Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 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 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
2025 MATHCOUNTS State Hub
SirAppel   625
N 17 minutes ago by Math-lover1
Previous Years' "Hubs": (2022) (2023) (2024)Please Read

Now that it's April and we're allowed to discuss ...
[list=disc]
[*] CA: 43 (45 44 43 43 43 42 42 41 41 41)
[*] NJ: 43 (45 44 44 43 39 42 40 40 39 38) *
[*] NY: 42 (43 42 42 42 41 40)
[*] TX: 42 (43 43 43 42 42 40 40 38 38 38)
[*] MA: 41 (45 43 42 41)
[*] WA: 41 (41 45 42 41 41 41 41 41 41 40) *
[*]VA: 40 (41 40 40 40)
[*] FL: 39 (42 41 40 39 38 37 37)
[*] IN: 39 (41 40 40 39 36 35 35 35 34 34)
[*] NC: 39 (42 42 41 39)
[*] IL: 38 (41 40 39 38 38 38)
[*] OR: 38 (44 39 38 38)
[*] PA: 38 (41 40 40 38 38 37 36 36 34 34) *
[*] MD: 37 (43 39 39 37 37 37)
[*] AZ: 36 (40? 39? 39 36)
[*] CT: 36 (44 38 38 36 35 35 34 34 34 33 33 32 32 32 32)
[*] MI: 36 (39 41 41 36 37 37 36 36 36 36) *
[*] MN: 36 (40 36 36 36 35 35 35 34)
[*] CO: 35 (41 37 37 35 35 35 ?? 31 31 30) *
[*] GA: 35 (38 37 36 35 34 34 34 34 34 33)
[*] OH: 35 (41 37 36 35)
[*] AR: 34 (46 45 35 34 33 31 31 31 29 29)
[*] NV: 34 (41 38 ?? 34)
[*] TN: 34 (38 ?? ?? 34)
[*] WI: 34 (40 37 37 34 35 30 28 29 29 29) *
[*] HI: 32 (35 34 32 32)
[*] NH: 31 (42 35 33 31 30)
[*] DE: 30 (34 33 32 30 30 29 28 27 26? 24)
[*] SC: 30 (33 33 31 30)
[*] IA: 29 (33 30 31 29 29 29 29 29 29 29 29 29) *
[*] NE: 28 (34 30 28 28 27 27 26 26 25 25)
[*] SD: 22 (30 29 24 22 22 22 21 21 20 20)
[/list]
Cutoffs Unknown

* means that CDR is official in that state.

Notes

For those asking about the removal of the tiers, I'd like to quote Jason himself:
[quote=peace09]
learn from my mistakes
[/quote]

Help contribute by sharing your state's cutoffs!
625 replies
SirAppel
Apr 1, 2025
Math-lover1
17 minutes ago
Doubt on a math problem
AVY2024   17
N an hour ago by LXC007
Solve for x and y given that xy=923, x+y=84
17 replies
1 viewing
AVY2024
Apr 8, 2025
LXC007
an hour ago
Find all functions $f$: \(\mathbb{R^+}\) \(\rightarrow\) \(\mathbb{R^+}\) such
guramuta   5
N 2 hours ago by jasperE3
Source: Balkan MO SL 2021
A5: Find all functions $f$: \(\mathbb{R^+}\) \(\rightarrow\) \(\mathbb{R^+}\) such that:
$$f(xf(x+y)) = xf(y) + 1 $$
5 replies
guramuta
3 hours ago
jasperE3
2 hours ago
number theory
frost23   3
N 2 hours ago by frost23
given any positive integer n show that there are two positive rational numbers a and b not equal to b which are such that a-b, a^2- b^2....................a^n-b^n are all integers
3 replies
frost23
2 hours ago
frost23
2 hours ago
partitioned square
moldovan   8
N 2 hours ago by cursed_tangent1434
Source: Ireland 1994
If a square is partitioned into $ n$ convex polygons, determine the maximum possible number of edges in the obtained figure.

