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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
and train with the best! Please note that early bird pricing ends August 19th!
Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

Our full course list for upcoming classes is below:
All classes start 7:30pm ET/4:30pm PT unless otherwise noted.

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0 replies
jwelsh
Jul 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i A Letter to MSM
Arr0w   23
N Sep 19, 2022 by scannose
Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ 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.
[list]
[*]Firstly, the case of $0^0$. It is usually regarded that $0^0=1$, 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.

[*]What about $\frac{\infty}{\infty}$? The issue here is that $\infty$ isn't even rigorously defined in this expression. What exactly do we mean by $\infty$? Unless the example in question is put in context in a formal manner, then we say that $\frac{\infty}{\infty}$ is meaningless.

[*]What about $\frac{1}{0}$? Suppose that $x=\frac{1}{0}$. Then we would have $x\cdot 0=0=1$, absurd. A more rigorous treatment of the idea is that $\lim_{x\to0}\frac{1}{x}$ does not exist in the first place, although you will see why in a calculus course. So the point is that $\frac{1}{0}$ is undefined.

[*]What about if $0.99999...=1$? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
\begin{align*}
\sum_{n=1}^{\infty} \frac{9}{10^n}&=9\sum_{n=1}^{\infty}\frac{1}{10^n}=9\sum_{n=1}^{\infty}\biggr(\frac{1}{10}\biggr)^n=9\biggr(\frac{\frac{1}{10}}{1-\frac{1}{10}}\biggr)=9\biggr(\frac{\frac{1}{10}}{\frac{9}{10}}\biggr)=9\biggr(\frac{1}{9}\biggr)=\boxed{1}
\end{align*}
[*]What about $\frac{0}{0}$? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
[/list]
Hopefully all of these issues and their corollaries are finally put to rest. Cheers.

2nd EDIT (6/14/22): Since I originally posted this, it has since blown up so I will try to add additional information per the request of users in the thread below.

INDETERMINATE VS UNDEFINED

What makes something indeterminate? As you can see above, there are many things that are indeterminate. While definitions might vary slightly, it is the consensus that the following definition holds: A mathematical expression is be said to be indeterminate if it is not definitively or precisely determined. So how does this make, say, something like $0/0$ 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 $0/0$ is an indeterminate form.

But we need to more specific, this is still ambiguous. An indeterminate form is a mathematical expression involving at most two of $0$, $1$ or $\infty$, 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
\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]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 $f:\mathbb{R}^{+}\cup\{0\}\rightarrow\mathbb{R}$ given by the mapping $x\mapsto \sqrt{x}$ is undefined for $x<0$. On the other hand, $1/0$ is undefined because dividing by $0$ is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context.

WHEN THE WATERS GET MUDDIED

So with this notion of indeterminate and undefined, things get convoluted. First of all, just because something is indeterminate does not mean it is not undefined. For example $0/0$ 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.

One example of making something undefined into something defined is the extended real number line, which we define as
\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]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 $-\infty\le a\le \infty$ for each $a\in\overline{\mathbb{R}}$ which means that via this order topology each subset has an infimum and supremum and $\overline{\mathbb{R}}$ is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In $\overline{\mathbb{R}}$ 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*}So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,
\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]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 $\infty \times 0=0$ as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by $0$ 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
\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, $\tilde{\infty}$ means complex infinity, since we are in the complex plane now. Here's the catch: division by $0$ is allowed here! In fact, we have
\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]where $\tilde{\infty}/\tilde{\infty}$ and $0/0$ are left undefined. We also have
\begin{align*}
z+\tilde{\infty}=\tilde{\infty}, \forall z\ne -\infty\\
z\times \tilde{\infty}=\tilde{\infty}, \forall z\ne 0
\end{align*}Furthermore, we actually have some nice properties with multiplication that we didn't have before. In $\mathbb{C}^*$ it holds that
\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]but $\tilde{\infty}-\tilde{\infty}$ and $0\times \tilde{\infty}$ 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 $\mathbb{C}^*$, by defining them as
\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]They behave in a similar way to the Riemann Sphere, with division by $0$ also being allowed with the same indeterminate forms (in addition to some other ones).
23 replies
Arr0w
Feb 11, 2022
scannose
Sep 19, 2022
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
The 24 Game, but with a twist!
PikaPika999   326
N a few seconds ago by PikaPika999
So many people know the 24 game, where you try to create the number 24 from using other numbers, but here's a twist:

You can only use the number 24 (up to 5 times) to try to make other numbers :)

the limit is 5 times because then people could just do $\frac{24}{24}+\frac{24}{24}+\frac{24}{24}+...$ and so on to create any number!

honestly, I feel like with only addition, subtraction, multiplication, and division, you can't get pretty far with this, so you can use any mathematical operations!

