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).
, let 
denote the greatest prime factor of 
. A strange pair is a pair of distinct primes 
and 
such that there is no integer for which 
.
?
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 
the set 
can be partitioned into triples in such a way that the numbers from each triple are the lengths of the sides of some obtuse triangle.
the set can be partitioned into triples in such a way that the numbers from each triple are the lengths of the sides of some obtuse triangle.
. The triples will have the form 
.
, and in our case, it means 
, which is equivalent to 
, but 
, so 
.
, 
.
(like for example 
) we have 
, so the triangle inequality is not satisfied. Trying to get the obtuse condition, you lost the ball out of your eyesight; it happens 
and 
if they're not used later?
, (all triplets used here assume 
), define 
and 
, then the problem is to divide 
into triplets, each ascending, where 
. good if 
. Note that if is good, so is 
, for any integral k.
can be partitioned into 
good triplets, the 
triplet having first element 
. For 
, it is clearly true since 
is good. Assume this is true for 
. Now let 
. Calculation shows that 
is good, as are 
. In total, these comprise all the elements of , except for 
and 
. To construct the missing triplets, take the solution triplets 
for 
, and create the triplets 
, all of which are good. The partition is now complete.
is obtuse for positive integers 
with 
.
which becomes 
which is true.
construct these triplets.
and 
for 
(these are all of the form as in the lemma which means they are sides of obtuse triangles) and the triplet 
which are also sides of an obtuse triangle for 
which is obvious.
are in different triples; let 
denote this statement.
is an obtuse triple, then so is 
for any 
.
is obviously true.
for all 
. and use the observation to get a construction for 
. Now consider the following table: ![\[ \left[ \begin{array}{cccc|cccc} 2 & 3 & \dots & n+1 & n+2 & n+3 & \dots & 2n+1 \\ \hline \multicolumn{4}{c|}{\text{Induction}} & 4n+2 & 4n+3 & \dots & 5n+1 \\ \multicolumn{4}{c|}{\text{hypothesis}} & 5n+2 & 5n+3 & \dots & 6n+1 \end{array} \right] \]](http://latex.artofproblemsolving.com/4/4/d/44db90b7e91e8d3ec765b4ac82e6ad06fe52241b.png)
We claim all the column are obtuse. Indeed, they are obviously the sides of a triangle; now let 
and note that ![\[ k^2 < 8n^2 \implies (n+k)^2 + (4n+k)^2 < (5n+k)^2 \]](http://latex.artofproblemsolving.com/9/1/2/91287a31a4ff0c4166e3068004a2d212e6228679.png)
as desired. 
for all 
. and use the observation to get a construction for 
. Now consider the following table: ![\[ \left[ \begin{array}{cccc|cccc} 2 & 3 & \dots & n+1 & n+2 & n+3 & \dots & 2n \\ \hline \multicolumn{4}{c|}{\text{Induction}} & 4n+1 & 4n+2 & \dots & 5n-1 \\ \multicolumn{4}{c|}{\text{hypothesis}} & 5n & 5n+1 & \dots & 6n-2 \end{array} \right] \]](http://latex.artofproblemsolving.com/4/0/9/409e8a9f1f7dc226a78ce108c61e90af05c87f26.png)
We claim all the columns are obtuse again. Indeed, they are obviously the sides of a triangle; now let 
and note that ![\[ (k-2)^2 < 8n^2-12n+4 \implies (n+1+k)^2 + (4n+k)^2 < (5n+k-1)^2 \]](http://latex.artofproblemsolving.com/8/1/4/814c628976e3fb91a26778e522afe1d023425a4c.png)
as desired. , we obtain for all .
are the sides of an obtuse triangle for positive integers 
and 
.![\[(n+d)^2-n^2-(d+1)^2=2nd-2d-1\ge 2d^2-1>0,\]](http://latex.artofproblemsolving.com/2/0/0/200a303eb34daa5300c8c661e22bb20e0dab645d.png)
so the triangle is obtuse. . is even.![$[2n]\setminus\{3n/2,2n\}$](http://latex.artofproblemsolving.com/3/8/e/38e6127bb67d935d6c29f1f92adceb6e04fa3c8b.png)
into 
pairs with differences ranging from to . To do this, we pair up 
for 
, and 
for 
. Let the pair with difference be 
, so 
.
