,
,
,
,
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 
where 
are integers such that 
for all 
.
![\[
r_1 \ge r_a,\quad r_2 \ge r_b,\quad r_3 \ge r_c
\]](http://latex.artofproblemsolving.com/8/2/1/8214c052ab9ad2c979414f92b6707fa6de33664f.png)
because if any of the radii were smaller than these, the circles would not be able to reach and touch the incircle.![\[
\frac{r_1 + r_2 + r_3}{r} \ge \frac{r_a + r_b + r_c}{r}
\]](http://latex.artofproblemsolving.com/b/b/a/bbadb40e6aa197207abfa907981e0a4a88926969.png)
![\[
\frac{r_a + r_b + r_c}{r} = \frac{h_a - 2r}{h_a} + \frac{h_b - 2r}{h_b} + \frac{h_c - 2r}{h_c}
\]](http://latex.artofproblemsolving.com/5/3/2/5329fec1f02a728b5f34817e0ed43ded663516ae.png)
![\[
= \left(1 - \frac{2r}{h_a}\right) + \left(1 - \frac{2r}{h_b}\right) + \left(1 - \frac{2r}{h_c}\right)
= 3 - 2r \left( \frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c} \right)
\]](http://latex.artofproblemsolving.com/1/e/0/1e087ea87566fb4cf8cd2ae0cda3389247095eea.png)
![\[
\frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c} = \frac{a + b + c}{2S}
\]](http://latex.artofproblemsolving.com/a/4/a/a4ab49131584cf2028270e7248fcbb5d2c6a7adb.png)
![\[
3 - 2r \cdot \frac{a + b + c}{2S} = 3 - \frac{4S}{a + b + c} \cdot \frac{a + b + c}{2S} = 3 - 2 = 1
\]](http://latex.artofproblemsolving.com/0/f/3/0f3ada5ca22deeecc1644d79138942caf04d5134.png)
![\[
\frac{r_1 + r_2 + r_3}{r} \ge 1
\]](http://latex.artofproblemsolving.com/7/1/e/71ea14c8fb8e3860a1db427a0afcdb3b2fd6254b.png)
a prime number with 
and 
when 
.
else 
.
be real numbers such that 
.
to be the sum of all permutations of the digits of 
such that no digit appears in its original spot. For example, 
and 
.
?
are
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The equation
is written on the board, with 
linear factors on each side. What is the least possible value of 
for which it is possible to erase exactly of these 
linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?
is written on the board, with 
linear factors on each side. What is the least possible value of 
for which it is possible to erase exactly of these 
linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?This post has been edited 2 times. Last edited by sseraj, Jul 12, 2016, 5:44 PM
If one erases all the 
in LHS and 
in RHS, then it can be proved that the remaining equation doesn't have real roots.
in 
where 
is a positive integer. If 
,
.
.
,
。
,![\[\prod_{i=-p}^{q}f(a-4i)=1\]](http://latex.artofproblemsolving.com/1/5/2/152bfa0d35a607fe6eb8bf2e3f0fe8c2b409b1e9.png)
where 
are non-negative integers (I'm too lazy to write down its exactly form).![\[\prod_{i=-p}^{-1} f(a-4i)=\prod_{i=1}^{p} f(a+4i) = \prod_{i=1}^{p} (1-\frac{2}{(a+4i)^2-0.25})\geq 1-\sum_{i=1}^{p} \frac{2}{(a+4i)^2-0.25}\]](http://latex.artofproblemsolving.com/e/c/2/ec22e739ab8c81b98a1058ed73bae7629adb02a0.png)
![\[>1-\frac{1}{6}-\sum_{i=2}^{\infty}\frac{2}{(a+4i)^2-0.25}>1-\frac{1}{6}-\frac{1}{8}\sum_{i=1}^{\infty}i^{-2}=\frac{5}{6}-\frac{\pi ^2}{48}>\frac{1}{2}\]](http://latex.artofproblemsolving.com/6/9/8/698794ee8808704f73d76c3a6e8c9fbef3e17044.png)
Similarly, 
.
,
,
,
,
.