Banned functions
326 replies
PikaPika999
Jul 1, 2025
PikaPika999
a few seconds ago
Cutting square into three congruent rectangles
Silverfalcon   13
N 20 minutes ago by Happycat2
A square is cut into three rectangles along two lines parallel to a side, as shown. If the perimeter of each of the three rectangles is 24, then the area of the original square is

IMAGE

$\text{(A)} \ 24 \qquad \text{(B)} \ 36 \qquad \text{(C)} \ 64 \qquad \text{(D)} \ 81 \qquad \text{(E)} \ 96$
13 replies
Silverfalcon
Dec 9, 2005
Happycat2
20 minutes ago
Factorial fraction
Silverfalcon   32
N 22 minutes ago by PikaPika999
$\frac{(3!)!}{3!} =$

$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 6\qquad \text{(D)}\ 40\qquad \text{(E)}\ 120$
32 replies
Silverfalcon
Nov 21, 2005
PikaPika999
22 minutes ago
Exponent of 1 and -1
Silverfalcon   32
N an hour ago by deeptisidana
$(-1)^{5^2} + 1^{2^5} =$

$\textbf{(A)}\ -7 \qquad
\textbf{(B)}\ -2 \qquad
\textbf{(C)}\ 0 \qquad
\textbf{(D)}\ 1 \qquad
\textbf{(E)}\ 57$
32 replies
Silverfalcon
Oct 22, 2005
deeptisidana
an hour ago
|4^m-7^n| a prime number
dangerousliri   10
N 3 hours ago by Maths_VC
Source: JBMO Shortlist 2024, N1
Find all pairs of positive integers $(m,n)$ such that $|4^m-7^n|$ is a prime number.

Proposed by Dorlir Ahmeti, Albania
10 replies
dangerousliri
Jun 26, 2025
Maths_VC
3 hours ago
Simple (EZ?) NT
obihs   0
3 hours ago
Source: own
Prove that there exists an integer $x$ such that for any prime $p$ and any positive integers $a,b,c,$ both $\dfrac{x-a}{p}$ and $\dfrac{x^p-x-p^bc}{p^{b+1}}$ are integers.
0 replies
obihs
3 hours ago
0 replies
2017 N4 type Problem
zqy648   0
Today at 9:57 AM
Source: 2023 New Year 谜之竞赛-3
Let prime number $p>3$ and $t$ be a positive integer. Show that
\[\binom{p^{t+1}}{p^t}\equiv\binom{p^t}{p^{t-1}}\pmod {p^{3t+2}}.\]
0 replies
zqy648
Today at 9:57 AM
0 replies
Difficult NT Problems
EthanWYX2009   1
N Today at 6:40 AM by EthanWYX2009
Source: 2024 June 谜之竞赛-6&7
Given prime number \( p > 2024^{2024} \). For a vector \(\vec{a} = (a_1, a_2, \cdots, a_p) \in \mathbb{F}_p^p \), define $f(\vec{a})$ as the number of permutations \(\sigma\) of \(\{1, 2, \cdots, p\}\) satisfying $ p \mid \sum\limits_{i=1}^p a_i \sigma(i).$

(1) Determine the number of vectors \(\vec{a}\) such that $ f(\vec{a}) = 0.$

(2) For vectors \(\vec{a}\) with $f(\vec{a}) \neq 0$, find the minimum possible value of $ f(\vec{a})$ and when equality holds.

