Post by: Bri MT on July 04, 2018, 05:26:48 pm
Quite a few of you have expressed interest in pursuing maths at a more difficult level than is currently offered to you, so I thought I'd make this thread as:
a) a place to share how you have extended your mathematical studies
b) a discussion thread for anything & everything related to your math journeys
c) a place to share interesting problems/resources that you've come across
Here are some introductory questions to kick things off:
1) What's your favourite aspect of maths?
2) Do any key moments stand out to you so far?
3) How far do you plan on going with maths? Do you see it playing a key role in your future?
4) What's your favourite form of mathematical learning?
Post by: turtlesforeveryone on July 05, 2018, 06:40:33 pm
I guess I'll start. Math is appealing to me in many ways. In the technical sense, it's a useful tool as it is essentially the formal application of logic. This means math can be universally used and understood, as we humans have an innate sense of logic. Thus, math is also a way to express our logical thoughts and arguments. This also leads to another neat application: math provides a platform to translate our world into just a few equations. It allows us to predict the world and understand it better (for example, someone from the US and china could both look at the same set of equations, and immediately know what it is describing). Even more amazingly, we have had many discoveries in math that had seemed just theoretical, but later (sometimes even centuries later) would be found to be the exact tool that fits a scenario in physics, engineering, or the life sciences. In some mystical way, the mathematical application is often discovered before the physical application, but always seems to predict the usage.
Personally, I like math mostly because math is beautiful. Attempting math problems gives you a sense of completion, because in many ways math is very black and white: it is either right or wrong. This allows us to prove or disproves claims efficiently (however there are quite a lot of problems that are too complicated to solve). In school, math is a list of steps, while in higher math, it is more of an art - an exploration. So I look forward to when my math education changes from rigid, uncreative processes to a more interesting and fun task.
I'm not sure what role math will play in my future yet, I'm just eager to learn and explore. So looking forward to how this thread grows!
(This year I tried starting up a math and science club in my school but sadly not a lot of people were interested and the teachers I tried to work with stopped communication, so it never got up and running :( I just wanted some people to talk with...)
Post by: Bri MT on July 20, 2018, 10:37:18 am
snip
Love your username!
I'm glad that you've really connected to this thread & in a broader sense that you have such a keen appreciation and interest in maths - as you've recognised, it truly is adaptive and relevant and relevant to every one.
Congratulations for actually taking the initiative to create a new club and invest in the communication necessary for that. It sucks that this time you didn't get to achieve the vision you had in mind, but I hope that this doesn't discourage you from reaching out in the future. In my course we talk about collecting as many "no"s as possible, because if you never fail and no one ever says "no" that just means you should be aiming higher and you aren't stretching your limits enough.
Nevertheless, I hope that you continue to connect here on atarnotes and that we can help you find some of the community you seek :)
Post by: turtlesforeveryone on July 21, 2018, 07:02:10 pm
(will be continually updated)
Youtube Channels
Vihart: Creative math through doodles
Numberphile: Videos featuring mathematicians talking about many interesting areas and problems in math.
3Blue1Brown: 3Blue1Brown provides explanations driven by animations, making difficult problems simple to understand, with changes in perspective.
PBS Infinite Series: "Ambitious content for viewers that are eager to attain a greater understanding of the world around them . . .With each episode you’ll begin to see the math that underpins everything in this puzzling, yet fascinating, universe."
Yaymath: Math videos filmed in a live classroom. Covers many high school math topics in a fun, inviting school atmosphere.
Eddie Woo: Math videos filmed in a classroom in a public school in Sydney, Australia. "I think learning is awesome, and love being able to share what I've learned with others!" Here's an introduction to who Eddie Woo is and what he does.
Think Twice: Elegant geometric proofs shown visually through animations.
Mathloger: "Enter the world of the Mathologer in which beautiful math(s) rules." Fun little explorations into math.
PatrickJMT: Our favourite online math teacher. "Free math videos for the world".
Educational Websites
Maths Is Fun: "We offer mathematics in an enjoyable and easy-to-learn manner, because we believe that mathematics is fun". The site is a great resource for simply and intuitively explained topics from Kindergarten to Year 12.
Brilliant: "Math and Science done right". Great for interactive learning, problem solving, and creative thinking.
NRICH: Provides free interesting mathematical games, problems and articles.
Paul's Online Math Notes: Provides a complete set of free online (and downloadable) notes and/or tutorials for classes that the author teaches at Lamar University. Topics cover Algebra, Calculus I, II and III, Differential Equations, and reviews. This site also provides cheat sheets such as Algebra, Trig, Calculus, and Laplace Transform cheat sheets.
Mathigon: "Textbooks come to life!" Mathigon provides a fun and interactive way to learn new mathematics. At every step students have to actively participate, explore, and discover new ideas. Mathigon encourages engaging through problem solving, reasoning and creativity. Every course is filled with colourful illustrations, puzzles, animations and real-life applications.
Cut The Knot: "An encyclopedic collection of math resources for all grades. Arithmetic games, problems, puzzles, and articles."
Purplemath: A good resource for learning many aspects of math. Purplemath is often cited, and is frequently referenced in google searches.
Plus magazine: "Plus is an internet magazine which aims to introduce readers to the beauty and the practical applications of mathematics. Plus provides articles and podcasts on many aspects of math, a news section, showing how recent news stories were often based on some underlying piece of maths that never made it to the newspapers, reviews of popular maths books, and puzzles for you to sharpen your wits.
Stat Trek: Online tutorials and tools to help you learn statistics.
Art of Problem Solving: This is a must for people who wish to undertake mathematical thinking, or are preparing for a math competition.
Better Explained: Math concepts explained in often a new or intuitive way. Gives you something to think about, covers everything from number systems, to calculus, to computer science.
Puzzles/Problems
Project Euler: Project Euler is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Although mathematics will help, the use of a computer and programming skills will be required to solve most problems.The intended audience includes students for whom the basic curriculum is not feeding their hunger to learn.
mathpuzzle: A site dedicated to posting math puzzles and mathematical recreations, inspired by Martin Gardner (who you may recognise from his section in the Scientific American, titled Mathematical games).
Crux Mathematicorum: "Crux Mathematicorum is an internationally respected source of unique and challenging mathematical problems published by the CMS. Designed primarily for the secondary and undergraduate levels, and also containing some pre-secondary material, it has been referred to as 'the best problem solving journal in the world'."
Entertainment Websites
Coolmath Games: A bunch of kid-friendly non-violent logic and math related games. Categories include strategy, skill, numbers, logic, and trivia. This used to be the go-to game website us kids would log on at school during primary.
Math with bad drawings: Math blog by a person who loves math but is bad at drawing.
r/badmathematics: "A place for sharing the bad math that plagues reddit and the internet as a whole." A good laugh at how people misuse or misunderstand mathematics.
What's Special About This Number?: A distinctive fact about each number from 1 to 9999.
The Geometry Junkyard - Origami: Lots of mathematical origami designs.
PDFs and books
How To Solve It by George Pólya: A short volume underlining different approaches to solving a mathematical problem. The ideas in this volume are so useful that they can be used not just for solving mathematical problems, but for solving any problem in any field.
How to write proofs: a quick guide by Eugenia Cheng: Short volume covering topics such as: What does writing a proof look like, the general shape of a proof, and common bad ways to write proofs.
The Art and Craft of Problem Solving by Paul Zeitz: Good for a breather in competition mathematics, but is aimed a bit higher than the lower level competitions. Lots of interesting mind exercises and mathematical problems.
A bit of everything
What's New: Terrence Tao's blog. The side bar has links to other math related and non-math related websites. Terrence Tao covers such a wide range of topics that it's hard to describe simply. This is an entry on career advice in math, covering primary school to post-doctoral level.
Math3ma: Tai-Danae Bradley's blog, originally created as a tool to help her transition from undergraduate to graduate level mathematics. Some topics covered include category theory, complex analysis, topology, set theory, and much more. Most entries are either a brief (and mostly non-technical) introduction to the topic, or an elaboration of the basic idea via mathematics.
Tao Manifesto: Why choose Tau.