,
.
inequalities
Notice that in all these inequalities, at most one of them has non-positive numbers in it, and we never have both zero. So it is OK to multiply them all together, and we get the conclusion. Thus we have a construction for 
,
the LHS and RHS must have two coinciding linear terms.
)
sides, it is on the left and we can multiply all together. Thus the only case that remains is if 
for some 
.![\[ \prod_{k \ge 0} \frac{(4k+2)(4k+3)}{(4k+1)(4k+4)} < e. \]](http://latex.artofproblemsolving.com/b/5/4/b54646631ad989e48589233f6ae72f311a4f2455.png)
To see this, note that it's equivalent to prove ![\[ \sum_{k \ge 0} \log \left( 1 + \frac{2}{(4k+1)(4k+4)} \right) < 1. \]](http://latex.artofproblemsolving.com/a/4/2/a4240564e98d58606835805a5eb1399f3ea7d784.png)
To this end, we use the deep fact that 
,![\[ \sum_{k \ge 0} \frac{1}{(4k+1)(4k+4)} < \frac12 \]](http://latex.artofproblemsolving.com/c/d/a/cda961fe5889adda61bfee040f49515bfbf34eae.png)
which one can obtain for example by noticing it's less than 
.
,
factors. There will be 
left, say 
and 
on the left and right hand side respectively. Since 
,
on both sides of the equation, and this equation would then have root 
,
and all RHS terms are 
:![\[(x - 1)(x - 4)\ldots(x - 2013)(x - 2016) = (x - 2)(x - 3)\ldots (x - 2014)(x - 2015)\]](http://latex.artofproblemsolving.com/4/2/f/42f115f1dddd37fb4e0604a5d847c5e8e86497e6.png)
which we claim to have no real roots. Note that for all 
in intervals of the form ![$[4i + 1, 4i + 2]$](http://latex.artofproblemsolving.com/4/8/8/4882d094b00b2d4611c5d733944d8788c0c4071e.png)
and ![$[4i + 3, 4i + 4]$](http://latex.artofproblemsolving.com/6/e/5/6e572e7c42b8a64e6d3a602fe0afd72d90f4d65f.png)
,
or 
both sides of the equality are positive but the inequality 
holds for all 
so taking the product of all these yields LHS 
RHS. Hence it remains to consider ![$[4i + 2, 4i + 3]$](http://latex.artofproblemsolving.com/c/4/7/c4783503a9854c491cc9a0b9b489784c512863be.png)
and ![$[4i, 4i + 1]$](http://latex.artofproblemsolving.com/5/4/1/54135869e837d592a8eb92e3e4c7e65e2014f7de.png)
.
for all 
intervals of the form 
.
and 
we multiply and get LHS 
RHS, and we are done. We have considered all intervals covering all real numbers, and there are no real roots to this equation, as desired. 
since erasing any 
(or less) factors will leave a common linear factor on both sides.
have all the 
roots and 
the 
roots:![\[
\begin{array}{cccc}
p(x) := (x-1)(x-4)&(x-5)(x-8)\ \cdots \ (x-2013)(x-2016) \\
q(x) := (x-2)(x-3)&(x-6)(x-7)\ \cdots \ (x-2014)(x-2015).
\end{array}
\]](http://latex.artofproblemsolving.com/e/b/9/eb912e0cf57830e228ca0a8ac9bdd90f8a16afbd.png)
We need to check that 
has no real solutions. We claim 
for all 
.![\[
\begin{array}{c|cccc|cccc|c|c}
(-\infty,1) & (1,2)&(2,3) & (3,4) & (4,5)&(5,6)&(6,7)&(7,8)&(8,9)&\cdots &(2016,\infty) \\
&a & b & c & d & a & b & c &d &\cdots
\end{array}
\]](http://latex.artofproblemsolving.com/1/9/b/19b23e288462f8a2280bea54c242c78f0de2a8fb.png)
It suffices to use only open intervals since on the boundaries, exactly one of 
or 
.
and that for 
and 
,
.
interval of type 
,
of 
of 
.
.
