,
,
,
,
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.
,
isn't even rigorously defined in this expression. What exactly do we mean by 
.
,
does not exist in the first place, although you will see why in a calculus course. So the point is that 
?
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 
,
or ![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](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 
.
is undefined because dividing by ![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](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 ![\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 ![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](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 ![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and 
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 ![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by 
be a set of integers with the following properties:
.
and 
,
.
,
is composite, then all positive divisors of 
such that 
is an integer for every positive integer 
has how many digits?
positive integers have no single-digit prime factors?
Y by Centralorbit, Nelu2003, mcn2003, samrocksnature, SSaad, megarnie, centslordm, jhu08, RedFlame2112, Lamboreghini, Adventure10, Mango247, Sedro
Let 
be the set of all integers. Find all pairs of integers 
for which there exist functions 
and 
satisfying
![\[ f(g(x))=x+a \quad\text{and}\quad g(f(x))=x+b \]](//latex.artofproblemsolving.com/3/2/8/328146c3331a66cdc5cfc4a01b0609e91d278ec5.png)
for all integers 
.
Proposed by Ankan Bhattacharya
be the set of all integers. Find all pairs of integers 
for which there exist functions 
and 
satisfying![\[ f(g(x))=x+a \quad\text{and}\quad g(f(x))=x+b \]](http://latex.artofproblemsolving.com/3/2/8/328146c3331a66cdc5cfc4a01b0609e91d278ec5.png)
for all integers 
.Proposed by Ankan Bhattacharya
This post has been edited 2 times. Last edited by v_Enhance, Apr 18, 2019, 12:51 AM
or 
.
and 
.
and 
.
and 
are bijections.
then 
,
If either 
is zero or 
is zero, we immediately get the other is zero, and hence done. So assume 
.
,![\[ \{ f(0), f(1), \dots, f(b-1) \} \pmod a \]](http://latex.artofproblemsolving.com/d/3/c/d3c3023aa17bb03745b143545924ee3c1ffc3898.png)
coincide, which together with repeatedly applying the first equation above will then give a contradiction to injectivity of 
swapping the roles of 
.
,
,
,
.![[asy] size(11cm); usepackage("amsmath"); usepackage("amssymb"); draw( (-3,2)--(13,2), Arrows(TeXHead) ); draw( (-3,-2)--(13,-2), Arrows(TeXHead) ); label("$\mathbb Z$", (13, 2), dir(0)); label("$\mathbb Z$", (13, -2), dir(0));
pair bm1 = (-5,2); pair b0 = (0,2); pair b1 = (5,2); pair b2 = (10,2); pair b3 = (15,2); pair am1 = (-1,-2); pair a0 = (2.5,-2); pair a1 = (6,-2); pair a2 = (9.5,-2); pair a3 = (13,-2);
dot("$0$", b0, dir(90)); dot("$b$", b1, dir(90)); dot("$2b$", b2, dir(90)); dot("$f(0)-a$", am1, dir(-90)); dot("$f(0)$", a0, dir(-90)); dot("$f(0)+a$", a1, dir(-90)); dot("$f(0)+2a$", a2, dir(-90));
draw(midpoint(bm1--am1)--am1, red, EndArrow, Margins); label("$f$", midpoint(bm1--am1)--am1, dir(50), red); draw(b0--a0, red, EndArrow, Margins); label("$f$", b0--a0, dir(50), red); draw(b1--a1, red, EndArrow, Margins); label("$f$", b1--a1, dir(50), red); draw(b2--a2, red, EndArrow, Margins); label("$f$", b2--a2, dir(50), red);
draw(am1--b0, blue, EndArrow, Margins); label("$g$", midpoint(am1--b0)--b0, dir(-50), blue); draw(a0--b1, blue, EndArrow, Margins); label("$g$", a0--b1, dir(-50), blue); draw(a1--b2, blue, EndArrow, Margins); label("$g$", a1--b2, dir(-50), blue); draw(a2--midpoint(a2--b3), blue, EndArrow, Margins); label("$g$", a2--midpoint(a2--b3), dir(-50), blue); [/asy]](http://latex.artofproblemsolving.com/7/9/1/7918f07d096e886b5606b0176e8dc52ad5f1ec47.png)
.