Proposed by Zhenyu Dong from Hangzhou Xuejun High School
1 reply
EthanWYX2009
Today at 6:39 AM
EthanWYX2009
Today at 6:40 AM
Product of corners of rectangles not divisible by p
gghx   2
N Yesterday at 5:16 PM by Tkn
Source: SMO open 2024 Q5
Let $p$ be a prime number. Determine the largest possible $n$ such that the following holds: it is possible to fill an $n\times n$ table with integers $a_{ik}$ in the $i$th row and $k$th column, for $1\le i,k\le n$, such that for any quadruple $i,j,k,l$ with $1\le i<j\le n$ and $1\le k<l\le n$, the number $a_{ik}a_{jl}-a_{il}a_{jk}$ is not divisible by $p$.

Proposed by oneplusone
2 replies
gghx
Aug 3, 2024
Tkn
Yesterday at 5:16 PM
Using Dilworth to Deal with Antichains
zqy648   0
Yesterday at 10:04 AM
Source: 2024 November 谜之竞赛-6
For prime number \( p \) and positive integer \( a \in \{1, 2, \cdots, p-1\} \), denote \( (a^{-1}\bmod{p}) \) by the unique integer \( m \) satisfying \( am \equiv 1 \pmod{p} \) with \( 0 \leq m \leq p-1 \).

Let \( k \) be a positive integer, and let \( p \) be a prime such that \( p > 100k^2 \). Prove that there exist positive integers \( x_1, x_2, \cdots, x_k \) satisfying:[list]
[*] \( 1 \leq x_1 < x_2 < \cdots < x_k \leq p-1 \);
[*]\( (x_1^{-1}\bmod{p}) > (x_2^{-1}\bmod{p}) > \cdots > (x_k^{-1}\bmod{p}). \) [/list]
Proposed by Cheng Jiang and Tianqin Li
0 replies
zqy648
Yesterday at 10:04 AM
0 replies
Sums of 1/i with Resticted Legrendre Symbol
EthanWYX2009   0
Jul 12, 2025
Source: 2023 September 谜之竞赛-5
Let prime number $p\equiv 1\pmod 8$, show that
$$\sum_{\substack{1\le i\le\frac{p-1}2\\\left(\frac ip\right)=1}}\frac 1i\equiv \sum_{\substack{1\le i\le\frac{p-1}2\\\left(\frac ip\right)=-1}}\frac 1i\pmod{p^2}.$$Created by Mucong Sun, Tianjin Experimental Binhai School
0 replies
EthanWYX2009
Jul 12, 2025
0 replies
Number theory - Iran
soroush.MG   35
N Jul 12, 2025 by NicoN9
Source: Iran MO 2017 - 2nd Round - P1
a) Prove that there doesn't exist sequence $a_1,a_2,a_3,... \in \mathbb{N}$ such that: $\forall i<j: gcd(a_i+j,a_j+i)=1$

b) Let $p$ be an odd prime number. Prove that there exist sequence $a_1,a_2,a_3,... \in \mathbb{N}$ such that: $\forall i<j: p \not | gcd(a_i+j,a_j+i)$
35 replies
soroush.MG
Apr 20, 2017
NicoN9
Jul 12, 2025
Functional Number Theory
Iveela   6
N Jul 12, 2025 by ihategeo_1969
Source: 2025 IRN-MNG Friendly Competition
Let $\mathbb{N}$ be the set of positive integers. Find all unbounded functions $f : \mathbb{N} \to \mathbb{N}$ such that
\[f(n + f(m) - 1) \mid f(n) + m - 1\]for all $m, n \in \mathbb{N}$.
6 replies
Iveela
Jun 8, 2025
ihategeo_1969
Jul 12, 2025
Number of positive divisors of 2p^2+2p+1 where p is prime
electrovector   6
N Jul 10, 2025 by Feita
Source: Turkey National Mathematical Olympiad 2020 P4
Let $p$ be a prime number such that $\frac{28^p-1}{2p^2+2p+1}$ is an integer. Find all possible values of number of divisors of $2p^2+2p+1$.
6 replies
electrovector
Mar 8, 2021
Feita
Jul 10, 2025
A Letter to MSM
G H J
G H BBookmark kLocked kLocked NReply
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Arr0w
2908 posts
#1 • 332 Y
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Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ 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.
  • Firstly, the case of $0^0$. It is usually regarded that $0^0=1$, 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.
  • What about $\frac{\infty}{\infty}$? The issue here is that $\infty$ isn't even rigorously defined in this expression. What exactly do we mean by $\infty$? Unless the example in question is put in context in a formal manner, then we say that $\frac{\infty}{\infty}$ is meaningless.
  • What about $\frac{1}{0}$? Suppose that $x=\frac{1}{0}$. Then we would have $x\cdot 0=0=1$, absurd. A more rigorous treatment of the idea is that $\lim_{x\to0}\frac{1}{x}$ does not exist in the first place, although you will see why in a calculus course. So the point is that $\frac{1}{0}$ is undefined.
  • What about if $0.99999...=1$? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
    \begin{align*}
\sum_{n=1}^{\infty} \frac{9}{10^n}&=9\sum_{n=1}^{\infty}\frac{1}{10^n}=9\sum_{n=1}^{\infty}\biggr(\frac{1}{10}\biggr)^n=9\biggr(\frac{\frac{1}{10}}{1-\frac{1}{10}}\biggr)=9\biggr(\frac{\frac{1}{10}}{\frac{9}{10}}\biggr)=9\biggr(\frac{1}{9}\biggr)=\boxed{1}
\end{align*}
  • What about $\frac{0}{0}$? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
Hopefully all of these issues and their corollaries are finally put to rest. Cheers.