The key is that for each row, both sides are positive. So multiplying gives 
,
and 
.
for some 
and 
for some 
,
construction, 
does not have degree 
,
for various 
.
doesn't work.
such that 
is on both sides of the equation after erasing, which implies that ![\[(x-1)(x-4)(x-5)(x-8)(x-9)(x-12)\cdots(x-2013)(x-2016)=(x-2)(x-3)(x-6)(x-7)(x-10)(x-11)\cdots (x-2014)(x-2015).\]](http://latex.artofproblemsolving.com/0/d/d/0dd5535f6f02beae034fa863a9f6f61d3f5698f9.png)
.![\[(x-1)(x-4)=x^2-5x+4<(x-2)(x-3)=x^2-5x+6, (x-5)(x-8)<(x-6)(x-7),\]](http://latex.artofproblemsolving.com/e/b/5/eb5efa237ba2fb09176d25b0b3d3d1130bd7b27b.png)
and so on until 
.
are all positive, which implies the right-hand side is strictly greater than the left-hand side.
.
don't have any solutions, where 
and 
,
don't have any solutions, where 
and 
)
don't have any solutions, where 
or 
)
don't have any solutions, where 
(we will put them back later). So the equation becomes ![\[(x-4)(x-5)(x-8)(x-9)(x-12)(x-13)\cdots(x-2012)(x-2013)=(x-3)(x-6)(x-7)(x-10)\cdots(x-2011)(x-2014).\]](http://latex.artofproblemsolving.com/b/4/2/b421ca8e9239ed0885374dcc3a5318147d90942d.png)
.![\[(x-4)(x-5)(x-8)(x-9)(x-12)(x-13)\cdots(x-2012)(x-2013)>(x-3)(x-6)(x-7)(x-10)\cdots(x-2011)(x-2014)>0.\]](http://latex.artofproblemsolving.com/6/5/0/65069459c656c30fa73bd02b3c2e36192d4969b8.png)
and 
,
.
(large positive number) = small negative number and RHS=(large negative number) 
remains on different sides of equality. We claim, that 
which suffices.
for 
By multiplying we are done.
is zero, hence the conclusion.![$x\in [1;2016]\backslash \mathbb{Z}.$](http://latex.artofproblemsolving.com/a/7/a/a7a1953b867fc01bad8f181a098804cd5532487e.png)
Here we consider two additional cases.
is odd.
we have 
Moreover 
so multiplying altogether we obtain 
will have no solution, there obviously must be no repeats on the 2 sides.
,
for the LHS and 
for the RHS.![$[1, 2, 3, \dotsm, 2016]$](http://latex.artofproblemsolving.com/1/f/2/1f2cdb78adc11d951aa91c376d78a11a86623d20.png)
,
![\[\prod_{k=1}^{504}{(x-(4k-3))(x-4k)}=\prod_{k=1}^{504}{(x-(4k-2))(x-(4k-1))}\]](http://latex.artofproblemsolving.com/3/9/6/396b43328a1baf91bef1adbac71439231fcc4eec.png)
has no real solutions. Clearly, there are no integer solutions between ![\[(x-(4k-3))(x-4k)=(x-(4k-2))(x-(4k-1))-2 < (x-(4k-2))(x-(4k-1)).\]](http://latex.artofproblemsolving.com/5/b/5/5b55107af582c6e0f3b885993bed40aa451ecc59.png)
Suppose 
then every factor on the LHS is positive and less than every factor on the RHS, so the LHS is less than the RHS. If 
for some integer 
then every factor on the LHS is positive except one, so it is negative. Every factor on the RHS is still positive, so it is positive. Again LHS 
.
and 
are the only negative terms on each side. We have 
.