Pi Day and The Pi Manifesto: Why choose Pi.
The Math Forum: The Math Forum has a rich history as an online hub for the mathematics education community. It contains such sections like Ask Dr. Math, the year game, and problems of the week.
Project Euclid: Project Euclid's mission is to provide powerful, low-cost online hosting and publishing services for theoretical and applied mathematics. It provides free online access to scholarly articles and published journals.
Mathematics Stack Exchange: A question and answer forum. You can browse through answers, or look for answers to your own questions. Covers a wide range of topics.
Gower's Weblog: Sir Timothy Gower - a fields medal recipient - 's blog.
The unfinished PDE coffee table book: "During 2000-2001 a group project based in the Oxford University was begun to write this book. Many people at Oxford and around the world contributed drafts, which were then extensively rewritten and edited to help bring about a uniform style and mathematical level. Unfortunately, the project is stalled, with no plans at present to complete it." This project is still very beautiful, despite being unfinished.
Mathblog: Mathblog.com is dedicated to promoting the beauty of Mathematics at every level. It covers many topics, including applied math, basic math (arithmetic, geometry, algebra, calculus etc), math education, and statistics.
Mathvault: "Resource hub for people pursuing higher mathematics through digital publishing and other nerdy gimmicks."
Desmos: Graph functions, plot data, evaluate equations, explore transformations, and more. It's like the ultimate graphing tool.
Math pages by Stan Brown: Articles covering how to use a graphing calculator, how to show your work and succeed as a student in math, how to teach math, and more topics such as algebra, trig, statistics, and calculus.
Mathsnacks: Collection of pdfs containing small packets of beautiful mathematics. Perfect for posters.
Post by: turtlesforeveryone on August 17, 2018, 09:29:03 pm
Well recently I had gone to a bridge building competition and managed to win a prize, despite this being our school's first year participating. I'm quite happy - we took a look at the different types of bridges and analysed them mathematically to decide on our design.
Post by: SChMurpel on August 17, 2018, 09:37:37 pm
In maths, she said 2 + 7 = 8 -_-
And always say "never question my judgment, it's always right"
Post by: Bri MT on August 18, 2018, 08:19:35 am
Well this thread has gone a bit dead :0. Still happy to share my interests in math with anyone else!
Well recently I had gone to a bridge building competition and managed to win a prize, despite this being our school's first year participating. I'm quite happy - we took a look at the different types of bridges and analysed them mathematically to decide on our design.
I've been to one of these contests before, could you elaborate on how they work?
My maths teacher is so trash, Idek why she's the year 9 co-ordinator
In maths, she said 2 + 7 = 8 -_-
And always say "never question my judgment, it's always right"
Everyone makes careless errors now and then, sounds exactly like the type of careless thing I would do in a test :P
It seems a bit odd that she'd say that, could she have been joking?
Post by: turtlesforeveryone on August 18, 2018, 11:34:12 am
I've been to one of these contests before, could you elaborate on hire they work?
I don't want to elaborate too much because it can be traced back to me, but basically every school could register a team of three to participate. We would be mailed a set of materials and could only use those materials (balsa wood, cardboard, string, and glue). Then we had about 6 weeks to make the final project. It would be tested using a machine placed in the middle of the bridge (btw we got a full day to scienceworks for free, because the competition was hosted there).
Post by: SChMurpel on August 18, 2018, 09:58:26 pm
Everyone makes careless errors now and then, sounds exactly like the type of careless thing I would do in a test :PShe's like so serious
It seems a bit odd that she'd say that, could she have been joking?
She said to my class you guys are all gonna fail VCE maths (just because one person doesn't do homework or paste in 20 pages of notes)
Post by: aspiringantelope on December 04, 2018, 03:46:45 pm
Here are some introductory questions to kick things off:
1) What's your favourite aspect of maths?
2) Do any key moments stand out to you so far?
3) How far do you plan on going with maths? Do you see it playing a key role in your future?
4) What's your favourite form of mathematical learning?
1) My favourite aspect of Maths is knowing that you've gotten the answer correct, giving you the satisfaction.
2) Um, one key moment stood out when I knew I had got some of the last 5 AMC Intermediate Questions correct.
3) I just feel that Math is something I am strong at, but don't see it going to play a major in my future life.
4) Does it mean my favourite topic in math or how I learn it?
My favourite topics are Algebra, Trigonometry and Quadratics (easiest for ME)
And I learn simply by practising and consolidating formulas.
Post by: S200 on December 04, 2018, 10:44:42 pm
this is particularly useful for Trig identities and Integration...
Post by: AlphaZero on December 04, 2018, 11:24:31 pm
I just finished my first year at University of Melbourne. Like many of you here, I've always loved maths and it's been a passion of mine ever since I can remember.
Since a few of you have taken interest in really extending/challenging yourselves, upon request, I'm more than happy to write some challenging questions for you all to work on together as a group (then, this will really become a maths club ;D).
Let me know if this is something you guys are interested in. I'm happy to post, say, one or two challenging questions per week :)
Post by: aspiringantelope on December 05, 2018, 05:13:43 pm
Hey 2021ers,Yes! Please post the challenging questions!!
I just finished my first year at University of Melbourne. Like many of you here, I've always loved maths and it's been a passion of mine ever since I can remember.
Since a few of you have taken interest in really extending/challenging yourselves, upon request, I'm more than happy to write some challenging questions for you all to work on together as a group (then, this will really become a maths club ;D).
Let me know if this is something you guys are interested in. I'm happy to post, say, one or two challenging questions per week :)
^May I suggest instead learning from first principles?Hey I just saw your post, may I know what you mean by first principles? Does it cover it in the Cambridge Math Methods 1/2 Textbook? Or do I have to do some online research on it. Thanks.
this is particularly useful for Trig identities and Integration...
Mod edit: Merged double post; please make sure not to double post as it is against forum rules.
Post by: AlphaZero on December 06, 2018, 12:15:55 am
Solve the following questions without the use of a calculator.
Question 1
Derive the quadratic equation. That is, show that the solutions of the equation \(ax^2+bx+c=0\) for \(x\) are given by \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.\]
Question 2
Given that \(\log_2\left(\dfrac{1}{10}\right)=-3.322\), correct to three decimal places, evaluate \(\log_2\left(\dfrac{2}{5}\right)\) correct to three decimal places.
Question 3
Three identical circles of diameter \(12\ \text{cm}\) are placed tangent to each other as shown in the diagram below.
(https://i.imgur.com/crkY6tE.jpg)
Find the exact area, in \(\text{cm}^2\), of the shaded region. Express your answer in the form \(a\sqrt{b}+c\pi\), where \(a\), \(b\) and \(c\) are integers \(\left(a,b,c\in\mathbb{Z}\right)\).
(https://i.imgur.com/crkY6tE.jpg)
Find the exact area, in \(\text{cm}^2\), of the shaded region. Express your answer in the form \(a\sqrt{b}+c\pi\), where \(a\), \(b\) and \(c\) are integers \(\left(a,b,c\in\mathbb{Z}\right)\).
Post by: aspiringantelope on December 06, 2018, 10:25:04 am
Alrighty. Look like we're doing this. First, I'm going to give you a few short questions. I just want to find out where you guys are at in maths. I have no idea if these are too easy, too hard or just right, so let me know :)Ok using LaTex is complicating so I will just try with a equation copier LOL
Solve the following questions without the use of a calculator.
1. \(x^2+\frac{bx}{a}+\frac{c}{a}=\(
2. \(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}{a}=0\)
3. \(\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac{c}{a}\)
4. \(\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}\)
5. \(\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\)
6. \(x+\frac{b}{2a}=+\ or\ -\sqrt{\frac{b^2-4ac}{4a^2}}\) *anyone know how to write the + with a - sign underneath? thanks*
7. \(x+\frac{b}{2a}=\frac{\sqrt{b^2-4ac}}{2a}\)
8. \(x=-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}\)
9. \(x=\frac{-b+or\ -\sqrt{b^2-4ac}}{2a}\)
Please anyone know how to do +/- sign?