. By our lemma, the first triples work. It suffices to check that the last one works. The triangle inequality is clearly satisfied, and![\[(3n+1)^2-(n+1)^2-(5n/2+1)^2=7n^2/4-n-1>0\]](http://latex.artofproblemsolving.com/d/e/d/ded833adc333b898eb261bee28f417f891ea8bd6.png)
for 
, as desired. is odd.![$[2n]\setminus\{3n/2-1/2,2n\}$](http://latex.artofproblemsolving.com/4/c/c/4cccb029424c00c481926cd215cf46a22fefae0d.png)
into pairs with differences ranging from to . Pair up 
for 
and 
for 
. Let the pair with difference be , so .
. By our lemma, the first triples work. It suffices to check that the last one works. The triangle inequality is clearly satisfied, and![\[(3n+1)^2-(n+1)^2-(5n/2+1/2)^2=7n^2/4+3n/2-1/4>0\]](http://latex.artofproblemsolving.com/0/b/4/0b4b458fa4e8b636126508bb87320cbcd6221164.png)
for 
, as desired..
a nice triple if 
are sides of an obtuse triangle, i.e. 
and 
. We wish to show that 
can be partitioned into nice triples. Call such a set 
a nice set. We apply prefix induction (look at the variants of induction given here) on , i.e. we try to prove that if is nice, then so is 
and 
, in which case we will be done. The base case 
is trivial (Take as the required nice triple). So now assume that the result is true for some .
. Now, if is a nice triple, then so are 
and 
. The first triple is easily seen to be nice, while the second one is nice using the following- 
This gives that if is a nice subset, then so is 
and 
. Using this fact, we get that, as is nice, so 
and 
are also nice. Since the union of two disjoint nice sets is also nice, so we get that is a nice set., which can be thought of as a union of the three sets of triples given below- 
It is easy to see that all of these triples are distinct and nice (Too lazy to show the calculations 
). This gives that is also a nice set, as desired. Hence, done. is nice (Luckily the first thing I tried was induction; and when the normal induction didn't seem to work, it made sense to use prefix induction). However, proving that is also nice if is nice came out as a difficult task, which I wasn't able to do even after trying for nearly an hour and a half (which is why I finally turned to finding out a construction, which is easy and doable in around an hour or so 
). So overall, I think this problem was a really nice one (pun intended ), in its formation, as well as in the risk of getting stuck on it for a long time.
, inductively construct a sequence of numbers in the following manner: Write 
, and then repeat the process for 
, until we get to . For example, the sequence we construct for 
is: 
. Now, if our sequence is 
backwards, we split ![$[n+2,3n+1]$](http://latex.artofproblemsolving.com/b/8/b/b8b8df6a531695b07188cefcc548578515b5d735.png)
into the following groups: 
, and in a group of size 
, we pair its first element with the 
st element, etc.. In this way, we split the largest 
numbers into pairs. Now, pair the pair with the th smallest smaller element with , to create triplets. We claim that each triplet forms an obtuse triangle.th triplet as 
such that 
are in the th group formed by the 
. In our construction for 
, we have 
, meaning that if are in the th group, we have that 
, meaning triangle inequality is satisfied. Furthermore, we have 
, so 
, so the triangle is obtuse. So, this is a valid construction, as desired.
stupid if and 
form an obtuse triangle, so we wish to partition 
into stupid triples. We begin with the following key lemma. is stupid, then so is 
for any 
.
for , so combining these gives the desired. being trivial (the only option of , which works). We will add the additional condition that our stupid triples should "separate" 
, i.e. each of these numbers is present (obviously as the least element) in a different stupid triple.
is even, then form the stupid triples 
. Note that we are left with the numbers 
. Evidently, each of these triples satisfies the triangle inequality. Further, for any triple listed, we have
so all these triangles are in fact obtuse. On the other hand, by inductive hypothesis we can form 
stupid triples with least parts 
from the set 
, so by our lemma we should also be able to do this for , completing the inductive step.