be from 
to 
then ![\[\frac{(x-(4(k-i)-2))(x-(4(k-i)-1))(x-(4(k+i)-2))(x-(4(k+i)-1))}{(x-(4(k-i)-3))(x-4(k-i))(x-(4(k+i)-3))(x-4(k+i))}=\frac{s^2}{s^2-4}\]](http://latex.artofproblemsolving.com/2/c/b/2cb5ddfeeaabbf768201cf4850e656433bee4a09.png)
and it is easy to see from here that no amount of this being multiplied together with distinct positive 
'
where 
and let the RHS contain the rest of the factors. Our equation is then ![\[\prod_{k=0}^{503}\left(x-(4k+1)\right)\left(x-(4k+4)\right)=\prod_{k=0}^{503}\left(x-(4k+2)\right)\left(x-(4k+3)\right)\]](http://latex.artofproblemsolving.com/2/e/6/2e6df08202e0e8f6147a10884f6c4ad1ae5d592e.png)
Notice ![\[\left(x-(4k+1)\right)\left(x-(4k+4)\right)+2=\left(x-(4k+2)\right)\left(x-(4k+3)\right)\quad(*)\]](http://latex.artofproblemsolving.com/7/b/0/7b03e0c83af240159f505ecfc3416fd2060c6ed0.png)
Hence,
then one side is zero and the other side nonzero.
for any 
then one side is negative and the other is positive.
,
for 
,
is a solution to the equation. Consider ![\[P=\prod_{0\le k\le 503,k\neq j}\frac{\left(x-(4k+1)\right)\left(x-(4k+4)\right)}{\left(x-(4k+2)\right)\left(x-(4k+3)\right)}=\frac{\left(x-(4j+2)\right)\left(x-(4j+3)\right)}{\left(x-(4j+1)\right)\left(x-(4j+4)\right)}=\frac{\epsilon^2-\epsilon}{\epsilon^2-\epsilon-2}<\frac{1}{9}\]](http://latex.artofproblemsolving.com/b/d/4/bd48586b33cbdf370df34c6d16b16a8920ed5e85.png)
Let 
be a term in 
.![\[f(j-\ell)\ge \frac{(4\ell)\cdot (4\ell-1)-2}{(4\ell)\cdot (4\ell-1)}\]](http://latex.artofproblemsolving.com/b/9/b/b9b9ea50496a571614af7f8de72422157f35a552.png)
and ![\[f(j+\ell)\ge \frac{(4\ell)\cdot (4\ell-1)-2}{(4\ell)\cdot (4\ell-1)}\]](http://latex.artofproblemsolving.com/6/a/d/6ad0f08a571c63666db5eb1ee6cf5b92b15ac091.png)
by 
.
increases as we get further from ![\begin{align*}
P&=[f(j-1)f(j-2)f(j-3)\dots f(j-1008)][f(j+1)f(j+2)\dots f(j+1008)]
\\&\ge\left(\prod_{\ell=1}^{1008}\frac{(4\ell)\cdot (4\ell-1)-2}{(4\ell)\cdot (4\ell-1)}\right)^2
\end{align*}](http://latex.artofproblemsolving.com/c/3/6/c360b36f104b15981c8d161e29246b8389721ecd.png)
It suffices to show ![\[\prod_{\ell=1}^{1008}\left(1+\frac{2}{(4\ell)(4\ell-1)-2}\right)<3\]](http://latex.artofproblemsolving.com/6/6/3/66338c417e0470bc501708aecf7892dbb97587af.png)
in order to obtain a contradiction. We know 
so
as desired. 
.
and RHS as 
.
(the key being for ANY x), so if both sides are negative LHS>RHS, if both sides are positive LHS<RHS, and both sides can't be zero, done.
,
,
.
must be erased, otherwise, 
,
from RHS and erase factors 
from LHS. Then we will prove that the equation 
has no real solution.![$x \not\in [1, 2016]$](http://latex.artofproblemsolving.com/f/b/f/fbf38a0b3dfffe78bbed89d46d1391cadc33b0b1.png)
,
,
.![$x \in [1, 2016]$](http://latex.artofproblemsolving.com/2/2/8/2281be6b459c825d1d38ca58871d44a6e40d4640.png)
.![$x \in [4s, 4s+1]$](http://latex.artofproblemsolving.com/b/b/2/bb2ff85969688d88f6e9064980e146a168bb11b4.png)
for some 
.
for all ![$x \in [4s+1, 4s+2]$](http://latex.artofproblemsolving.com/9/4/c/94cdc8d442836a610221dbe804c57648a61eeb72.png)
or ![$x \in [4s+3, 4s+4]$](http://latex.artofproblemsolving.com/a/d/3/ad37e197226bc73d0110770c70c81c33aeb9766c.png)
for some 
and for all 
,
.![$x \in [4s+2, 4s+3]$](http://latex.artofproblemsolving.com/7/b/a/7ba958ecbedd4d183ab3e6c3381ad3d9323ba8ff.png)
for some 
,
.
from 2 sides and using the inequality 
,
.