Thanks
q2. haven't learn logarithms yet.. (I have but fully forgotten because I am doing the textbook form the start to end)
q3 i dont know either omg this is embarrassing :-[
Post by: Bri MT on December 06, 2018, 10:45:15 am
Ok using LaTex is complicating so I will just try with a equation copier LOL
1. \(x^2+\frac{bx}{a}+\frac{c}{a}=\(
2. \(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}{a}=0\)
3. \(\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac{c}{a}\)
4. \(\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}\)
5. \(\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\)
6. \(x+\frac{b}{2a}=+\ or\ -\sqrt{\frac{b^2-4ac}{4a^2}}\) *anyone know how to write the + with a - sign underneath? thanks*
7. \(x+\frac{b}{2a}=\frac{\sqrt{b^2-4ac}}{2a}\)
8. \(x=-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}\)
9. \(x=\frac{-b+or\ -\sqrt{b^2-4ac}}{2a}\)
Please anyone know how to do +/- sign?
Thanks
You might want to see this LaTex guide by RuiAce if you are interested in learning to type out LaTex on the forums :) . Unfortunately I couldn't see the +/- symbol there, but hopefully someone else is able to help you with that
One of my issues with maths is that I don't explain things & show my working as well as I should a lot of the time, to help you avoid my bad habits, I'd encourage you to explain your first line more :)
Nothing embarrassing about not understanding right away at all
small hints
For 2, I'd encourage you to look at the log laws (there should be a list of them in your textbook/notes) and think about the relationship between 1/10 and 2/5
For 3, what does having the diameter allow you to find? Can you use subtraction?
For 3, what does having the diameter allow you to find? Can you use subtraction?
Post by: AlphaZero on December 06, 2018, 12:03:36 pm
Also, if you're on Google chrome (not sure about other browsers), you should be able to right-click any LaTeX display and show its source code under "show as... TeX commands".
Also, there is nothing embarrassing about not understanding or not knowing. It is embarrassing however if one doesn't at least make the effort to investigate, research and try to understand. Not knowing something just means there's something to learn.
I expect the questions I give to stump most of you. I'd be upset of you guys got these on first try. My goal is to make you curious about maths so that you can become an independent learner.
Post by: aspiringantelope on December 06, 2018, 07:31:50 pm
You might want to see this LaTex guide by RuiAce if you are interested in learning to type out LaTex on the forums :) . Unfortunately I couldn't see the +/- symbol there, but hopefully someone else is able to help you with thatThanks, for first step, I divided everything by "a" but I think the 0 got deleted, then I used the completing the square method for second step.
One of my issues with maths is that I don't explain things & show my working as well as I should a lot of the time, to help you avoid my bad habits, I'd encourage you to explain your first line more :)
Nothing embarrassing about not understanding right away at allsmall hintsFor 2, I'd encourage you to look at the log laws (there should be a list of them in your textbook/notes) and think about the relationship between 1/10 and 2/5
For 3, what does having the diameter allow you to find? Can you use subtraction?
Question two log one
\(\log _2\left(\frac{2}{5}\right)=\log _2\left(2\right)-\log _2\left(5\right)\)
\(1-\log _2\left(5\right)\) Given that loga(a) = 1
How do I do the rest without calculator?
Post by: Bri MT on December 06, 2018, 08:12:32 pm
Thanks, for first step, I divided everything by "a" but I think the 0 got deleted, then I used the completing the square method for second step.No worries :)
Question two log one
\(\log _2\left(\frac{2}{5}\right)=\log _2\left(2\right)-\log _2\left(5\right)\)
\(1-\log _2\left(5\right)\) Given that loga(a) = 1
How do I do the rest without calculator?
For any "given that" question, you should use the provided information to find your answer. In this case, that's the equation with the other log in it.
things to think about
How are 1/5 and 1/10 similar?
From that, how can you use the provided log to help you find the answer?
From that, how can you use the provided log to help you find the answer?
Hope this helps :)
Post by: fun_jirachi on December 06, 2018, 08:16:00 pm
Thanks, for first step, I divided everything by "a" but I think the 0 got deleted, then I used the completing the square method for second step.
Question two log one
\(\log _2\left(\frac{2}{5}\right)=\log _2\left(2\right)-\log _2\left(5\right)\)
\(1-\log _2\left(5\right)\) Given that loga(a) = 1
How do I do the rest without calculator?
Spoiler
Just a hint; start with the 1/10 and manipulate from there, not the 2/5. This is probably one of the cases where you don't want to be working towards something you already have. You can, but it's a lot harder to see. ;)
EDIT: miniturtle is already in, but this is going up anyway (the more ways to think about something the better :))
Post by: aspiringantelope on December 07, 2018, 09:04:23 pm
\(\log_2\left(\frac{2}{5}\right)=\log_2\left(\frac{4}{10}\right)\)
idk -____-
I'm only starting the Math Methods 1/2 book, and haven't prone deeply into logarithms so I cannot answer this question I'm afraid. Maybe in late January I will be able :)
Post by: Aaron on December 07, 2018, 09:06:53 pm
Remember everyone - to encourage users to demonstrate understanding with their questions.... it will help them in the long run! 8) If all they post is a question, prompt/challenge them: "what do you think the first step is?", "how would you start it off", "what do you have so far?" (examples).... and if you're somebody who posts questions and is asking something, consider posting your working out/what you have so far (working out/thoughts/explanation etc) to help diagnose your actual issue.
Post by: Bri MT on December 07, 2018, 09:44:54 pm
\(\log_2\left(\dfrac{1}{10}\right)=-3.322\)
\(\log_2\left(\frac{2}{5}\right)=\log_2\left(\frac{4}{10}\right)\)
idk -____-
I'm only starting the Math Methods 1/2 book, and haven't prone deeply into logarithms so I cannot answer this question I'm afraid. Maybe in late January I will be able :)
There's no shame in not knowing how to do something - the pity is when people don't recognise that & learn.
I think it's fantasic that you're persevering and working through this, and I'm glad you're taking a growth approach to this :)
you're almost there
You now have what you're looking for (\(\log_2\left(\dfrac{2}{5}\right)\)) in terms of something you know (\(\log_2\left(\dfrac{1}{10}\right)\)).
Since we don't know what \(\log_2\left(\dfrac{4}{10}\right)\) is, we want to find a log law that lets us write
\(\log_2\left(\dfrac{4}{10}\right)\) = \(\log_2\left(\dfrac{1}{10}\right)\) (a plus or minus or multiplication or something) (another log)
From there, there are 2 different ways you can approach the next step
Since we don't know what \(\log_2\left(\dfrac{4}{10}\right)\) is, we want to find a log law that lets us write
\(\log_2\left(\dfrac{4}{10}\right)\) = \(\log_2\left(\dfrac{1}{10}\right)\) (a plus or minus or multiplication or something) (another log)
From there, there are 2 different ways you can approach the next step
Love your work miniturtle.Thanks Aaron :)
Remember everyone - to encourage users to demonstrate understanding with their questions.... it will help them in the long run! 8) If all they post is a question, prompt/challenge them: "what do you think the first step is?", "how would you start it off", "what do you have so far?" (examples).
Definitely agree that it's more beneficial for users to show their thoughts & what they've tried, than only getting a suggested solution
Post by: turtlesforeveryone on December 07, 2018, 10:05:12 pm
For question 3:
We are given three identical circles with diameter 12. From the center of each circle we draw a line to the center of each other circle, forming an equilateral triangle with side length 12cm, and thus height 6√3 (As per the 1:2:√3 ratio of a 30:60:90 triangle, which this is because the triangle formed from the centers of three identical circles has an interior angle of 60 degrees.) To find the area we subtract three identical sectors from the overall area of the triangle, which is 6*6√3 = 36√3. The area of each sector is π(6^2)*1/6 = 36π*1/6 = 6π. 6π*3 = 18π. Therefore the area of the shaded region is 36√3 - 18π.
