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 an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
is the largest positive integer that divides both 
and 
. Integers and are coprime if 
.
, which doesn't have right angle, let 
be its circumcenter. Circle 
intersects 
and 
at 
and 
for the second time, respectively. Similarly, circle 
intersects 
and 
at 
and 
, and circle 
intersects 
and 
at 
and 
for the second time, respectively. Let 
and 
be circumcenters of triangle 
and 
, respectively. Prove that 
are collinear.
be an acute triangle and 
its orthocenter. Let 
be a point on the parallel through 
to such that 
. Define 
and 
as points on the parallels through 
to and through 
to similarly. If 
are positioned around the sides of as in the given configuration, prove that are collinear.
has been partitioned into disjoint pairs 
(
) so that for all 
, 
equals or 
. Prove that the sum ![\[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \]](http://latex.artofproblemsolving.com/0/4/4/04422a7d2266b92a6b3f215b698f078cda85091f.png)
ends in the digit 
.
or we can just switch 
and 
.
.
is odd. is odd as well. is the sum of 
's or 's and is odd, we know there must be an even number of 's. Let 
be the number of 's.
as desired. QED.
and 
. For 
, we say that 
links onto if if 
or 
. links onto if and only if links onto , there are an even number of ordered pairs of integers 
such that pair 
links onto pair .
, there are an odd number of integers between and , so the number of pairs that link onto is odd. Thus, the total number of ordered pairs of integers such that pair links onto pair is congruent to![\[ \sum_{i=1}^k(\text{number of pairs that link onto pair \textit{i}}) \equiv \sum_{i=1}^k 1\equiv k \bmod 2. \]](http://latex.artofproblemsolving.com/d/3/2/d329c536fdd045d921eb518e92a266b261a917f2.png)
Therefore, since we already know that this number is even, we find that 
is even, and the value of the given sum is![\[ 6k+(999-k) = 999+5k\equiv 9\bmod10. \]](http://latex.artofproblemsolving.com/5/6/7/567b4b4748515f02829e222347cb938debad384f.png)
modulo 
part. Why can we add the numbers like that? (I don't really get the 
part).
implying the desired sum is congruent to 
Combining gives the result.
comes 'even-even' or 'odd-odd'. We claim that the number of 's must be even. If odd then we get greater than odd or even number, this is a contradiction because 
have even and odd. So
and 
For 
note that 
since absolute value doesn't matter in 
So our sum is just 
For 
just notice that its always 
for all of 
so our answer is just 
Using CRT, we get an answer of 
for all integers 
.
and 
, so by CRT we have that the sum is 
, as desired. 
the sum of 999 terms has a remainder of 1 when divided by 2 similarly we see that 
so the sum of 999 terms has a remainder of 4 when divided by 5 so by CRT we see that the sum has a remainder of 9 when divided by 10 which implies that the units digit is 9
. Note that 
. Also note that has the same parity as 
which clearly odd, so 
. Hence the remainder of 
is unique by the Chinese Remainder Theorem, and it is easy to check that it is indeed.
the sum of 999 terms has a remainder of 1 when divided by 2 similarly we see that so the sum of 999 terms has a remainder of 4 when divided by 5 so by CRT we see that the sum has a remainder of 9 when divided by 10 which implies that the units digit is 9
???
and differ by 
, the ending digit is whenever there is an odd number of such that 
and is 
otherwise. There are odds and evens so there can't be an even number of such that , because that would imply an even number of odds and an even number of evens. Therefore, the ending digit must be .
has been partitioned into disjoint pairs () so that for all , equals or . Prove that the sum ends in the digit .
, so it remains to show that it’s 
.
for any integers and , this means![\[|a_1-b_1|+|a_2-b_2|+\dots+|a_{999}-b_{999}|\equiv 1+2+\dots + 1998\equiv 1\pmod 2\]](http://latex.artofproblemsolving.com/f/4/2/f42a811a8eb834c662a16847d8f14d3074516c22.png)
and we are done. 
for each .![\[\sum a_i + \sum b_i = 999 \cdot 1999,\]](http://latex.artofproblemsolving.com/c/5/c/c5cfa60d865257af1f2074885e56c96b03777da1.png)
is odd. Thus, the number of 's, denote as 
, is even. We have![\[\sum a_i - \sum b_1 = 6 \cdot 2k + (999-2k) = 10k+999 \equiv 9 \pmod{10}. \ \square\]](http://latex.artofproblemsolving.com/e/f/d/efd5f612eb6c209b6e41929f00282ac3ca0737de.png)
or . Say for of these pairs. The sum would then be equal to 
. Note that a pair with difference has same parity and a pair with difference does not. All pairs of difference would give us an even number of even numbers and an even number of odd numbers. If was even, that suggests there are even numbers of odd numbers and even numbers of even numbers. However, since there is a total of even numbers and odd numbers, this cannot be true. So thus, is odd, and the sum would end with digit .
such that 
.
six-pairs 
such that 
. Then, there are 
one-pairs such that . The given expression evaluates to 
For this to be 
, it suffices to show that is even.