is odd, then form the stupid triples 
. Note that we are left with the numbers 
. As before, each of these triples satisfies the triangle inequality, and also
hence these triples are actually stupid. By inductive hypothesis, we can form 
stupid triples with least parts 
from 
, so we can also do this for by the lemma. This again completes the inductive step., we are done.
would be 
.
case it seemed highly likely to me that some construction where we take as the smallest elements and throw together the other elements (at the very least, it's worth a try). The most obvious way to do this would be to take 
, and so on, which is what's suggested by the construction, but when gets bigger the difference of between the squares isn't enough. On the other hand, if we make the differences really big to the point where the triangle is obtuse if it exists, the triangle inequality might not be satisfied. To overcome these "extreme issues" we probably want to do something in the middle. Looking at numbers near the top, we would probably like to get 
all the way to 
for some suitable (approximate) choice of . It turns out that 
works nicely, since the triangle inequality should just barely hold at the lower end and the obtuse condition should also hold. The rest is just correctly choosing boundary conditions.
be triples of the following form:![\[(a-k+1,b-2k+1,b-k+1), (a-k+2,b-2k+2,b-k+2), \dots, (a,b-k, b)\]](http://latex.artofproblemsolving.com/b/f/f/bff483c135f2d72cac52c80547d745fd1134c4d9.png)
We impose the precondition that all of those triangles are non-degenerate and obtuse. Call an obtuse triple if , 
and 
. Then, if is obtuse then 
, 
and ![\[a^2+(b+k)^2=a^2+b^2+2kb+k^2 < c^2+2kb+k^2 < c^2+ 2kc + k^2 = (c+k)^2\]](http://latex.artofproblemsolving.com/f/5/c/f5c528bb6944b24be15561ee0fe1bf5b42b3e695.png)
so is also obtuse. Note that if and then ![\[(a-1)^2+(b-1)^2 = a^2+b^2-2(a+b)+1<a^2+b^2-2c+1<c^2-2c+1=(c-1)^2\]](http://latex.artofproblemsolving.com/6/f/3/6f310925d41cdfc8e244afceffbd0dc499905dc4.png)
so if is obtuse and 
then so is 
. Note that since 
for all , we have 
and we have 
, so 
are all obtuse. It also uses every number like 
.
in a way that separates 
then if you add to each 
and 
in the obtuse triples, they stay obtuse and take up 
. Together with we completed the partition. Thus, if the claim is true for then it is true for . implying 
we can simply use 
for the same effect and we will not go into detail.
the set can be partitioned into triples in such a way that the numbers from each triple are the lengths of the sides of some obtuse triangle and none of the triangles have two side lengths less than or equal to 
.
good if they are the length of the sides of an obtuse triangle. Notice that is good iff (WLOG 
) 
and 
. We want to split 
into good triples. is good with , then 
is also good for any positive integer . was good, then 
and 
. It suffices to show that 
and 
. The second inequality holds true obviously since 
and the first inequality holds true because and 
. 
is good, , and 
, then 
is also good. was good. Clearly![\[(a-1)^2 + (b-1)^2 = a^2 + b^2 - 2(a+b) + 2 < c^2 - 2(c + 1) + 2 < (c-1)^2\]](http://latex.artofproblemsolving.com/f/8/7/f8794ab642bf9d346cf6e1d2861b8db0701c4d2d.png)
and 
, so is also good. , if the statement is true, then the statement is true for also.
in the following way:![$[n+2, 4n + 1]$](http://latex.artofproblemsolving.com/d/a/a/daa50c469f77bcf419f313fb41a7a106d25b688c.png)
as![\[\{n + 2, 2n + 2, 3n + 2\}, \{n + 3, 2n + 3, 3n + 3\}, \ldots, \{2n+1, 3n+1, 4n+1\}\]](http://latex.artofproblemsolving.com/b/1/4/b146e495250e73a5c61b47d7da998cebeee5ec55.png)
Obviously no two of these sets intersect. To see they are good, notice that 
, and then applying claim 2 gives the desired result.