Post by: AlphaZero on December 07, 2018, 11:09:50 pm
Question 1 Solution
Question 2 Solution
Question 3 Solution
Well done guys. I'll post 3 more questions soon :)
Post by: GMT. -_- on December 07, 2018, 11:36:40 pm
log2(2/5)= log2(4)-log2(10)
= 2- log2(10)
= 2 + log2(10^-1)
= 2 - 3.322
= -1.322
Post by: AlphaZero on December 08, 2018, 01:28:26 am
Some Background Theory
A very large part of mathematics is being able to prove statements. But what are proofs? A mathematical proof is a rigorous argument made to convince other people of the absolute truth of a statement.
A big part of being able to understand and prove statements is understanding logic at its core. Unfortunately, the English language can be vague and often, words can have multiple meanings, leading to ambiguity. Thus, we need to give very precise meanings to the words we use when presenting a proof.
In mathematics, we study statements. They are either true, or false, but never both. For example, the statement "6 is an even integer" is true, and the statement "4 is an odd integer" is false. Often, '\(p\) and '\(q\)' are used to denote statements.
To combine and modify statements, we use what are called logical operators. The basic ones are NOT, AND, OR and IF... THEN. Here, we will give precise definitions of these.
NOT: If \(p\) is a statement, then the statement 'not \(p\)' is defined to be
> true, when \(p\) is false;
> false, when \(p\) is true.
AND: If \(p\) and \(q\) are statements, then the statement '\(p\) and \(q\)' is defined to be
> true, when \(p\) and \(q\) are both true;
> false, when \(p\) is false, or \(q\) is false, or both \(p\) and \(q\) are false.
OR: If \(p\) and \(q\) are statements, then the statement '\(p\) or \(q\)' is defined to be
> true, when \(p\) is true, or \(q\) is true, or \(p\) and \(q\) are both true (or in other words, at least one of \(p\) and \(q\) are true);
> false, when \(p\) and \(q\) are both false.
Note: in English, we sometimes use 'or' in the sense that the statement '\(p\) or \(q\)' is true when either \(p\) is true, or \(q\) is true, but not both. In mathematics, this will never be the case. In fact, we give a new word to that modifier, but we won't get into that here.
IF... THEN: If \(p\) and \(q\) are statements, then the statement 'If \(p\) then \(q\)' is defined to be
> true, when \(p\) and \(q\) are both true, or \(p\) is false;
> false, when \(p\) is true and \(q\) is false.
Note: the 'If... then' statement can be quite confusing at first, especially in trying to understand why the 'If... then' is always true whenever \(p\) is false. However, when \(p\) is false, we say that the 'If... then' statement is vacuously true. (Feel free to ask me anything about that).
Okay, now we just need a few more tools, and then you should be ready to do the questions.
Let's develop some number and set theory. A set is a collection of objects. The objects in a set are called elements. Sets are always denoted by curly brackets. For example, \(\{2,\ 4,\ 6, 8\}\) is a set. It contains the elements 2, 4, 6 and 8. To say that an object is contained in a set, we use the element symbol: \(\in\). For example, \(6\in \{2,\ 4,\ 6, 8\}\), but \(5\notin \{2,\ 4,\ 6, 8\}\).
There are some very specific number sets we would like to make use of. You have probably heard of these before, but not seen definitions of these.
Natural Numbers: \(\mathbb{N}=\{1,\ 2,\ ...\}\) These are the counting numbers starting from 1.
Integers: \(\mathbb{Z}=\{...\ -1,\ 0,\ 1,\ ...\}\) These include the natural numbers, their negatives and 0.
Rational Numbers: \(\mathbb{Q}=\left\{a/b\ |\ a,b\in\mathbb{Z}\ \text{and}\ b\neq 0\right\}\) These are all the possible numbers \(a/b\) such that \(a\) and \(b\) are integers and \(b\) is not 0.
Real Numbers: \(\mathbb{R}\) (its construction is too difficult to develop here) These are all numbers on the real number line, including ones that are irrational like \(\pi\) and \(\sqrt{2}\).
This should be enough. I've ordered the questions in such a way so that you are led to the answer :)
A big part of being able to understand and prove statements is understanding logic at its core. Unfortunately, the English language can be vague and often, words can have multiple meanings, leading to ambiguity. Thus, we need to give very precise meanings to the words we use when presenting a proof.
In mathematics, we study statements. They are either true, or false, but never both. For example, the statement "6 is an even integer" is true, and the statement "4 is an odd integer" is false. Often, '\(p\) and '\(q\)' are used to denote statements.
To combine and modify statements, we use what are called logical operators. The basic ones are NOT, AND, OR and IF... THEN. Here, we will give precise definitions of these.
NOT: If \(p\) is a statement, then the statement 'not \(p\)' is defined to be
> true, when \(p\) is false;
> false, when \(p\) is true.
AND: If \(p\) and \(q\) are statements, then the statement '\(p\) and \(q\)' is defined to be
> true, when \(p\) and \(q\) are both true;
> false, when \(p\) is false, or \(q\) is false, or both \(p\) and \(q\) are false.
OR: If \(p\) and \(q\) are statements, then the statement '\(p\) or \(q\)' is defined to be
> true, when \(p\) is true, or \(q\) is true, or \(p\) and \(q\) are both true (or in other words, at least one of \(p\) and \(q\) are true);
> false, when \(p\) and \(q\) are both false.
Note: in English, we sometimes use 'or' in the sense that the statement '\(p\) or \(q\)' is true when either \(p\) is true, or \(q\) is true, but not both. In mathematics, this will never be the case. In fact, we give a new word to that modifier, but we won't get into that here.
IF... THEN: If \(p\) and \(q\) are statements, then the statement 'If \(p\) then \(q\)' is defined to be
> true, when \(p\) and \(q\) are both true, or \(p\) is false;
> false, when \(p\) is true and \(q\) is false.
Note: the 'If... then' statement can be quite confusing at first, especially in trying to understand why the 'If... then' is always true whenever \(p\) is false. However, when \(p\) is false, we say that the 'If... then' statement is vacuously true. (Feel free to ask me anything about that).
Okay, now we just need a few more tools, and then you should be ready to do the questions.
Let's develop some number and set theory. A set is a collection of objects. The objects in a set are called elements. Sets are always denoted by curly brackets. For example, \(\{2,\ 4,\ 6, 8\}\) is a set. It contains the elements 2, 4, 6 and 8. To say that an object is contained in a set, we use the element symbol: \(\in\). For example, \(6\in \{2,\ 4,\ 6, 8\}\), but \(5\notin \{2,\ 4,\ 6, 8\}\).
There are some very specific number sets we would like to make use of. You have probably heard of these before, but not seen definitions of these.
Natural Numbers: \(\mathbb{N}=\{1,\ 2,\ ...\}\) These are the counting numbers starting from 1.
Integers: \(\mathbb{Z}=\{...\ -1,\ 0,\ 1,\ ...\}\) These include the natural numbers, their negatives and 0.
Rational Numbers: \(\mathbb{Q}=\left\{a/b\ |\ a,b\in\mathbb{Z}\ \text{and}\ b\neq 0\right\}\) These are all the possible numbers \(a/b\) such that \(a\) and \(b\) are integers and \(b\) is not 0.
Real Numbers: \(\mathbb{R}\) (its construction is too difficult to develop here) These are all numbers on the real number line, including ones that are irrational like \(\pi\) and \(\sqrt{2}\).
This should be enough. I've ordered the questions in such a way so that you are led to the answer :)
Question 1
Prove that the statements 'If \(p\) then \(q\)' and 'If [not \(q\)] then [not \(p\)]' are logically equivalent. You should use a table in your answer.
Question 2
An integer \(x\) is said to be even if there exists an integer \(y\) such that \(x=2y\). An integer \(x\) is said to be odd if there exists an integer \(y\) such that \(x=2y+1\).
Prove that for \(x\in\mathbb{Z}\), if \(x^2\) is even then \(x\) is even.
Prove that for \(x\in\mathbb{Z}\), if \(x^2\) is even then \(x\) is even.
Question 3
Prove that \(\sqrt{2}\notin \mathbb{Q}\).
Hint: first, assume that \(\sqrt{2}\) is rational and try to derive a contraction. Note that if 'not \(p\)' is false, then by definition, \(p\) must be true.