(You may wish to use the following theorem of Euler: If a polygon is partitioned into $ n$ polygons with $ v$ vertices and $ e$ edges in the resulting figure, then $ v-e+n=1$.)
8 replies
moldovan
Jun 29, 2009
cursed_tangent1434
2 hours ago
Geometry
Lukariman   0
2 hours ago
Given acute triangle ABC ,AB=b,AC=c . M is a variable point on side AB. The circle circumscribing triangle BCM intersects AC at N.

a)Let I be the center of the circle circumscribing triangle AMN. Prove that I always lies on a fixed line.

b)Let J be the center of the circle circumscribing triangle MBC. Prove that line segment IJ has a constant length.
0 replies
Lukariman
2 hours ago
0 replies
Finding positive integers with good divisors
nAalniaOMliO   3
N 3 hours ago by nAalniaOMliO
Source: Belarusian National Olympiad 2025
For every positive integer $n$ write all its divisors in increasing order: $1=d_1<d_2<\ldots<d_k=n$.
Find all $n$ such that $2025 \cdot n=d_{20} \cdot d_{25}$.
3 replies
nAalniaOMliO
Mar 28, 2025
nAalniaOMliO
3 hours ago
Geometry Help
ILOVECATS127   2
N 3 hours ago by Soupboy0
Hello, I needed some help understanding this concept from Chapter 12, Geometry:

Points P, Q and R are on circle O such that

Arc PQ = 78°, arc QR = 123°, and arc PQR = 201°.

1. Find ∠PQO
2. Find ∠POR

Please help me understand HOW to solve these 2 problems.
2 replies
ILOVECATS127
Yesterday at 1:07 PM
Soupboy0
3 hours ago
Concurrent lines
MathChallenger101   4
N 3 hours ago by oVlad
Let $A B C D$ be an inscribed quadrilateral. Circles of diameters $A B$ and $C D$ intersect at points $X_1$ and $Y_1$, and circles of diameters $B C$ and $A D$ intersect at points $X_2$ and $Y_2$. The circles of diameters $A C$ and $B D$ intersect in two points $X_3$ and $Y_3$. Prove that the lines $X_1 Y_1, X_2 Y_2$ and $X_3 Y_3$ are concurrent.
4 replies
MathChallenger101
Feb 8, 2025
oVlad
3 hours ago
Find the value
sqing   7
N 3 hours ago by giangtruong13
Source: Own
Let $a,b,c$ be distinct real numbers such that $ \frac{a^2}{(a-b)^2}+ \frac{b^2}{(b-c)^2}+ \frac{c^2}{(c-a)^2} =1. $ Find the value of $\frac{a}{a-b}+ \frac{b}{b-c}+ \frac{c}{c-a}.$
Let $a,b,c$ be distinct real numbers such that $\frac{a^2}{(b-c)^2}+ \frac{b^2}{(c-a)^2}+ \frac{c^2}{(a-b)^2}=2. $ Find the value of $\frac{a}{b-c}+ \frac{b}{c-a}+ \frac{c}{a-b}.$
Let $a,b,c$ be distinct real numbers such that $\frac{(a+b)^2}{(a-b)^2}+ \frac{(b+c)^2}{(b-c)^2}+ \frac{(c+a)^2}{(c-a)^2}=2. $ Find the value of $\frac{a+b}{a-b}+\frac{b+c}{b-c}+ \frac{c+a}{c-a}.$
7 replies
sqing
Mar 17, 2025
giangtruong13
3 hours ago
2025 HMIC-5
EthanWYX2009   0
3 hours ago
Source: 2025 HMIC-5
Compute the smallest positive integer $k > 45$ for which there exists a sequence $a_1, a_2, a_3, \ldots ,a_{k-1}$ of positive integers satisfying the following conditions:[list]
[*]$a_i = i$ for all integers $1 \le i \le 45;$
[*] $a_{k-i} = i$ for all integers $1 \le i \le 45;$
[*] for any odd integer $1 \le n \le k -45,$ the sequence $a_n, a_{n+1}, \ldots  , a_{n+44}$ is a permutation of
$\{1, 2, \ldots  , 45\}.$[/list]
Proposed by: Derek Liu
0 replies
EthanWYX2009
3 hours ago
0 replies
I need the technique
DievilOnlyM   14
N 3 hours ago by Entei
Let a,b,c be real numbers such that: $ab+7bc+ca=188$.
FInd the minimum value of: $5a^2+11b^2+5c^2$
14 replies
DievilOnlyM
May 23, 2019
Entei
3 hours ago
Cyclic Quads and Parallel Lines
gracemoon124   15
N 3 hours ago by Adywastaken
Source: 2015 British Mathematical Olympiad?
Let $ABCD$ be a cyclic quadrilateral. Let $F$ be the midpoint of the arc $AB$ of its circumcircle which does not contain $C$ or $D$. Let the lines $DF$ and $AC$ meet at $P$ and the lines $CF$ and $BD$ meet at $Q$. Prove that the lines $PQ$ and $AB$ are parallel.
15 replies
gracemoon124
Aug 16, 2023
Adywastaken
3 hours ago
Combinatorics
Ro.Is.Te.   2
N 4 hours ago by SpeedCuber7
$\binom{24}{24} + \binom{24}{23} + \binom{24}{22} + \binom{24}{21} + \binom{24}{20}+...+\binom{24}{2}$
2 replies
Ro.Is.Te.
5 hours ago
SpeedCuber7
4 hours ago
two solutions
τρικλινο   10
N Apr 14, 2025 by Safal
in a book:CORE MATHS for A-LEVEL ON PAGE 41 i found the following