2nd EDIT (6/14/22): Since I originally posted this, it has since blown up so I will try to add additional information per the request of users in the thread below.

INDETERMINATE VS UNDEFINED

What makes something indeterminate? As you can see above, there are many things that are indeterminate. While definitions might vary slightly, it is the consensus that the following definition holds: A mathematical expression is be said to be indeterminate if it is not definitively or precisely determined. So how does this make, say, something like $0/0$ 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 $0/0$ is an indeterminate form.

But we need to more specific, this is still ambiguous. An indeterminate form is a mathematical expression involving at most two of $0$, $1$ or $\infty$, 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
\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]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 $f:\mathbb{R}^{+}\cup\{0\}\rightarrow\mathbb{R}$ given by the mapping $x\mapsto \sqrt{x}$ is undefined for $x<0$. On the other hand, $1/0$ is undefined because dividing by $0$ is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context.

WHEN THE WATERS GET MUDDIED

So with this notion of indeterminate and undefined, things get convoluted. First of all, just because something is indeterminate does not mean it is not undefined. For example $0/0$ 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.

One example of making something undefined into something defined is the extended real number line, which we define as
\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]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 $-\infty\le a\le \infty$ for each $a\in\overline{\mathbb{R}}$ which means that via this order topology each subset has an infimum and supremum and $\overline{\mathbb{R}}$ is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In $\overline{\mathbb{R}}$ 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*}So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,
\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]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 $\infty \times 0=0$ as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by $0$ 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
\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, $\tilde{\infty}$ means complex infinity, since we are in the complex plane now. Here's the catch: division by $0$ is allowed here! In fact, we have
\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]where $\tilde{\infty}/\tilde{\infty}$ and $0/0$ are left undefined. We also have
\begin{align*}
z+\tilde{\infty}=\tilde{\infty}, \forall z\ne -\infty\\
z\times \tilde{\infty}=\tilde{\infty}, \forall z\ne 0
\end{align*}Furthermore, we actually have some nice properties with multiplication that we didn't have before. In $\mathbb{C}^*$ it holds that
\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]but $\tilde{\infty}-\tilde{\infty}$ and $0\times \tilde{\infty}$ 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 $\mathbb{C}^*$, by defining them as
\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]They behave in a similar way to the Riemann Sphere, with division by $0$ also being allowed with the same indeterminate forms (in addition to some other ones).
This post has been edited 13 times. Last edited by Arr0w, Sep 17, 2022, 11:43 PM
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greenturtle3141
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#12 • 49 Y
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My several cents:
  • I think we should all just accept $0^0 = 1$ and move on with it, really no reason not to (the rationale being, "Because it is").
  • $\infty$ does not necessarily refer to the limit of a function. Honestly just reject any "arithmetic" with $\infty$ unless you have properly defined it. So we should not bother computing even "obvious" quantities like $\infty+\infty$ or $\infty + 7$. If you don't define what $\infty$ is then there is no point doing anything with it. As a side effect you get un-definedness of $\infty/\infty$ etc. for free, because... we literally have not defined $\infty$. Tada.
  • To wit, I will complain that $\infty$ should not necessarily represent the limit of a function. But if you are interpreting it as such, then $\infty/\infty$ is not undefined, but rather indeterminate.
  • At some point in math (which is not anywhere within 5 years if you're in middle school), we do start messing with $+\infty$ as a perfectly valid number, because it starts becoming useful. Particularly, we do define $0 \cdot \infty = 0$ in contexts such as measure theory. This differs from the "limit of a function" interpretation, in which $0 \cdot \infty$ would be indeterminate.