Hint: first, assume that \(\sqrt{2}\) is rational and try to derive a contraction. Note that if 'not \(p\)' is false, then by definition, \(p\) must be true.
Post by: turtlesforeveryone on December 08, 2018, 12:06:05 pm
(Also someone please teach me how to write in LaTeX!)
Set 2
Spoiler
Q1) I am not familiar with solving this by truth table, so I will solve it by definition.
p -> q = ~p v q (by the definition of the if... then statement.) By the commutative laws:
~p v q = q v ~p,
which can then be turned into
~q -> ~p.
QED.
~p v q = q v ~p,
which can then be turned into
~q -> ~p.
QED.
Spoiler
Q2) Le'ts assume x is an odd number (proof by contradiction).
This gives:
∴ x must be even.
This gives:
(2y+1)^2
= 4y^2 +4y +1
= 2(2y^2 + 2y) + 1
which is an odd number. Because the question states that x^2 is even, this means x can not be odd.= 4y^2 +4y +1
= 2(2y^2 + 2y) + 1
∴ x must be even.
Spoiler
Q3) Let's say √2 is rational, as in, it can be represented by the ratio m/n where m and n have no common factors.
If m and n are both even, then m/n will not be the smallest ratio that √2 can he expressed as, because 2 and 2 are common factors.
∴√2 is irrational.
√2 = m/n
2 = m^2/n^2
2n^2 = m^2.
Thus m^2 must be an even number, which is only true if m is an even number. However, n^2 = m^2/2, and as m^2 is established to be an even number, n^2 must be even, and n must be even. 2 = m^2/n^2
2n^2 = m^2.
If m and n are both even, then m/n will not be the smallest ratio that √2 can he expressed as, because 2 and 2 are common factors.
∴√2 is irrational.
Edited as per @dantraicos's request.
Post by: AlphaZero on December 08, 2018, 12:38:26 pm
Your solution to question 1 is correct. Now try to use truth tables (this method by the way is called perfect induction).
Your solution to question 2 is also correct, however you need to be careful with the words. Try to describe exactly why your proof works. It helps in convincing others that you're correct :)
Your solution to question 3 is incomplete. You haven't actually derived a contradiction. Initially, you said \(m\) and \(n\) are integers and you concluded that they are even. There is no contradiction in that (even numbers are indeed integers). You're missing a very key phrase in your first sentence.
Edit after you made some changes: Well done. Your answers are now correct. :)
Post by: fun_jirachi on December 08, 2018, 06:27:45 pm
Adding on to what Dan said, those are nice solutions!
If you want to type in LaTeX (this has actually been asked and addressed a whole lot (I know that because I did it too!)), here is the link to Rui's AMAZING guide.
https://atarnotes.com/forum/index.php?topic=165627.0
Post by: AlphaZero on December 18, 2018, 11:08:59 pm
Question 1 Solution
(https://i.imgur.com/zEdjEjg.png)
Hence, by the principle of perfect induction, the statements are equivalent.
Hence, by the principle of perfect induction, the statements are equivalent.
Question 2 Solution
Here, it is sufficient to prove the statement "If \(x\) is odd, then \(x^2\) is odd" (result from Question 1).
If \(x\) is odd, then \(x=2k+1\) for some \(k\in\mathbb{Z}\), and so, \[x^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1.\] If we write \(m=2k^2+2k\in\mathbb{Z}\), the required result follows.
If \(x\) is odd, then \(x=2k+1\) for some \(k\in\mathbb{Z}\), and so, \[x^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1.\] If we write \(m=2k^2+2k\in\mathbb{Z}\), the required result follows.
Question 3 Solution
We will proceed to prove this statement by contradiction.
Suppose that \(\sqrt{2}\in\mathbb{Q}\). Then there exists \(a,b\in\mathbb{Z}\), where \(a\) and \(b\) are coprime, such that \(\sqrt{2}=\dfrac{a}{b}\).
Then, we obtain \(a^2=2b^2\), and so \(a^2\) is even. Since \(a\in\mathbb{Z}\), \(a\) must be even (as proved in Question 2), and so we can write \(a=2m\) for some \(m\in\mathbb{Z}\).
However, the same equation yields \(b^2=2m^2\). Therefore, \(b^2\) is even, and since \(b\in\mathbb{Z}\), \(b\) is even (result from Question 2), meaning we can write \(b=2n\) for some \(n\in\mathbb{Z}\).
But then, \(\sqrt{2}=\dfrac{a}{b}=\dfrac{2m}{2n}=\dfrac{m}{n}\), contradicting the fact that \(a\) and \(b\) are coprime (they both have a common factor - mainly, 2).
Thus, \(\sqrt{2}\notin\mathbb{Q}\).
Suppose that \(\sqrt{2}\in\mathbb{Q}\). Then there exists \(a,b\in\mathbb{Z}\), where \(a\) and \(b\) are coprime, such that \(\sqrt{2}=\dfrac{a}{b}\).
Then, we obtain \(a^2=2b^2\), and so \(a^2\) is even. Since \(a\in\mathbb{Z}\), \(a\) must be even (as proved in Question 2), and so we can write \(a=2m\) for some \(m\in\mathbb{Z}\).
However, the same equation yields \(b^2=2m^2\). Therefore, \(b^2\) is even, and since \(b\in\mathbb{Z}\), \(b\) is even (result from Question 2), meaning we can write \(b=2n\) for some \(n\in\mathbb{Z}\).
But then, \(\sqrt{2}=\dfrac{a}{b}=\dfrac{2m}{2n}=\dfrac{m}{n}\), contradicting the fact that \(a\) and \(b\) are coprime (they both have a common factor - mainly, 2).
Thus, \(\sqrt{2}\notin\mathbb{Q}\).
Let me know guys when/if you want another set :)
Post by: turtlesforeveryone on December 20, 2018, 02:01:14 pm
Set 2 Solutions:Question 1 Solution(https://i.imgur.com/zEdjEjg.png)
Hence, by the principle of perfect induction, the statements are equivalent.Question 2 SolutionHere, it is sufficient to prove the statement "If \(x\) is odd, then \(x^2\) is odd" (result from Question 1).
If \(x\) is odd, then \(x=2k+1\) for some \(k\in\mathbb{Z}\), and so, \[x^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1.\] If we write \(m=2k^2+2k\in\mathbb{Z}\), the required result follows.Question 3 SolutionWe will proceed to prove this statement by contradiction.
Suppose that \(\sqrt{2}\in\mathbb{Q}\). Then there exists \(a,b\in\mathbb{Z}\), where \(a\) and \(b\) are coprime, such that \(\sqrt{2}=\dfrac{a}{b}\).
Then, we obtain \(a^2=2b^2\), and so \(a^2\) is even. Since \(a\in\mathbb{Z}\), \(a\) must be even (as proved in Question 2), and so we can write \(a=2m\) for some \(m\in\mathbb{Z}\).
However, the same equation yields \(b^2=2m^2\). Therefore, \(b^2\) is even, and since \(b\in\mathbb{Z}\), \(b\) is even (result from Question 2), meaning we can write \(b=2n\) for some \(n\in\mathbb{Z}\).
But then, \(\sqrt{2}=\dfrac{a}{b}=\dfrac{2m}{2n}=\dfrac{m}{n}\), contradicting the fact that \(a\) and \(b\) are coprime (they both have a common factor - mainly, 2).
Thus, \(\sqrt{2}\notin\mathbb{Q}\).
Let me know guys when/if you want another set :)
Your solutions are very elegant! I would like another set of problems please, thanks for taking the time to make these :)
Post by: AlphaZero on December 20, 2018, 05:25:46 pm
Some Background Theory
For this set of questions, we are going to discuss mathematical induction.
Mathematical induction is a very powerful tool when it comes to proving statements, especially for proving that a property, \(P(n)\), holds for all \(n\in\mathbb{N}\).
A proof by mathematical induction requires two cases to be proved. The first case is called the base case where we need to prove that the statement holds for the smallest element, usually \(n=1\). The second case is called the induction step where we need to prove that, if the statement holds for some number \(k\), then it holds for the next number \(k+1\).