1st solution


$x^2-5x=0$



$ x(x-5)=0$



hence x=0 or x=5



2nd solution



$x^2-5x=0$

$x-5=0$ dividing by x



hence the solution x=0 has been lost



is the above correct?
10 replies
τρικλινο
Apr 12, 2025
Safal
Apr 14, 2025
two solutions
G H J
G H BBookmark kLocked kLocked NReply
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
τρικλινο
502 posts
#1
Y by
in a book:CORE MATHS for A-LEVEL ON PAGE 41 i found the following


1st solution


$x^2-5x=0$



$ x(x-5)=0$



hence x=0 or x=5



2nd solution



$x^2-5x=0$

$x-5=0$ dividing by x



hence the solution x=0 has been lost



is the above correct?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Safal
169 posts
#2
Y by
2nd Solution is basically wrong. Why? Here is the explanation.

$$x(x-5)=0$$Then there are two cases either $x=0$ or $x\neq 0$.When we are admitting the case $x=0$ we cannot divide by $0$. So, in the case we apply divison by $x$ then $x\neq 0$ is a solid prerequisite to do so.Thus, $x-5=0$ from $x(x-5)=0$ we must take the assumption in hand that $x\neq 0$. For example take the extension of the same problem in $\mathbb{F}_5$ then the same problem reads $$x^2=0$$,implying only one solution with optimistic repetation of root $0$, two times that is the multiplicity of $0$ in $x^2$. Thankfully, we are lucky enough that we are in the field of $\text{char}$ $0$.The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of $\text{char}$ $0$.
This post has been edited 10 times. Last edited by Safal, Apr 12, 2025, 7:52 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
τρικλινο
502 posts
#3
Y by
Safal wrote:
2nd Solution is basically wrong. Why? Here is the explanation.

$$x(x-5)=0$$Then there are two cases either $x=0$ or $x\neq 0$.When we are admitting the case $x=0$ we cannot divide by $0$. So, in the case we apply divison by $x$ then $x\neq 0$ is a solid prerequisite to do so.Thus, $x-5=0$ from $x(x-5)=0$ we must take the assumption in hand that $x\neq 0$. For example take the extension of the same problem in $\mathbb{F}_5$ then the same problem reads $$x^2=0$$,implying only one solution with optimistic repetation of root $0$, two times that is the multiplicity of $0$ in $x^2$. Thankfully, we are lucky enough that we are in the field of $\text{char}$ $0$.The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of $\text{char}$ $0$.

so how do we get x=0 or x=5 ,since we assumed $x\neq 0$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
maxamc
568 posts
#4
Y by
Move this to MSM, reported
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Safal
169 posts
#5
Y by
τρικλινο wrote:
Safal wrote:
2nd Solution is basically wrong. Why? Here is the explanation.

$$x(x-5)=0$$Then there are two cases either $x=0$ or $x\neq 0$.When we are admitting the case $x=0$ we cannot divide by $0$. So, in the case we apply divison by $x$ then $x\neq 0$ is a solid prerequisite to do so.Thus, $x-5=0$ from $x(x-5)=0$ we must take the assumption in hand that $x\neq 0$. For example take the extension of the same problem in $\mathbb{F}_5$ then the same problem reads $$x^2=0$$,implying only one solution with optimistic repetation of root $0$, two times that is the multiplicity of $0$ in $x^2$. Thankfully, we are lucky enough that we are in the field of $\text{char}$ $0$.The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of $\text{char}$ $0$.

so how do we get x=0 or x=5 ,since we assumed $x\neq 0$.