Digression
This post has been edited 2 times. Last edited by greenturtle3141, Feb 12, 2022, 4:05 AM
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Kempu33334
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#17 • 12 Y
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I think it would be good for the 0/0 one to be said as indeterminate
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Arr0w
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#26 • 8 Y
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greenturtle3141 wrote:
At some point in math (which is not anywhere within 5 years if you're in middle school), we do start messing with $+\infty$ as a perfectly valid number, because it starts becoming useful. Particularly, we do define $0 \cdot \infty = 0$ in contexts such as measure theory.
You seem to mention this quite a bit in your past posts as well. Could you clarify why this would be used and is necessary in measure theory? I think it would be a cool addition to the original post. Thanks!
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greenturtle3141
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#39 • 20 Y
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Arr0w wrote:
greenturtle3141 wrote:
At some point in math (which is not anywhere within 5 years if you're in middle school), we do start messing with $+\infty$ as a perfectly valid number, because it starts becoming useful. Particularly, we do define $0 \cdot \infty = 0$ in contexts such as measure theory.
You seem to mention this quite a bit in your past posts as well. Could you clarify why this would be used and is necessary in measure theory? I think it would be a cool addition to the original post. Thanks!

Measure Zero Differences

When you change very few points of a function, its integral should not change. For example, we know that $\int_0^1 x^2\,dx = \frac13$. But what about e.g. $\int_0^1 f(x)\,dx$ where $f(x) := \begin{cases}x^2, & x \ne \frac12 \\ 7, & x = \frac12\end{cases}$? I have changed a single point. Of course, this shouldn't matter because a single point is negligible for changing the area under the curve. If you think about it, I'm essentially adding in a rectangle with dimensions $0 \times 7$, which is zero. So $\int_0^1 f(x)\,dx = \frac13$.

In general, any measure zero difference cannot change an integral.

Infinity

Let $\overline{\mathbb{R}} := \mathbb{R} \cup \{\pm \infty\}$. This is the extended reals, and is a useful abstraction.

One way in which this makes mathematicians happy is in making limits exist. For example, consider the monotone convergence theorem, which states that if a sequence of functions $f_n:E \to \overline{\mathbb{R}}$ is increasing (i.e. $f_n \leq f_{n+1}$ everywhere in $E$, for all $n$), then $f_n$ converges pointwise to a function $f:E \to \overline{\mathbb{R}}$ and moreover
$$\lim_{n \to \infty} \int_E f_n\,dx = \int_E f\,dx.$$Notice how I allow my functions to take values of $\infty$. If I didn't let them do that, then I have to take separate cases in the statement of the theorem as to whether limits diverge or whatnot. But as you can see, there really is no issue if I let functions take values of $\infty$.

This also implies that I can integrate functions that take values of $\infty$... indeed I certainly can. What happens?

Why $\infty \times 0$ shows up

Ok, let's define this function:
$$f(x) := \begin{cases}x^2, & x \ne \frac12 \\ +\infty, x = \frac12\end{cases}$$What is $\int_0^1 f(x)\,dx$?

In the definition of Lebesgue integration, you'll find the possibility that you have to consider the "rectangle" $\{1/2\} \times [0,\infty)$. The dimensions of this are $0 \times \infty$. Well? Does this change the integral?

Here's the thing: If it did, it would be really stupid and annoying. Remember, measure-zero differences should NOT do anything to an integral. So we force $0 \times \infty := 0$ to make this work, so that $\int_0^1 f(x)\,dx = 1/3$ still.

