note that if 
is a constant function then . Assume that is a non-constant function.
gives 
. such that 
. Then, 
gives 
. But 
. So, 
. Now, 
gives 
. We conclude that either 
. Note that 
gives 
. But 
gives . Take such that . We have 
. Then, 
gives 
. It follows that and 
. There is no restriction on , so we can take any real and . So, , contradicting 
.. 
denote the assertion., . Proof: Choose 
.
such that 
, since 
for all is not a solution. Hence 
and so . Now suppose exists such that 
,
so 
. So .
, so .
. So 
at the same time and since 
, 
which is not okay such that . So 
for all .
so is constant. Now checking only works.
denote the assertion. gives so .
gives 
.
gives 
so plugging gives 
but so 
so . Plugging this back into gives .
gives 
so 
.. Assume FTSOC that there exists some 
in the image of . Then, 
so 
is also in the image of . However, this means that 
. Since 
, this is a contradiction, as desired.
, which obviously works. We claim that is constant. Let denote the assertion.
But since , from () and (), we have 
, thus ![\[
f(f(x))+f(x-f(0))=0.
\]](http://latex.artofproblemsolving.com/f/e/1/fe116fa8b2fc4b7a83c53181eb3444dd539c1770.png)
Put in above, we have 
. Thus from ()., and 
gives . This implies is constant. because for any real 
, Letting 
to get ![\[
f(t) = f(x+y)=f(x-y)=f(0).
\]](http://latex.artofproblemsolving.com/0/9/c/09c59d3592cfb4b0e2b0db0dfa13e150e3bc5f0d.png)
And so we're done.
.
min
.
is 
we obtain 
Substituting 
proving our desired claim.
we obtain 
and 
Substituting 
be assertion to 
at 
and therefore 
.
and 
.
this obviously works.
,
?
Finally,
We see that 
gives us 