This explanation can be quite confusing, so it's best explained with a metaphor. Suppose we have an infinitely long ladder and we would like to prove that we can climb up it infinitely. To do this, we consider two cases. We start by proving that we can climb onto the first rung of the ladder (the base case). Then, we prove that from any rung, we can always climb to the next rung (induction step). And then we're done! We've proved we can always climb the ladder infinitely.
Here's a simple example of how we use mathematical induction. Let's prove that \[1+2+\dots+n=\frac{n(n+1)}{2},\quad n\in\mathbb{N}.\]That is, we want to prove the formula for the sum of the first \(n\) natural numbers.
Let's start with the base case: \(n=1\)
\(\text{LHS}=1\), and \(\text{RHS}=\dfrac{1(1+1)}{2}=\dfrac{2}{2}=1\).
Great. We get \(1=1\). Although, not part of the proof, just for our curiosity, let's try it for some other numbers.
\(n=2\):
\(\text{LHS}=1+2=3\), and \(\text{RHS}=\dfrac{2(2+1)}{2}=\dfrac{6}{2}=3\).
\(n=3\):
\(\text{LHS}=1+2+3=6\), and \(\text{RHS}=\dfrac{3(3+1)}{2}=\dfrac{12}{2}=6\).
So, at this point, we might be convinced that this formula is correct. It is, but proving that a property holds for \(n=1,2,3\) isn't a proof. We now need to do the induction step.
Suppose that the statement is true for some \(k\in\mathbb{N}\). That is, let's assume it's true that \[1+2+\dots+k=\frac{k(k+1)}{2}.\]
Now, does the property hold for \(k+1\)? Let's find out.
\begin{align*}\text{LHS}&=1+2+\dots+k+(k+1)\\
&=[1+2+\dots+k]+(k+1)\\
&=\frac{k(k+1)}{2}+(k+1)\quad\quad (\text{the expression in the square brackets is exactly what we assumed!})\\
&=\frac{k(k+1)}{2}+\frac{2(k+1)}{2}\quad\quad (\text{common denominator})\\
&=\frac{(k+1)(k+2)}{2}\quad\quad (\text{factor out }(k+1) )\\
&=\frac{(k+1)\Big((k+1)+1\Big)}{2}\quad\quad (\text{small algebraic manipulation})\\
&=\text{RHS}\mid_{n=k+1},\quad\quad (\text{which is the RHS of the equation if we replaced }n\ \text{with}\ k+1)\end{align*}
And we are done! Let's just write a concluding statement:
Hence, by the principle of mathematical induction, the statement is true.
Anyway, this should be enough background information. The questions are standalone, so each question doesn't require results from the previous question. I've put them in order of easiest to hardest (in my opinion). Best of luck!
Mathematical induction is a very powerful tool when it comes to proving statements, especially for proving that a property, \(P(n)\), holds for all \(n\in\mathbb{N}\).
A proof by mathematical induction requires two cases to be proved. The first case is called the base case where we need to prove that the statement holds for the smallest element, usually \(n=1\). The second case is called the induction step where we need to prove that, if the statement holds for some number \(k\), then it holds for the next number \(k+1\).
This explanation can be quite confusing, so it's best explained with a metaphor. Suppose we have an infinitely long ladder and we would like to prove that we can climb up it infinitely. To do this, we consider two cases. We start by proving that we can climb onto the first rung of the ladder (the base case). Then, we prove that from any rung, we can always climb to the next rung (induction step). And then we're done! We've proved we can always climb the ladder infinitely.
Here's a simple example of how we use mathematical induction. Let's prove that \[1+2+\dots+n=\frac{n(n+1)}{2},\quad n\in\mathbb{N}.\]That is, we want to prove the formula for the sum of the first \(n\) natural numbers.
Let's start with the base case: \(n=1\)
\(\text{LHS}=1\), and \(\text{RHS}=\dfrac{1(1+1)}{2}=\dfrac{2}{2}=1\).
Great. We get \(1=1\). Although, not part of the proof, just for our curiosity, let's try it for some other numbers.
\(n=2\):
\(\text{LHS}=1+2=3\), and \(\text{RHS}=\dfrac{2(2+1)}{2}=\dfrac{6}{2}=3\).
\(n=3\):
\(\text{LHS}=1+2+3=6\), and \(\text{RHS}=\dfrac{3(3+1)}{2}=\dfrac{12}{2}=6\).
So, at this point, we might be convinced that this formula is correct. It is, but proving that a property holds for \(n=1,2,3\) isn't a proof. We now need to do the induction step.
Suppose that the statement is true for some \(k\in\mathbb{N}\). That is, let's assume it's true that \[1+2+\dots+k=\frac{k(k+1)}{2}.\]
Now, does the property hold for \(k+1\)? Let's find out.
\begin{align*}\text{LHS}&=1+2+\dots+k+(k+1)\\
&=[1+2+\dots+k]+(k+1)\\
&=\frac{k(k+1)}{2}+(k+1)\quad\quad (\text{the expression in the square brackets is exactly what we assumed!})\\
&=\frac{k(k+1)}{2}+\frac{2(k+1)}{2}\quad\quad (\text{common denominator})\\
&=\frac{(k+1)(k+2)}{2}\quad\quad (\text{factor out }(k+1) )\\
&=\frac{(k+1)\Big((k+1)+1\Big)}{2}\quad\quad (\text{small algebraic manipulation})\\
&=\text{RHS}\mid_{n=k+1},\quad\quad (\text{which is the RHS of the equation if we replaced }n\ \text{with}\ k+1)\end{align*}
And we are done! Let's just write a concluding statement:
Hence, by the principle of mathematical induction, the statement is true.
Anyway, this should be enough background information. The questions are standalone, so each question doesn't require results from the previous question. I've put them in order of easiest to hardest (in my opinion). Best of luck!
Question 1
Prove, using mathematical induction, that the sum of the first \(n\) odd natural numbers is equal to \(n^2\). That is prove that, \[1+3+\dots+(2n-1)=n^2,\quad n\in\mathbb{N}.\]
Question 2
Prove, using mathematical induction, that \(7^n-4^n\) is divisible by \(3\) for all \(n\in\mathbb{N}\).
Question 3
Prove, using mathematical induction, that \[2\cdot 2^0\;+\; 3\cdot 2^1\;+\;4\cdot 2^2\;+\;\dots\; +\; (n+1)\cdot 2^{n-1}=n\cdot 2^n,\quad n\in\mathbb{N}.\]
Best of luck :)
Post by: turtlesforeveryone on December 20, 2018, 11:04:41 pm
Set 3 Questions: 20 Dec 2018 to 2 Jan 2019Some Background TheoryFor this set of questions, we are going to discuss mathematical induction.
Mathematical induction is a very powerful tool when it comes to proving statements, especially for proving that a property, \(P(n)\), holds for all \(n\in\mathbb{N}\).
A proof by mathematical induction requires two cases to be proved. The first case is called the base case where we need to prove that the statement holds for the smallest element, usually \(n=1\). The second case is called the induction step where we need to prove that, if the statement holds for some number \(k\), then it holds for the next number \(k+1\).
This explanation can be quite confusing, so it's best explained with a metaphor. Suppose we have an infinitely long ladder and we would like to prove that we can climb up it infinitely. To do this, we consider two cases. We start by proving that we can climb onto the first rung of the ladder (the base case). Then, we prove that from any rung, we can always climb to the next rung (induction step). And then we're done! We've proved we can always climb the ladder infinitely.
Here's a simple example of how we use mathematical induction. Let's prove that \[1+2+\dots+n=\frac{n(n+1)}{2},\quad n\in\mathbb{N}.\]That is, we want to prove the formula for the sum of the first \(n\) natural numbers.
Let's start with the base case: \(n=1\)
\(\text{LHS}=1\), and \(\text{RHS}=\dfrac{1(1+1)}{2}=\dfrac{2}{2}=1\).
Great. We get \(1=1\). Although, not part of the proof, just for our curiosity, let's try it for some other numbers.
\(n=2\):
\(\text{LHS}=1+2=3\), and \(\text{RHS}=\dfrac{2(2+1)}{2}=\dfrac{6}{2}=3\).