was matched with two other numbers from 
to 
in the original partition of 
. In our new partition, match it with those two same numbers except each added by 
. These two numbers will be each at least 
originally, so when added by , they will at least 
, so they cannot intersect in the interval ![$[n + 2, 4n + 1]$](http://latex.artofproblemsolving.com/5/b/e/5bed9ef0341b47df3d34b20c0a35d86b3ae4af76.png)
. Furthermore, each of these triplets are good by claim 1, and cannot intersect each other. Therefore this partition works, so the claim is proven. 
if the statement is true, then the statement is true for also.
in the following way:![$[n+2, 4n -2]$](http://latex.artofproblemsolving.com/2/c/9/2c967612b3fd277611846368c7c7d3e1282e92df.png)
as![\[\{n+2, 2n+1, 3n\}, \{n+3, 2n+2, 3n+1\}, \ldots, \{2n , 3n - 1, 4n - 2\}\]](http://latex.artofproblemsolving.com/a/7/b/a7b21dfcbd89993665470aaabe5e52baf27ac728.png)
Obviously no two of these sets intersect. To see they are good, notice 
, and then applying claim gives the desired result.
was matched with two other numbers from to in the original partition of 
. In our new partition, match it with those two same numbers except added by 
. These two numbers will be originally at least , so when added by , at least 
so they cannot intersect in the interval ![$[n + 2, 4n - 2]$](http://latex.artofproblemsolving.com/3/d/9/3d9656a1e130bb72eb75236659a1391133c24f47.png)
. Furthermore, each of these triplets are good by claim , and cannot intersect each other. Therefore, this partition works, so the claim is proven. 
and 
, by
respectively. to show it is true for all . Note it is true for 
. Now suppose it was true for everything less than for some 
. is even, the statement is true for 
, so using claim , it is true for . is odd, the statement is true for 
, so using claim 
, it is true for .
we have the base case of 
. We now induct on the hypothesis that for a given , the elements 
are paired with two elements each from 
. gives us one for .
. We can shift the latter two elements of the earlier triplets to get obtuse triangles on 
with 
.
with 
.
for 
which satisfies the triangle condition and are obtuse. 
.
with 
.
. 
![$[2,n+1]$](http://latex.artofproblemsolving.com/d/f/7/df781e082302f0605440fdb3fec91b036f44bb70.png)
<l and secondly that for the pair each of these distinct numbers is associated with, we want to choose a pair from where the pair is as spread out as possible and its size corresponds to the size of the small number it is associated with.. The main idea is we basically want to partition into intervals of length 
, where then we can choose pairs from that interval with the same difference. For example, for 
, we have the distinct numbers 
, and we partition ![$[5,10]$](http://latex.artofproblemsolving.com/1/5/9/15989f0fb397b7748f3a1d695a363efd5a78bc9e.png)
into ![$[5,6]$](http://latex.artofproblemsolving.com/f/f/4/ff4562c006c708088ea1694078c1b8fceaff6ecb.png)
and ![$[7,10]$](http://latex.artofproblemsolving.com/9/1/3/913d38d18c5fa61d3818835295cdf714319626cd.png)
which would give triples 
. with a dot, and continue recursively marking 
until is marked (e.g. for , we mark 
). Then label all these marks in ascending order, 
where 
. We have that for 
that 
or 
, except in case where 
.
, and let 
. For each 
we form:![\[ (s_i + j, n + 2s_i - 2 + j, n + s_{i+1} + s_i - 2 + j) \]](http://latex.artofproblemsolving.com/8/0/7/807d244f3d406772df409d2690cf85be2ad19e06.png)
We can verify that every single number is used distinctly in this construction of triples, then by some algebraic bash (
by the way) we can prove every triple satisfies the inequalities for obtuse triangles.
Prove that
,
,
is obtuse.
obtuse for all 
.
exactly once.
,
to be the pairing done in the method described below 
not in any pair. Bond it with the number 
to form a pair. Repeat until finishing step 
numbers remain (since 
is even and each time the number of unpaired numbers in 
decreases by 
denote the non-decreasing sequence of all positive pairwise differences between members bonded by 
when we set 
.
.
and are the largest in some 
by its floor function). Arrange all bonds induced by 
from smallest to largest differences.
and we are done. The numbers are in the same pair because of the preceding two lemmas.