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 a set of integers with the following properties:
.
and 
, then 
.
, 
is composite, then all positive divisors of are in . contains all positive integers.
and 
be positive integers with 
. Let 
be a polynomial of degree with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers 
such that the polynomial 
divides , the product 
is zero. Prove that has a nonreal root.
such that 
is an integer for every positive integer 
has how many digits?
positive integers have no single-digit prime factors?
and 
be real numbers with 
and 
such that 
Find the least possible value of 
and foci at 
. Another ellipse is centered at 
with semi-major axis 
and foci 
. We claim that minimizing 
means that these ellipses are tangent - indeed, assume they are not. Then, we can reduce the major axis of the first extremely slightly, by some 
, such that the circles still intersect, but decreased, a contradiction.
and 
, with foci 
and 
are tangent at 
. is either 
or 
.![[asy]
import geometry;
ellipse ellipse(point A, point B, point C) {
return ellipse(A, B, (abs(A-C)+abs(B-C))/2);
}
size(9cm);
point E1=(-4,0),E2=(4,0),F2=(10,12),F1=(10,5.6),T=(6,4);
draw(ellipse(E1,E2,T));
draw(ellipse(F1,F2,T));
draw(E1--F1);
draw(E2--F2);
label("$E_1$",E1,S);
label("$E_2$",E2,S);
label("$F_1$",F1,E);
label("$F_2$",F2,E);
label("$T$",T,E);
dot(E1);
dot(E2);
dot(F1);
dot(F2);
dot(T);
[/asy]](http://latex.artofproblemsolving.com/f/d/6/fd6a014a5e43bc44c63e9142cc8a9d576b2e0c34.png)
, and we show that if 
(given that 
lies on both line segments; otherwise look at ), then the ellipses are tangent. Clearly, there is at most one ellipse with foci at 
and 
that is tangent to a given ellipse, so it suffices to prove that is the one.
on both ellipses. This means that![\[E_1Q+E_2Q=E_1P+E_2P\]](http://latex.artofproblemsolving.com/b/9/b/b9b5fad352f09fefbd6e36610bd6ff0ce7042de0.png)
![\[F_1Q+F_2Q=F_1P+F_2P\]](http://latex.artofproblemsolving.com/a/7/c/a7c463802afb25c362fed3dae081c82364ae0dff.png)
or upon addition![\[E_1Q+E_2Q+F_1Q+F_2Q=E_1P+E_2P+E_1P+E_2Q\tag{1}\]](http://latex.artofproblemsolving.com/f/d/6/fd672be93623e38dd5b75904cbbd559c80f66e38.png)
However, by the triangle inequality in 
and 
, we get![\[E_1Q+F_1Q\geq E_1F_1=E_1P+PF_1\]](http://latex.artofproblemsolving.com/d/9/2/d92c781baa8ccc1e6722f6396d097fd896727870.png)
![\[E_2Q+F_2Q\geq E_2F_2=E_2P+PF_2\]](http://latex.artofproblemsolving.com/4/d/d/4ddab415a5047b636c69ac7ea7f99e3d3060a1ef.png)
with both equalities impossible, as that means 
, impossible. Thus, we know that![\[E_1Q+E_2Q+F_1Q+F_2Q>E_1P+E_2P+E_1P+E_2Q\]](http://latex.artofproblemsolving.com/0/1/1/0112a38638bd43e79ddd0af815ff6bb358797c29.png)
a contradiction to (1).
with 
, and 
with 
. We can compute this point to be 
. Thus, the semi-major axis, and , are just half of the sums of the distances from the foci. In particular,
Thus, we get the minimum value of 
.
and 
and the second ellipse has foci 
and 
. Now consider the point 
. We have that 
and since 
and 
, we have that 
, where equality occurs when is the intersection of 
and 
. Then 
, and the answer is 
.
and the foci of the second ellipse are 
. By the definition of an ellipse, we have:
So
Equality can occur, just draw a picture or solve for the intersection of the lines if you aren't convinced yet.
and 
are the foci of the first ellipse and and are the foci of the second ellipse, then the lines 
and 
intersect at the tangency point. This can be proven intuitively since AM-GM equality gives that 
minimizes any sum in general. Therefore we want them to be as small as possible while not too large, giving the assertion. and 
which gives the tangency point 
by basic ellipse properties and coordinates. We then plug this point back into each equation to solve for and respectively using the substitution 
and 
. Miraculously, solving each quadratic gives 
and 
, or 
and 
, which means 
. 
is centered at the origin with a major axis of 
and minor axis 
, while 
is centered at 
with a major axis of 
and minor axis of 
. Now, let 
denote the foci of and 
denote the foci of , respectively. Let 
be the intersection of and . Spamming the definition of an ellipse gives that 
. This problem is boiled down to finding the minimum value of 
.
and are collinear, and 
are collinear. The rest is just simple computation:
is that of an ellipse centered at 
with foci at 
and 
. Similarly, the equation 
is that of an ellipse centered at with foci at 
and 
. Note that from the definition of an ellipse is the sum the distances from any point on the ellipse to the two foci (
is defined similarly). We know that 
lies on the two ellipses. Thus, we want to minimize the sum of the distances from to the two pairs of foci, which have coordinates , , , 
. In other words, we want to minimize the sum of the distances from to those four points. is the intersection of the segments connecting and , and (provable using triangle inequality). We can easily solve for 
, so 
(I skipped over some distance formula computation).
is the sum of 
from the four foci. The four foci are literally just 
. But minimizing 
is minimizing 
and 
and in fact these minima can happen at the same time at 
. Basic system of equations extracts 
and 
, and the foci for the second ellipse are 
and 
. For a pair 
to exist, the graphs must intersect. Let them intersect at 
. Then 
and 
. 
. To minimize this expression, we must have circles with centers at the 4 foci points intersect at the point . We must minimize the sum of all the radii of these four circles. Consider the circles 
and 
at centers and respectively. Clearly 
. Now consider circles 
and 
at centers and respectively. Similarly, 
. Both of these inequalities are equal when the two circles are tangent, and the pairs of circles are tangent at the same point, giving 
.
:
.
and semimajor axis 
and semimajor axis 
.
,
and 
.
.
The lines between foci intersect at (14,15/2) and so that is the minimum => 16+7 = 23