and 
for all 
and 
.
iff 
,![\[f(x + b) = f\Big(g\big(f(x)\big)\Big) = f(x) + a.\]](http://latex.artofproblemsolving.com/b/e/b/bebd65bec227fa4f3a00f8c1ddb4703149dc24a2.png)
Thus 
.
also works.
?
by considering 
.
,
.
,
.
.
into this gives 
.
,
.
,
.
,
.
.
,
.
.
We must have 
.
.
and 
works.
,
and 
works.
![\[a=b:f(x)=x, g(x)=x+a,\]](http://latex.artofproblemsolving.com/e/b/9/eb91fdaf5c25b0c4e3af63374bc6b20e553428d1.png)
![\[a=-b:f(x)=-x+a ,g(x)=-x.\]](http://latex.artofproblemsolving.com/9/7/a/97a9eb8c2fbec9e92ad1615e1388e7730cb3cd54.png)
.![\[g(f(u))=u+a=v+a=g(f(v)),\]](http://latex.artofproblemsolving.com/7/b/2/7b2aabed189d6cd0d594f69aab043eee4b2dcb65.png)
so they are injective (it is the exact same approach for 
)![\[g(f(g(x)))=g(x+a)=g(x)+b,\]](http://latex.artofproblemsolving.com/0/f/6/0f6b8b16420172bc5d5237492ce1ed5cc6d96adf.png)
![\[f(g(f(x)))=f(x+b)=f(x)+a.\]](http://latex.artofproblemsolving.com/e/2/4/e2499c539aeb0d0b1e11c8dd1cc2a9d17889afa7.png)
If one of 
.
.
,
such that 
.
such that 
and such that 
and 
are different modulo 
,
and 
,
,
We now assume that 
to 
,
satisfying 
,
.
.
we have 
and since 
such that 
.
,
,
,
and 
and if 
,
and 
.
and 
for 
.![\[ f(x) = -x + \left\lfloor \frac{t}{2} \right\rfloor,\qquad\ g(x) = -x + \left\lfloor \frac{t+1}{2} \right\rfloor. \]](http://latex.artofproblemsolving.com/1/f/d/1fd17c2c9156d9f259e5b935843bc0063e230129.png)
is as follows,![\[ f(x) = x + \left\lfloor \frac{t}{2} \right\rfloor,\qquad\ g(x) = x+\left\lfloor \frac{t+1}{2} \right\rfloor. \]](http://latex.artofproblemsolving.com/4/0/a/40a47429d4bc73b710203ae18661f968163ad513.png)
.
.
.
.
.
elements. So by PHP, there exists two elements with the same modulo 
.
and 
.
.
and 
.
of numbers that are 
like this.
.
.
as 
.
which is a contradiction and we are done. 
and 
,
respectively. Now we prove 
and 
are the only possibilities.
and 
.
,
;
Observe that each residue of 
corresponds to a (not necessarily distinct) residue of 
we get that 
and similarly 
.
.
,
since if otherwise there exist k such that 
for which 
,
.
.
,
iff 
.
are injective and surjective functions. Therefore, 
.
to 
.
,
.
.
so 
must have at least two equal values by PHP. By applying injectivity, this has to be a contradiction.
then 
works
then 
works.
and 
respectively.
so assume otherwise. For the sake of contradiction assume 
.![\[f(x) + a = f(g(f(x))) = f(x + b). \]](http://latex.artofproblemsolving.com/2/a/6/2a661a4e761a3a6c941b10c8c9bcb35eb64f4071.png)
Therefore, 
over 
since 
