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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 2025
0 replies
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
Is your state listed?
Chatelet1   361
N 20 minutes ago by Cool12345678
Multiple states have announced their top students who will advance to the 2025 MATHCOUNTS National Competition in May:

• From Alabama: Henry Gladden of Mobile, Austin Lu of Birmingham, Jessie Shi of Vestavia, and Minlu Wang-He of Auburn.

• From Arkansas: Ryan Fan of Fayetteville, Vivek Kalyankar of Fayetteville, Evan Ning of Fayetteville and Charles Yao of Conway.

• From Connecticut: Hayden Hughes of Newtown, Ethan Shi of Riverside, Alex Svoronos of Greenwich and Elaine Zhou of Hamden.

• From the Department of Defense: Narmin Guliyeva of Ankara, Turkey; Taeyul Kim of Manana, Bahrain; Nathan Liang of Wiesbaden, Germany; and Lucas Sze of Okinawa, Japan.

• From Hawaii: Taehwan Jeon, Hilohak Kwak, Isaac Qian and Thien Tran, all from Honolulu.

• From Kansas: Haidan Anderson & Jayden Xue of Overland Park, Christopher Spencer of Manhattan, and Ruby Jiang of Lawrence.

• From Maine: Ana Kanitkar & Connor Kirkham of Falmouth, Anna McClary of Hermon and Poppy Sandin of Bar Harbor.

• From Massachusetts: Eric Huang of Acton, Shlok Mukund & Brandon Ni of Lexington, and Soham Samanta of Medford.

• From Missouri: Lucas Lai of Columbia, Kevin Shi of St. Louis, Charles Yong & Jay Zhou of Chesterfield.

• From Montana: Titus Gilder of Missoula, Otis Heggem of Billings, Kaleb Houtz of Great Falls and Evan Newcomer of Missoula.

• From Nevada: Solomon Dumont of Las Vegas, Aaron Lei of Reno, Leeoz Nebat of Henderson and Maxwell Tsai of Las Vegas.

• From New Mexico: Mark Goldman, Daniel He, Iris Huang and Patrick McArdle, all from Albuquerque.

• From New York: Derrick Chen of Great Neck, Victor Yang of Great Neck, Hanru Zhang of Jericho and Ryan Zhang of Jericho.

• From Rhode Island: Kahlan Anderson of the Wheeler School, Julian Bernhoft & Colin Hegstrom of Providence, and Theodora Watson of Barrington.

• From South Carolina: Yukai Hu of Elgin, Justin Peng of Clemson, Geonhoo Shim of Columbia, and Aaron Wang of Mount Pleasant.

• From South Dakota: Seth Chaplin & Maxwell Wang of Sioux Falls, Laukia Gundewar of Aberdeen, and Cohwen Heimann of Aberdeen.

• From Texas: Shaheem Samsudeen & Ayush Narayan of Plano, Nathan Liu of Richardson, and James Stewart of Southlake.

• From Vermont: Mohid Ali of South Burlington, Vivek Chadive of South Burlington, Joshua Kratze of St. Johnsbury and Albert Zhang of South Burlington.

• From Wisconsin: August Reeder & Lucy Chen of Fitchburg, Junhao Feng of Milwaukee, and Jiyan Singh of River Hills.

===
Updated on 3/15/2025:

• From Colorado: Noah Liu, Christopher Zhu, Neo Luo, and Andrew Zhao.

• From Florida: Arnav Bhatia, Gnaneswar Peddesugari, Edwin Gao, and Rananjay Parmar.

• From Indiana: Roland Li, Hrishabh Bhowmik, Sophia Chen, and Arjun Raman.

• From Kentucky: Sri Shubhaan Vulava, Joyce Liu, Victor Gong, and Brandon Tedja.

• From Maryland: Eric Xie, Angie Zhu, Roger Huang, and Leo Su.

• From Michigan: Arnav Vunnam, Eric Jin, Akshaj Malraj, and Chaithanya Budida.

• From Minnesota: Ahmed Ilyasov, Will Masanz, Anshdeep Singh, and Branden Qiao.

• From New Jersey: Ethan Imanuel, Advait Joshi, Jay Wang, and Easton Wei.

• From North Carolina: Shivank Chintalpati, Steven Wang, Lucas Li, and Leo Hong.

• From Ohio: Henry Lu, Andy Mo, Archishmen Dey, and Caleb Tan.

• From Oregon: Sophia Han, Kevin Cheng, Garud Shah, and Ryan Zhang.
361 replies
Chatelet1
Mar 8, 2025
Cool12345678
20 minutes ago
9 Pi or Tau
jkim0656   77
N 38 minutes ago by BananaBall00
Hey Aops!
Pi = Circumfrence/Diameter
Tau = Circumfrence/Radius
I have noticed a lot of sites, including Khan Academy, in support of tau over pi...
so what do you think?
https://www.scientificamerican.com/article/let-s-use-tau-it-s-easier-than-pi/
However i am still in support of the good ol pi :)
(btw this is my first aops poll) :-D

EDIT: 50 votes!!! :play_ball:
EDIT: 100 votes!!! :jump:
EDIT: 150 votes! :trampoline:

If u support pi pls upvote :)
77 replies
jkim0656
Mar 14, 2025
BananaBall00
38 minutes ago
MATHCOUNTS State Preparation
mithu542   21
N an hour ago by hashbrown2009
Hello!

I'm going to prepare for Mathcounts state soon. I want some advice on what to do. I am in 7th grade, and I want to make it to nationals. I know I should obviously take practice tests, but should I do something else other than that, or just grind all (or most) practice tests from previous years? Also, how much should I focus on Countdown round relative to the other tests?

(For reference, I got 43 on school, and 41 on chapter. Last year, I got 16/116 rank in state. Since then, I have done the following courses from aops:
Intro: algebra b, number theory, c&p, geometry
Intermediate: algebra, number theory, c&p)
21 replies
mithu542
Feb 14, 2025
hashbrown2009
an hour ago
Problem of the week
evt917   23
N an hour ago by DearPrince
Whenever possible, I will be posting problems twice a week! They will be roughly of AMC 8 difficulty. Have fun solving! Also, these problems are all written by myself!

First problem:

$20^{16}$ has how many digits?
23 replies
evt917
Mar 5, 2025
DearPrince
an hour ago
No more topics!
A Letter to MSM
G H J
G H BBookmark kLocked kLocked NReply
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Arr0w
2908 posts
#1 • 317 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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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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I think it would be good for the 0/0 one to be said as indeterminate
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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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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
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$\frac{0}{0}$ is indeterminate not undefined
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Arr0w
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#46 • 6 Y
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Facejo wrote:
$\frac{0}{0}$ is indeterminate not undefined
Thank you, this has been revised.
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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
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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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#68 • 5 Y
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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
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@above No. In general, you cannot say that.
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aayr
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#70 • 5 Y
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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
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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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#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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#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
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#140 • 3 Y
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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
Quote:
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
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#146 • 5 Y
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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
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#147 • 4 Y
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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
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#149 • 9 Y
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I'm pretty sure that 9x-x=8.999... was just a typo; they probably meant 10x-x=8.999...?
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