, in order to make successful pairings, we must have either exactly one of the six-pairs 
or one of the following sets of three six-pairs 
is even, as desired 
is 
, the sum is clearly . It suffices to show that the sum is odd. Clearly the absolute value when taking the parity, so the sum has the same parity as . Let 
. We have 
is odd, so 
has the opposite parity of 
, implying that is odd, as desired.
pairs that sum to 
and 
pairs that sum to 
Then 
Meanwhile, if WLOG for all 
then our sum equals 
so by CRT it ends in the digit 
so we are done. QED
(mod )
(mod )
(mod )
(mod )
(mod ) (mod 
) as desired.
.
as the number of pairs where 
,
as the number of pairs where 
.
.
,
and 
are positive integers, where the middle numbers weren't any 
or 
.
and 
from the lemma, we know that
.
.
)
.
.
,
.
,
is odd.
for all 
since 
for all integers 
is odd, or equivalently, 
and 
have different parities. Assume otherwise. This implies 
is even. However,![\[\displaystyle \sum_{i=1}^{999} a_i + \displaystyle \sum_{i=1}^{999} b_i = \displaystyle \sum_{i=1}^{1998} i = 999 \times 1999\]](http://latex.artofproblemsolving.com/b/5/f/b5f1d2f0aec740f64bf91caa0089ff75c0ebeffe.png)
is odd. Since the only possible values 
is even. Let the number of integers 
We wish to show that 
which is true since we have already shown that 
.
.
such that ![\[\sum_{i=1}^{999} |a_i-b_i| \equiv k + (999-k)\cdot 6 \equiv 4-5k\pmod {10}.\]](http://latex.artofproblemsolving.com/b/9/0/b90e9dfcff5c8390e8e5554a9561e2e29d3d399c.png)
However, since 
is even.
evens, hence 
as desired. 
![\[ S = \sum |a_i-b_i| \equiv \sum (a_i+b_i) = 1 + 2 + \dots + 1998 \equiv 1 \pmod 2. \]](http://latex.artofproblemsolving.com/2/b/2/2b2d29e81ec5446574365666892450113fcce7a5.png)
Modulo ![\[ S = \sum |a_i-b_i| = 1 \cdot 999 \equiv 4 \pmod 5. \]](http://latex.artofproblemsolving.com/3/5/6/3569ffd364e1e1e1b1e457ab0cee1d62330f8982.png)
So 
.
for all integers 
regardless of whether 
.
.
equals 
Notice that 
and we are done. 
and we have 
,
![\[(999-x)+6x = 999+5x,\]](http://latex.artofproblemsolving.com/5/f/4/5f4a6b12990e8b7e3635bd8c5e38bb58cb379b5e.png)
which has units digit either 
,![\[\lvert a_i - b_i \rvert \equiv a_i - b_i \equiv a_i + b_i \mod 2.\]](http://latex.artofproblemsolving.com/d/e/1/de1a1a014ccc2723f786957c0c1b60f128ba099a.png)
we need to prove that this is 
note that 
has the same parity as 
so we need to prove 
so 
hence proved 
![\[ \displaystyle\sum |a_i-b_i| = \displaystyle\sum a_i - b_i \equiv \displaystyle\sum a_i + b_i = \dfrac{1998\cdot 1999}{2} \equiv 1 \pmod{2}. \]](http://latex.artofproblemsolving.com/c/a/c/cac30537422175850423b5e4cc253d9f555352b3.png)
![\[ \displaystyle\sum a_i - b_i \equiv \displaystyle\sum 1 = 999 \equiv 4 \pmod{5}. \]](http://latex.artofproblemsolving.com/6/f/4/6f4bbef6189b53a64a92b4a7057f57f4776631db.png)
.
,
.
.
,
,
Note that 
,
Now from the Chinese Remainder Theorem, 
.
,
,
.
,
.
.
The only thing left is to prove that 
However, 
.
modulo 5. But that is trivial since there are 999 terms in the sum, each of which are 1 modulo 5.
is odd.
,