This isn't contrived, this makes sense. A $0 \times \infty$ rectangle has no area. It doesn't cover any significant 2D space. So its area is zero. If you're pedantic, you can even prove it using the rigorous definition of area. In any case, this is a context in which it is absolutely, 100% clear what the value of $0 \times \infty$ should be. It is zero. No other value for it would be useful, no other value would make remotely any sense. This is the only possible value for it here.
This post has been edited 1 time. Last edited by greenturtle3141, Feb 23, 2022, 5:23 AM
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Nickelslordm
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#40 • 5 Y
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@bove but infinity has no concrete definition; therefore, how can we even tell that it follows rules such as multiplication by zero? I don't know for sure, but I would guess that infinity bends the whole concept of multiplication. Someone whose name I forgot said that zero times infinity equals infinity divided by 2.
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Facejo
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#44 • 7 Y
Y by HWenslawski, BlinkySalamander11, Kea13, ImSh95, WiseTigerJ1, DanielP111, TeamFoster-Keefe4Ever
$\frac{0}{0}$ is indeterminate not undefined
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Arr0w
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#46 • 6 Y
Y by HWenslawski, Kea13, ImSh95, WiseTigerJ1, TeamFoster-Keefe4Ever, chriscassano
Facejo wrote:
$\frac{0}{0}$ is indeterminate not undefined
Thank you, this has been revised.
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Facejo
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#47 • 7 Y
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Facejo wrote:
$\frac{0}{0}$ is indeterminate not undefined

wait what's the difference

Indeterminate means there is no specific value, while undefined means it doesn't exist

@above You're welcome
This post has been edited 1 time. Last edited by Facejo, Apr 14, 2022, 11:58 PM
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ssbgm9002
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#48 • 5 Y
Y by HWenslawski, Kea13, ImSh95, WiseTigerJ1, TeamFoster-Keefe4Ever
jmiao wrote:
Facejo wrote:
$\frac{0}{0}$ is indeterminate not undefined

wait what's the difference
Great explanation here
"The big difference between undefined and indeterminate is the relationship between zero and infinity. When something is undefined, this means that there are no solutions. However, when something in(is GET YOUR GRAMMAR RIGHT) indeterminate, this means that there are infinitely many solutions to the question." - http://5010.mathed.usu.edu/Fall2018/LPierson/indeterminateandundefined.html#:~:text=The%20big%20difference%20between%20undefined,many%20solutions%20to%20the%20question.
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NightFury101
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#61 • 6 Y
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Just a picky note on the $\frac10$ thing:

If we have $f(x) = \frac1x$, then $\lim_{x \to 0^+}f(x) = \infty$ and $\lim_{x \to 0^-}f(x) = -\infty$. In other words, $f$ approaches positive infinity as $x$ approaches $0$ from the right hand side, and $f$ approaches negative infinity as $x$ approaches $0$ from the left hand side.

This is why the limit does not exist and $\frac10$ is undefined.
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YaoAOPS
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#66 • 5 Y
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Arr0w wrote:
Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ 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'd be nice to have the other indeterminate forms mentioned, i.e. $\infty - \infty, 1^{\infty}$. Very cool post
This post has been edited 3 times. Last edited by YaoAOPS, May 30, 2022, 3:15 PM
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Turtle09
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#68 • 5 Y
Y by Kea13, ImSh95, WiseTigerJ1, TeamFoster-Keefe4Ever, mithu542
YaoAOPS wrote:
Arr0w wrote:
Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ 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'd be nice to have the other indeterminate forms mentioned, i.e. $\infty - \infty, 1^{\infty}$. Very cool post

wait isn't $1^\infty = 1$?
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Facejo
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#69 • 3 Y
Y by Kea13, ImSh95, WiseTigerJ1
@above No. In general, you cannot say that.
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aayr
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#70 • 5 Y
Y by mighty_champ, Kea13, ImSh95, WiseTigerJ1, Kawhi2
wait why not
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mahaler
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#71 • 3 Y
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Turtle09 wrote:
wait isn't $1^\infty = 1$?
Facejo wrote:
@above No. In general, you cannot say that.
aayr wrote:
wait why not

Yeah, I don't know either...

