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).
(where 
are the digits and 
) we define the \textit{permutation sum} as the value 
For example the permutation sum of 20253 is 
Let 
and 
be two fivedigit integers with the same permutation sum.
.
be non-negative real numbers such that 
. Prove that![\[a\sqrt{1+b^{3}}+b\sqrt{1+c^{3}}+c\sqrt{1+a^{3}}\le 5.\]](http://latex.artofproblemsolving.com/f/f/a/ffaeec844a6bdf10a96e8e9d298ef97ac839d713.png)
such that there exists a sequence of positive integers 
, 
,
, 
satisfying: ![\[a_{k+1}=\frac{a_k^2+1}{a_{k-1}+1}-1\]](http://latex.artofproblemsolving.com/7/b/4/7b4ab849cdc409906514e2ebb688802f17df41ea.png)
for every 
with 
.
has how many digits?
![$\sqrt[2]{2}$](http://latex.artofproblemsolving.com/1/8/6/1869ac018fb72a26a12ab14499cbf1b1c13b4a10.png)
, ![$\sqrt[3]{3}$](http://latex.artofproblemsolving.com/1/7/a/17a6bb5dc11696b8650b6d3a11e770f861f4c326.png)
, and ![$\sqrt[5]{5}$](http://latex.artofproblemsolving.com/a/b/9/ab90f2d42dc24065e2a64988fe76f22eb9359799.png)
.
. I couldn't solve this and was wondering if any of you can find the pattern
such that: 
be an odd prime number. Prove that there exist sequence such that: 
)=(
)
=1(mod p) then 
=2(mod p) with all odd prime p
such that: 
?
such that: ?
such that: ?
. The other part is your
such that: be an odd prime number. Prove that there exist sequence such that: 
. Choose 
where 
. We get that 
which is impossible if both 
are even and so one of is odd, let us say without loss of generality that 
is odd. Then choose 
and we see that 
which means that necessarily 
is always odd, now choose 
and see that 
which is a contradiction and hence our assumption is wrong as desired. is an odd prime, take 
such that 
. We see that 
which is a contradiction, implying that 
for all 
which is the desired contradiction.
such that: be an odd prime number. Prove that there exist sequence such that: 
. is odd then as we need 
So 
must be even.
then must be odd. Now we need 
such that 
but observe if is even then 
and similarly if is odd then 
. Hence contradiction. is even then 
so 
is even and 
so is odd. But then 
Hence contradiction. be an odd prime number. Prove that there exist sequence such that: we can always have a sequence 
which indeed satisfy 
. 
's I hope my sequence works 
where is fixed and 
varies We have![\[ \text{gcd}(a_{2m}+(2t+1), a_{2t+1}+(2m) ) = 1\]](http://latex.artofproblemsolving.com/4/d/e/4defa4fe4fba22b7997b0cdf25ab82f1c026ec9c.png)
where we take 
![\[ (a_1, a_2, a_3, a_4, ... ) = (0,0,0,0, ... ) \]](http://latex.artofproblemsolving.com/7/5/0/75054a4dc8a22b74906743ae36bc6b59a98b5021.png)
![\[ (a_1,a_2,a_3,a_4, ... ) = (1,1,1,1, ... ) \]](http://latex.artofproblemsolving.com/5/c/2/5c2537793c4063e43cf125732eb5e04efa1f8d47.png)
and we have 
, contradiction.
and we have 
, contradiction.
exists.
with an even index , as otherwise, if there is another integer with an even index, say , then 
. Similarly, there can be at most one odd number 
with an odd index 
. Hence, for all the other numbers, the paritity of and 
are different. Among those, choose an even and any odd 
. This also immediatily implies that 
.
works for any odd prime . First note if does not divide 
, then can't divide both of 
and 
, and hence cannot divide their gcd. Now,
so we are done.