,
.
,
.
We show that removing all with constant either 
from the left side and 
from the right side works. Note that for all 
we have that
We can simply multiply all the inequalities of this form together whenever 
However, when 
we have that for exactly one inequality the left and right sides are both negative, where the left is maximum the product of 
Note that 
Note that we have two of these products because 
so because 
we're don
we make the following construction:
If 
or 
then 
and all of these pairs of products are positive, so the left hand side is clear less than the right hand side. If 
but 
for some integer 
then the LHS is less than 
while the right hand side is not.
then
Now, we get that everything else is at most
Thus, this ratio cannot be 
We will prove that this has no solutions. Indeed
Suppose there is a root 
or 
,
or 
.
.
when compared, since all terms are positive, can be multiplied together
(Note that the actual product without inversion in signs is negative so the sign gets flipped)
appears on both sides.![\[P(x) = \prod_{i=0}^{503}(x-4i-1)(x-4i-4) - \prod_{i=0}^{503} (x-4i-2)(x-4i-3).\]](http://latex.artofproblemsolving.com/9/e/c/9ecdf5356715ea5b5d1ffebe66310607da10fe98.png)
We will prove that 
for all real numbers 
and 
denote the two products, such that 
.
when 
or 
for each 
.![$x \in [4k+1, 4k+2]$](http://latex.artofproblemsolving.com/c/7/d/c7df36c7ab9cdb6eb130fad2d53f0370c34c367b.png)
or ![$x \in [4k+3, 4k+4]$](http://latex.artofproblemsolving.com/c/3/3/c33e7408a8d5fd50b7f4abfe71d74767ad8df259.png)
for any integer 
,
is nonnegative and 
is nonpositive, with equality not holding simultaneously. When 
,
where each inequality follows again by multiplying the same inequalities in the claim.
.
and 
,
for these values of 
Hence we get the result.
,
on one side and 
on the other side works.
.
,
,
integer 
,
,
,
all succumb to both terms being positive or the inequalities 
,
.
and 
and let 
,
![$<\frac{1}{9}(1+\frac{2}{2\cdot 5})(1+\frac{2}{6\cdot 9})\dots(1+\frac{2}{(4k-2)(4k+1)})(1+\frac{2}{2\cdot 5})(1+\frac{2}{6\cdot 9})\dots(1+\frac{2}{(2010-4k)(2013-4k)})<1\longleftrightarrow \log [(1+\frac{2}{2\cdot 5})(1+\frac{2}{6\cdot 9})\dots(1+\frac{2}{(4k-2)(4k+1)})(1+\frac{2}{2\cdot 5})(1+\frac{2}{6\cdot 9})\dots(1+\frac{2}{(2010-4k)(2013-4k)})]<\frac{2}{2\cdot 5}\frac{2}{6\cdot 9}\dots \frac{2}{(4k-2)(4k+1)}\frac{2}{2\cdot 5}\frac{2}{6\cdot 9}\dots \frac{2}{(2010-4k)(2013-4k)}=\frac{2}{3}(\frac{1}{2}-\frac{1}{5}+\dots +\frac{1}{4k-2}-\frac{1}{4k+1}+\frac{1}{2}-\frac{1}{5}+\dots +\frac{1}{2010-4k}-\frac{1}{2013-4k})=\frac{2}{3}(1+(\frac{1}{6}-\frac{1}{5}+ \dots + \frac{1}{4k-2}-\frac{1}{4k-3}+\frac{1}{6}-\frac{1}{5}+ \dots + \frac{1}{2010-4k}-\frac{1}{2009-4k})-\frac{1}{4k+1}-\frac{1}{2013-4k})<\frac{2}{3}<\log 9$](http://latex.artofproblemsolving.com/4/5/5/455c248c0b51905edb2480933600a251e1ae9a00.png)
,
and 
with such 
would repeat in both polynomials 