Couldn't you just prove that $1^n = 1$ for any positive integer $n$ using induction?
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Lionking212
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#72 • 3 Y
Y by Kea13, ImSh95, WiseTigerJ1
mahaler wrote:
Turtle09 wrote:
wait isn't $1^\infty = 1$?
Facejo wrote:
@above No. In general, you cannot say that.
aayr wrote:
wait why not

Yeah, I don't know either...

Couldn't you just prove that $1^n = 1$ for any positive integer $n$ using induction?

because
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Facejo
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#81 • 4 Y
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Post #70 by aayr

Post #71 by mahaler

Consider the limit $\displaystyle \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\approx 2.718281828459045$
This post has been edited 1 time. Last edited by Facejo, May 30, 2022, 7:06 PM
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Arr0w
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#96 • 3 Y
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RaymondZhu wrote:
Can you please define indeterminate and undefined and the differences in your post? Thanks!
Howdy Raymond!

I have made sure to add some additional items like you requested. If there's anything more you guys want to see from this thread let me know so I can add/change it. Thanks!
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Arr0w
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#130 • 4 Y
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peelybonehead wrote:
Leo2020 wrote:
bump$            $
We should bump every single day so people can be reminded of this thread every single day
Please don't do this. If you would like to share this post for whatever reason, you can just link it using the following format:
Please see [url=https://artofproblemsolving.com/community/c3h2778686_a_letter_to_msm] here [/url].

In the meantime, I have made some additional edits to the letter as I have changed my mind on some conclusions I made previously. If there are any discrepancies, please let me know. Thank you.
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wamofan
6814 posts
#140 • 3 Y
Y by Kea13, ImSh95, WiseTigerJ1
thebluepenguin21 wrote:
but only 1=1, right? So you can't say that 0.99999999... = 1.
that's like saying only 2=2, so you can't say that 6-4=2; completely wrong
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Sure it makes sense because that is the closest that we can get, but it can not be. And 9x -x = 8.99999999999... So this can not be true.

why is 9x-x=8.99999?
9x=9, x=1 so 8x=8 so x=1
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ap246
1796 posts
#146 • 5 Y
Y by Kea13, ImSh95, s12d34, CatsMeow12, WiseTigerJ1
Basically, we can just use the Epsilon - Delta definition of a limit. If you give me an $\epsilon > 0$ such that $\epsilon = 1 - 0.99999\dots$ then I will provide a sufficient $\delta$ for the number of nines needed such that $1 - 0.99999\dots < \epsilon$ We can use this to form a proof by contradiction. More specifically, for a positive value $x,$ if $\epsilon > 10^{-x}$ then $\delta = x$ is sufficient. Therefore, there will always be a delta such that $1 - 0.99999\dots < \epsilon$

We can define the limit: $$\lim_{x\to \infty} f(x) = y$$if for every $\epsilon > 0$ there exists a $\delta$ such that $x > \delta$ which implies $$|f(x) - y| < \epsilon$$so $f(x) = \sum_{n = 1}^{n = x} 9 * 10^{-n}$ for a positive value of $n.$

If $2$ quantities aren't equal, then there must be a nonzero difference, but given that for every $\epsilon$ there exists a $\delta$ such that the given information is met, we prove that there doesn't exist a nonzero difference, so both quantities are equal.

$$Q.E.D$$
This post has been edited 2 times. Last edited by ap246, Sep 19, 2022, 8:03 PM
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scannose
1026 posts
#147 • 4 Y
Y by ultimate_life_form, ImSh95, megahertz13, MathematicalGymnast573
Also, if you agree that geometric series works:
$0.999\dots=9({1\over{10}}+{1\over{100}}+\dots)=9\cdot{1\over{1-{1\over{10}}}}=9\cdot{1\over9}=1$
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scannose
1026 posts
#149 • 10 Y
Y by mathfan2020, ultimate_life_form, lucaswujc, ImSh95, MathematicalGymnast573, megahertz13, WiseTigerJ1, TeamFoster-Keefe4Ever, Blue_banana4, jkim0656
I'm pretty sure that 9x-x=8.999... was just a typo; they probably meant 10x-x=8.999...?
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