is odd.![\[\gcd(a_1 + 2k + 1, a_{2k + 1} + 1) = 1\]](http://latex.artofproblemsolving.com/c/b/c/cbc33daa897dc568fa6aaae39636a754929432f3.png)
So, 
is odd. Consider is odd. Then,![\[\gcd(a_2 + 3, a_3 + 2) = 1\]](http://latex.artofproblemsolving.com/2/8/d/28d6eff02cd0d08c8b0cf5a5e3ce35db3b6272a2.png)
is assumed odd, 
. Hence, 
and 
. Contradiction. So, is even.![\[\gcd(a_2 + 4, a_4 + 2) = 1\]](http://latex.artofproblemsolving.com/c/4/a/c4a43e374290b67db615a9974fe399562103f18b.png)
, so 
is odd 
is odd.![\[\gcd(a_4 + 5, a_5 + 4) = 1\]](http://latex.artofproblemsolving.com/8/2/1/8217bbb6e1b86c5a554bc3cf961303936d17abd1.png)
is even and so does 
. Check that, 
, contradiction! Now, we assumed to be odd. Now, we are left with the part to be even.![\[\gcd(a_1 + 2k, a_{2k} + 1) = 1\]](http://latex.artofproblemsolving.com/8/2/d/82dfe8fea3743c325ee45e85ba2ae3b01246eaa9.png)
must be odd. So, 
is even. Hence, is even. Let's assume is even too.![\[\gcd(a_3 + 1, a_1 + 3) = 1\]](http://latex.artofproblemsolving.com/8/b/f/8bfe1db16995502989be046bc5c7f7efc871b408.png)
Contradiction. So, is odd.![\[\gcd(a_3 + 2k + 1, a_{2k + 1} + 3) = 1\]](http://latex.artofproblemsolving.com/7/d/3/7d3847edcda41862b5d4a8abec64e37568218e82.png)
Hence, 
is even. Now, note that,![\[d \mid a_1 + 5 \text{ and } d \mid a_5 + 1\]](http://latex.artofproblemsolving.com/9/0/8/90852f05c0536ffb8f522751ad3f1bcee957688f.png)
![\[d \mid (a_5 - a_1) - 4\]](http://latex.artofproblemsolving.com/2/5/8/258fdcc469e711feec0ceefd2f82a4ea04ca02b6.png)
and both are even, hence 
, contradiction! 
Note that for the latter condition to hold, we must have at most value of for which 
is even. Therefore, 
From here, we can proceed with casework.
However, since is odd, 
is even, and hence 
must be odd, so 
is odd.
However, since 
is odd, 
is even, and hence 
must be odd, so 
is odd.
we cannot have both and be odd, since then there would be more than one value of for which is even, and hence 
a contradiction.
holds since is an odd prime.
and 
. Then, 
, so this means that at least one of 
and 
is odd. 
and this means that at least one of 
and 
is odd. Now, by comparing 
that is odd and even, we arrive at a contradiction as gcd has to be 2.
works where is a variable that makes positive.
such that: be an odd prime number. Prove that there exist sequence such that: as an odd number we will get a contradiction because we could pair all fo and we will see that all of 
will be odd so we could not pair two odd because we will get an 
of a multiple of 2. Therefore, is even, and also because of the previous temp, we get all even terms odd and every odd ters even and we will get a contradiction if we pair 1 even term and 1 odd term.
and 
. Then, 
and 
can't both be even, so one of and 
is clearly odd. Say it is . Then, for all odd integers , 
is even. Thus, 
cannot be even, which implies that 
is odd for all odd integers . But then, considering two odd integers 
and 
we have that 
and 
are both even, and thus there 
is greater than 1. Contradiction! Thus, there cannot exist such a sequence.
for all positive integers . In particular, here 
for all indices . Then, say there exists two indices and which violate the desired condition. Then, 
and 
. But, if![\[p \mid (p+1)-i+j \implies p \mid j-i +1 \text{ and } p \mid (p+1)-j+i \implies p\mid i-j+1\]](http://latex.artofproblemsolving.com/a/8/3/a83fceba5835669531d647dea141ea3c8c9c61b6.png)
summing implies that 
which is a clear contradiction. Thus, this sequence indeed satisfies the given condition and we are done. (This actually solves the extension in #5 as well).
, it follows that or is odd. WLOG is odd. Then 
so is odd if is odd. But 
and 
are both even, a contradiction. 
such that 
. Then 
is not divisible by since 
. 
: 
implies that either or is odd. WLOG let to be odd. Then: 
implies 
is odd for all integers 
. Notice that: 
implies 
to be even, contradiction. There doesnt exists such a sequence.
evidently works and we are done.)
thus: 
. Taking any sequence which satisfies these conditions would work and we are done.
: implies that either or is odd. WLOG let to be odd. Then: implies is odd for all integers . Notice that: implies to be even, contradiction. There doesnt exists such a sequence. evidently works and we are done.) thus: . Taking any sequence which satisfies these conditions would work and we are done.
for some odd . If is even, take even (any even), which gives that must be even. Note 
a contradiction since 
. If is odd, consider any other odd, . This gives 
odd which means even, and we finish like earlier.
, which finishes.
,
or 
is odd. Let it is 
is odd for every 
- contradiction
and 
conclude 
odd. Take 
to get 
odd. Take 
odd to get 
this works, since 
.
s.t. 
)
s.t. 
.
.
must be even, therefore for the 
of the two terms to be 1, 
must be odd 
is odd.
we see that 
is also odd.
.
and 
works.
gives that 
and finish.
.
such that 
will be enough.
and easily steer clear out of this problem 
Similarly, we get 
Since 
S
For 
Clearly this works.
for any 
then 
or 
for all 
Then, 
a contradiction. Hence, we assume 
for all 
a contradiction.
and assume FTSOC that 
Then, 
or 
