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, 
,
, 
, 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 real numbers such that 
. Find the value of 
be real numbers such that 
and 
. Find the value of 
, let 
be a point on arc 
of circle which doesn't contain 
. 
and intersect at 
. Let 
and 
be the reflection of about to 
and 
, respectively. 
intersects at 
, and 
intersects at 
. Prove that circumcenter of triangle 
lies on .
. The complement of this would be 
or 
. The final thing we can do is to make sure it is simplified and add the numerator and the denominator which is 
and 
so 
This should be final answer. (Mathdash rating 800)
?
using 
to obtain 
.I should have written it the other way round for more clarity. Edit: It might seem that this step is invalid for 
but then 
which is impossible for (due to AM-GM) is not always true...
for example is an equality case for this problem.
and the case 
is easy. Sorry, my silly fault of not considering this case 
very small. take 
and we get 
and 
, so that we will prove that 
with 
using Lagrange Multipliers. Note that 
at all points, and moreover that 
and 
have continuous partial derivatives.
, then 
which is bounded, then the constraint set, 
is compact. Hence, it achieves a minimum value say 
. lies on the boundary, then at least one component of is zero, let's say , then we have ![\[bc=b+c \implies \frac{b+c}{2}\geq \sqrt{bc}, bc\geq 4 \iff f(0,b,c)=bc\geq 4\]](http://latex.artofproblemsolving.com/e/3/c/e3c118c68854cb2dbedb7e27da7afff331757e88.png)
as desired and the minimum value holds at 
.
. By Lagrange Multiplies, we have ![\[\left<bc(2a+2), c(a^2+2a+1), b(a^2+2a+1)\right>=\lambda \left<b+c-1, c+a-1, a+b-1\right>.\]](http://latex.artofproblemsolving.com/d/6/7/d670ded5587dae0888f4c874004afa9c6486f7a8.png)
If 
then 
, 
which don't satisfy the conditions that 
. So, 
, we will get that ![\[2bc(a+1)=\lambda (b+c-1) \ \ (1)\]](http://latex.artofproblemsolving.com/9/2/c/92c871b6b7592e42a40cccace2e82ffefc774167.png)
![\[c(a+1)^2=\lambda (c+a-1) \ \ (2)\]](http://latex.artofproblemsolving.com/3/2/7/327dcd0c206cb98b158cf2ced80fbce601e825a5.png)
![\[b(a+1)^2=\lambda (a+b-1) \ \ (3)\]](http://latex.artofproblemsolving.com/c/6/4/c648c131114c0054fa763483ffd2ac4bb0a299fa.png)
From 
, sub it into 
, we get ![\[a=-1+\frac{4b-2bc}{b-c+1}\]](http://latex.artofproblemsolving.com/2/5/0/250d5e87908e291abdc7bcf3c4c8f3470c6f3b4f.png)
and so from 
, ![\[b(a+1)^2=\frac{2bc(a+1)(b+a-1)}{b+c-1}\implies a=-1+\frac{2bc-4c}{b-c-1}.\]](http://latex.artofproblemsolving.com/b/5/7/b576ae6736e6a7bed3b87884569df2fe9e4abb10.png)
Equating these two equalities, ![\[\frac{4b-2bc}{b-c+1}=\frac{2bc-4c}{b-c-1}\implies (b-c)(b+c-bc-1)=0\]](http://latex.artofproblemsolving.com/4/2/d/42d5edf3c7eff737bbb6cb649a180022ebac515f.png)
![\[a=-1+\frac{2bc-4c}{b-c-1}=-2c^2+4c-1\]](http://latex.artofproblemsolving.com/4/0/8/4082c1fe05b29aa9aa03e1bb223c2610b6518a3a.png)
and from 
![\[c(-2c^2+4c-1)+c^2+c(-2c^2+4c-1)-(-2c^2+4c-1)-c-c=0 \implies (c-1)(4c^2-7c+1)=0\]](http://latex.artofproblemsolving.com/9/0/c/90c7490f9c3835ad9dec2496ba693e666dbb4066.png)
If 
, then 
and 
which means 
. If 
and solving for gives 
. Take 
, then we can check that ![\[f\left(\frac{-1+\sqrt{33}}{16}, \frac{7+\sqrt{33}}{8}, \frac{7+\sqrt{33}}{8}\right)\geq 4.\]](http://latex.artofproblemsolving.com/4/f/6/4f6e99282b539bbfd4ef706a65ec8b6ae319e2dd.png)
Case 2:
\implies 
or 
. In either case, we can get that also.
. Since this is the minimum point, 
. We have 
So 
.
. So 
.
For 
, we have 
. So 
For 
, 
For 
, we have 
Q.E.D
be fixed. Find
over all 
be fixed. Find
over all 
Prove that
Prove that
Let 
Prove that![$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$](http://latex.artofproblemsolving.com/f/5/e/f5e68204507aa9c3e71c28844bbee3a804546b94.png)
!
candy bars. Let N & M be positive integers, what is N+M?
be real numbers such that 
and 
and determine the equality cases.
, so... is it you dear sir? 
,
.
,
.
,
.
,
.
and 
.
.
if and only if 
.
.
.
,
.
.
.
or 
,
.
.
.
done
then denoting 
,
,
but 
so it remains to prove that
true.
or 
and 
note that 
and since 
we obtain that 
Now apply AM-GM 
such that 
Prove
When do we have equality?
does
hold for all 
satisfying 
and 
,
.
,
which is 
, impossible.
,
and 
.
,
It's false, of course