\(n=3\):
\(\text{LHS}=1+2+3=6\), and \(\text{RHS}=\dfrac{3(3+1)}{2}=\dfrac{12}{2}=6\).
So, at this point, we might be convinced that this formula is correct. It is, but proving that a property holds for \(n=1,2,3\) isn't a proof. We now need to do the induction step.
Suppose that the statement is true for some \(k\in\mathbb{N}\). That is, let's assume it's true that \[1+2+\dots+k=\frac{k(k+1)}{2}.\]
Now, does the property hold for \(k+1\)? Let's find out.
\begin{align*}\text{LHS}&=1+2+\dots+k+(k+1)\\
&=[1+2+\dots+k]+(k+1)\\
&=\frac{k(k+1)}{2}+(k+1)\quad\quad (\text{the expression in the square brackets is exactly what we assumed!})\\
&=\frac{k(k+1)}{2}+\frac{2(k+1)}{2}\quad\quad (\text{common denominator})\\
&=\frac{(k+1)(k+2)}{2}\quad\quad (\text{factor out }(k+1) )\\
&=\frac{(k+1)\Big((k+1)+1\Big)}{2}\quad\quad (\text{small algebraic manipulation})\\
&=\text{RHS}\mid_{n=k+1},\quad\quad (\text{which is the RHS of the equation if we replaced }n\ \text{with}\ k+1)\end{align*}
And we are done! Let's just write a concluding statement:
Hence, by the principle of mathematical induction, the statement is true.
Anyway, this should be enough background information. The questions are standalone, so each question doesn't require results from the previous question. I've put them in order of easiest to hardest (in my opinion). Best of luck!Question 1Prove, using mathematical induction, that the sum of the first \(n\) odd natural numbers is equal to \(n^2\). That is prove that, \[1+3+\dots+(2n-1)=n^2,\quad n\in\mathbb{N}.\]Question 2Prove, using mathematical induction, that \(7^n-4^n\) is divisible by \(3\) for all \(n\in\mathbb{N}\).Question 3Prove, using mathematical induction, that \[2\cdot 2^0\;+\; 3\cdot 2^1\;+\;4\cdot 2^2\;+\;\dots\; +\; (n+1)\cdot 2^{n-1}=n\cdot 2^n,\quad n\in\mathbb{N}.\]
Best of luck :)
My Solution 1
Claim: \(1+3+\dots+(2n-1)=n^2,\quad n\in\mathbb{N}.\)
For the case of \(n=1, 1=1^2\), which is true.
For \(n=k+1,1+3+\dots+(2(k+2)-1) = (k+1)^2 = k^2+2k+1\)
LHS \(= 1+3+\dots+(2k+1)\)
\(=1+3+\dots+(2k-1)+(2k+1) = k^2+2k+1\) << Cancel out 2k+1 from both sides
\(=1+3+\dots+(2k-1)=k^2\) << Which was what we postulated at the start
\(\therefore\) by the principle of mathematical induction, our claim is true for all \(n\in\mathbb{N}.\)
For the case of \(n=1, 1=1^2\), which is true.
For \(n=k+1,1+3+\dots+(2(k+2)-1) = (k+1)^2 = k^2+2k+1\)
LHS \(= 1+3+\dots+(2k+1)\)
\(=1+3+\dots+(2k-1)+(2k+1) = k^2+2k+1\) << Cancel out 2k+1 from both sides
\(=1+3+\dots+(2k-1)=k^2\) << Which was what we postulated at the start
\(\therefore\) by the principle of mathematical induction, our claim is true for all \(n\in\mathbb{N}.\)
My Solution 2
Claim: \(7^n-4^n\) is divisible by \(3\) for all \(n\in\mathbb{N}\)
We will rewrite this as \(7^n-4^n=3m\) where \(m\in\mathbb{N}.\)
For the case of \(n=1, 7-4=3\), which is true.
For \(n = k+1, \text{LHS} = 7^{k+1}-4^{k+1}\)
LHS \(=7(7^k)-4(4^k)\) << by index laws
\(=7(3m+4^k)-4(4^k)\) << using the fact that \(7^k=3m+4^k\)
\(=7(3m)+7(4^k)-4(4^k)\)
\(=7(3m)+(7-4)(4^(4k)\)
\(=3(7m+4^k)\)
As we have shown for the case of \(n=k+1\), the result is still divisible by 3.
\(\therefore\) by the principle of mathematical induction, the claim is true for all \(n\in\mathbb{N}.\)
We will rewrite this as \(7^n-4^n=3m\) where \(m\in\mathbb{N}.\)
For the case of \(n=1, 7-4=3\), which is true.
For \(n = k+1, \text{LHS} = 7^{k+1}-4^{k+1}\)
LHS \(=7(7^k)-4(4^k)\) << by index laws
\(=7(3m+4^k)-4(4^k)\) << using the fact that \(7^k=3m+4^k\)
\(=7(3m)+7(4^k)-4(4^k)\)
\(=7(3m)+(7-4)(4^(4k)\)
\(=3(7m+4^k)\)
As we have shown for the case of \(n=k+1\), the result is still divisible by 3.
\(\therefore\) by the principle of mathematical induction, the claim is true for all \(n\in\mathbb{N}.\)
My Solution 3
Claim: \(2\cdot 2^0\;+\; 3\cdot 2^1\;+\;4\cdot 2^2\;+\;\dots\; +\; (n+1)\cdot 2^{n-1}=n\cdot 2^n,\quad n\in\mathbb{N}.\) (1)
For \(n=1, 2 = (1+1)(2^{1-1}) = 2\), which is true.
For \(n=k+1, 2(2^0)+3(2^1)+\dots+(k+2)(2^k)=(k+1)(2^{k+1})\) (2)
To prove our claim this time, we want to prove that \(f(k+1) = f(k)\). More specifically, \(f(k+1)\) can be manipulated to the same result of \(f(k), k(2^k)\).
\(=2(2^0)+3(2^1)+\dots+(k+1)(2^{k-1})+(k+2)(2^k)=(k+1)(2^{k+1})\)
\(=2(2^0)+3(2^1)+\dots+(k+1)(2^{k-1})=(k+1)(2^{k+1})-(k+2)(2^k)\) << subtract \((k+2)(2^k)\) from both sides. Note how the LHS is the same as (1)
RHS \(=(k+1)(2^{k+1})-(k+2)(2^k)\)
\(=(k)(2^{k+1})+2^{k+1}-(k)(2^k)+2^{k+1}\)
\(=(k)(2^{k+1})-(k)(2^k)\)
\(=(k)(2^k\) << which equals the RHS of (1)
\(\therefore f(k+1)\) is true when \(f(k)\) is true. By the principle of mathematical induction, the claim is true.
For \(n=1, 2 = (1+1)(2^{1-1}) = 2\), which is true.
For \(n=k+1, 2(2^0)+3(2^1)+\dots+(k+2)(2^k)=(k+1)(2^{k+1})\) (2)
To prove our claim this time, we want to prove that \(f(k+1) = f(k)\). More specifically, \(f(k+1)\) can be manipulated to the same result of \(f(k), k(2^k)\).
\(=2(2^0)+3(2^1)+\dots+(k+1)(2^{k-1})+(k+2)(2^k)=(k+1)(2^{k+1})\)
\(=2(2^0)+3(2^1)+\dots+(k+1)(2^{k-1})=(k+1)(2^{k+1})-(k+2)(2^k)\) << subtract \((k+2)(2^k)\) from both sides. Note how the LHS is the same as (1)
RHS \(=(k+1)(2^{k+1})-(k+2)(2^k)\)
\(=(k)(2^{k+1})+2^{k+1}-(k)(2^k)+2^{k+1}\)
\(=(k)(2^{k+1})-(k)(2^k)\)
\(=(k)(2^k\) << which equals the RHS of (1)
\(\therefore f(k+1)\) is true when \(f(k)\) is true. By the principle of mathematical induction, the claim is true.
Whew, this was my first time writing in LaTeX and it sure took a bit of time to complete! There may be typos.