If you read carefully I haven't said that we cannot get $x=0$.The assumption whenever $x\neq 0$ we get $x=5$ else we get the case $x=0$.I can explain you more but the fact is I cannot use argument of field theory to explain it in total details.The reason why it's actually the case lies in field theory logics.
This post has been edited 1 time. Last edited by Safal, Apr 13, 2025, 6:23 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
τρικλινο
502 posts
#6
Y by
what is field theory logic.
IS the logic that suports the development of field theory?
THIS post should not be moved to Middle School Math
Because in the 2nd solution we have the answer : x different than zero this implies x=5
And according to logic this is equivelant to x=0 or x=5.Hence no solution is lost as the book claims
There for it should be removed back to at least college algebra although i doupt if even there anyone knew of that solution
WEmake use of the law of propositional calculus: ¬p implies q this is equivelant to p or q
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Safal
169 posts
#7
Y by
"If you judge a fish because it cannot climb a tree , it will be foolish"-Unknown.

I am not commenting further in this post thank you.

Thanks to aops for moving it to MSM and I support it.
This post has been edited 1 time. Last edited by Safal, Apr 13, 2025, 4:51 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SpeedCuber7
1837 posts
#8
Y by
@triklino dude that's an awesome username i didn't even know greek letters were allowed lol
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sadas123
1262 posts
#9
Y by
Proof: $x^2-5x=0$ Which means that the roots of this equation have to be real so we can use an method that is lost in the darkness called factoring. We can factor out the $x$ from each of the terms on the left hand side and get $x(x-5)=0$ which with more logic we can find that the possible outcomes is that if the Parantheses are 0 or the x=0 so first we can subsitute a value of x into that to make the value 0 so we get that x=5 and we finally get the solutions of $x=5$ and $x=0$ and to wrap up our proof we can prove that factoring is the best way to go because with quadratics you would only find 2 possibiliteis or 1 depending on the plus minus. And the other thing is that if you divide by x and just solve it with algebra then you will only get the solution of 5. Thus, proving that factoring is the best method out of all of them. We can use the remainder theorem to prove this which can be done easily. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
τρικλινο
502 posts
#10
Y by
please read the previous posts


The question here is not which is the best method to solve the problem,but if we lose a solution if we solve the problem by dividing the equation by a non zero x
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Safal
169 posts
#13
Y by
Dear Triklino , I am high yesterday I am sorry for being rude. Can you please explain your Question properly that is what the thing you exactly want to know. According to what I understand, you wanted to know why in 2nd Solution $x=0$ is lost? right. Well forget about field theory and all that, Let me explain it in layman's term what is actually happening. In second solution , the solution $x=0$ is not actually lost. The reason we are getting $x=5$ but not $x=0$ is beacuse when we are dividing by $x$ we making an assumption that $x\neq 0$ and since we are making this assumption the solution $x=0$ is lost. For example when we divide by $x-5$ the solution $x=5$ is lost why $x-5=y(say)$ and we are assuming $y\neq 0$ which is equivalent to $x\neq 5$. Now divison by zero is not possible which is not at all very easy to explain. Now $x=5$ and $x=0$ is not possible at the same time. Thus either $x=0$ or $x=5$.

Now why I am talking about field beacuse $0$ and $5$ can be same when we are in a field of $\text{char} 5$. If you are avoid knowing what is field that's perfectly fine to learn later, but just in layman's term note that $0=5$ is possible in finite fields of charecteristic $5$.

Well I like sour grapes but fox will be happy if he clear your doubt thanks.

I hope it is clear now. If it is not then text me in dm.
Z K Y
N Quick Reply
G
H
=
a