Post by: turtlesforeveryone on December 29, 2018, 04:51:22 pm
What I think is good to learn that directly follows specialist so far includes:
- Calculus (In American terms, Calc 2 and 3, as well as Multivariate Calculus and Applied Calculus)
- Differential Equations (applications as well?)
- Analysis 1 and 2, Real and Complex
- Linear Algebra
- some of the 'theories', Set Theory, Number Theory, Category Theory, Group Theory
- Further math logic and proofs
(I've been using MIT courses as a basis for some of these)
Any other suggestions? The perfect resource for me would be a mathematical tree showing which subjects naturally lead to which and how they are all linked together in terms of some reasonable progression, but after scouring the internet nothing like that seems to be out there.
Post by: RuiAce on December 29, 2018, 04:59:45 pm
In terms of self learning mathematics, what are some good pathways to follow up on, especially beyond the specialist curriculum? (I know this is hard to put into words haha).Whilst you don't need to go into depth with linear algebra (i.e. vector spaces, linear transformations, eigenvalues, ...) you should have a more refined understanding of vectors and matrices before jumping into multivariable calculus. I feel as though specialising into one direction so early on isn't the best idea - you should have foundations in a bit of everything, and then decide on what to specialise in.
What I think is good to learn that directly follows specialist so far includes:
- Calculus (In American terms, Calc 2 and 3, as well as Multivariate Calculus and Applied Calculus)
- Differential Equations (applications as well?)
- Analysis 1 and 2, Real and Complex
- Linear Algebra
- some of the 'theories', Set Theory, Number Theory, Category Theory, Group Theory
- Further math logic and proofs
(I've been using MIT courses as a basis for some of these)
Any other suggestions? The perfect resource for me would be a mathematical tree showing which subjects naturally lead to which and how they are all linked together in terms of some reasonable progression, but after scouring the internet nothing like that seems to be out there.
I prefer a reasonable understanding of vector geometry as well as knowing the cross product and matrices before beginning multivariable calculus (calc III stuff). Differential equations should be taken concurrently with calc II and III content because the techniques overlap, because the easy techniques in DEs only require calc II material whilst the more complicated ones require calc III. After that, you can fill in the gaps with linear algebra.
(Alternatively, do all of linear algebra before continuing with calculus and analysis.)
Discrete math topics (set theory, number theory, logic, ...) can be studied stand-alone. Whilst in practice set theory should be something you should always know, it's also one of the easier things to learn and along with everything else in discrete math, can be learnt whenever you want to.
Technically speaking though, you really shouldn't just put a whole bunch of topics down and be like "what should I do". (I also have a bad feeling that MIT topics are designed in an order suitable for the extremely gfited.) Instead, you can just look up some first year unit outlines (surely UniMelb and Monash have these somewhere) and just follow the structures they outline on their sem 1 courses first.
Post by: turtlesforeveryone on December 29, 2018, 07:22:02 pm
Whilst you don't need to go into depth with linear algebra (i.e. vector spaces, linear transformations, eigenvalues, ...) you should have a more refined understanding of vectors and matrices before jumping into multivariable calculus. I feel as though specialising into one direction so early on isn't the best idea - you should have foundations in a bit of everything, and then decide on what to specialise in.
I prefer a reasonable understanding of vector geometry as well as knowing the cross product and matrices before beginning multivariable calculus (calc III stuff). Differential equations should be taken concurrently with calc II and III content because the techniques overlap, because the easy techniques in DEs only require calc II material whilst the more complicated ones require calc III. After that, you can fill in the gaps with linear algebra.
(Alternatively, do all of linear algebra before continuing with calculus and analysis.)
Discrete math topics (set theory, number theory, logic, ...) can be studied stand-alone. Whilst in practice set theory should be something you should always know, it's also one of the easier things to learn and along with everything else in discrete math, can be learnt whenever you want to.
Technically speaking though, you really shouldn't just put a whole bunch of topics down and be like "what should I do". (I also have a bad feeling that MIT topics are designed in an order suitable for the extremely gfited.) Instead, you can just look up some first year unit outlines (surely UniMelb and Monash have these somewhere) and just follow the structures they outline on their sem 1 courses first.
Thank you! I'll definitely check out the course outlines for Melbourne Uni and Monash, however, they don't seem to be as comprehensive as the online MIT resources which come with free assignments, lectures, and problem sets. I'm already working on some of the broad areas I listed but I guess was just hoping for some other ideas to add breadth and some sort of direction to my mathematics journey. Predominantly I am working on Calc II material together with Differential equations, and will definitely take on the advice you gave in the first paragraph in the future (can you please elaborate on the concepts that overlap with Linear Algebra and Calc III?).
Guess my dream of having a flowchart to visualise mathematics and my progression is still far off :P
Post by: RuiAce on December 29, 2018, 07:48:41 pm
Thank you! I'll definitely check out the course outlines for Melbourne Uni and Monash, however, they don't seem to be as comprehensive as the online MIT resources which come with free assignments, lectures, and problem sets. I'm already working on some of the broad areas I listed but I guess was just hoping for some other ideas to add breadth and some sort of direction to my mathematics journey. Predominantly I am working on Calc II material together with Differential equations, and will definitely take on the advice you gave in the first paragraph in the future (can you please elaborate on the concepts that overlap with Linear Algebra and Calc III?).Basically you won't see any vector spaces and etc. in calc III - that's what makes linear algebra distinguishable from calculus. But as you start progressing into vector calculus (belongs in calc III) you're expected to know all the fundamentals (matrix multiplication, ...) to a more in-depth level than when you first start learning it. All of those concepts are treated as assumed knowledge here.
Guess my dream of having a flowchart to visualise mathematics and my progression is still far off :P
(i.e. Not the further-away concepts in linear algebra. Just a much more refined understanding of the basic tools. A convenient thing about specialist maths is that almost all of the basic tools get introduced, but there's still a lot more refining that can occur.)
A flowchart sounds good and all but once you get deeper into mathematics you actually begin to find that certain areas of mathematics don't go well with you. (Whereas others you'll grow to love.) For example whilst I didn't mind elementary group theory, the further I got into abstract algebra the further I started hating it and eventually was like "nah I'm dropping this unit".
Remember that after you look at all the level 1 (first year) units you can then jump to the level 2 (typically second year) units. After you've done all the level 1 stuff, you can then start building a flow chart with level 2.
Post by: aspiringantelope on December 29, 2018, 09:09:38 pm
In terms of self learning mathematics, what are some good pathways to follow up on, especially beyond the specialist curriculum? (I know this is hard to put into words haha).Wow!!
What I think is good to learn that directly follows specialist so far includes:
- Calculus (In American terms, Calc 2 and 3, as well as Multivariate Calculus and Applied Calculus)
- Differential Equations (applications as well?)
- Analysis 1 and 2, Real and Complex
- Linear Algebra
- some of the 'theories', Set Theory, Number Theory, Category Theory, Group Theory
- Further math logic and proofs
(I've been using MIT courses as a basis for some of these)
Any other suggestions? The perfect resource for me would be a mathematical tree showing which subjects naturally lead to which and how they are all linked together in terms of some reasonable progression, but after scouring the internet nothing like that seems to be out there.
Are you planning to do UMEP Maths?
Post by: turtlesforeveryone on January 23, 2019, 09:44:06 pm
Wow!!
Are you planning to do UMEP Maths?
Most likely :p. Any sort of university extension would be great. You?
Post by: aspiringantelope on February 11, 2019, 05:48:10 pm
Most likely :p. Any sort of university extension would be great. You?Sorry, I have just read this (O.O)
And I don't think I'm going to be doing any university extension courses because there are no universities close to me =[ (However this may change but unlikely because I'd already gotten a rough idea of the 6 subjects I would be doing [>.<]
Post by: MB_ on February 11, 2019, 06:03:16 pm
Sorry, I have just read this (O.O)UMEP maths isn't actually taken at the university, its conducted at various schools. I've attached an image with all the locations, maybe there is something closer for you.
And I don't think I'm going to be doing any university extension courses because there are no universities close to me =[ (However this may change but unlikely because I'd already gotten a rough idea of the 6 subjects I would be doing [>.<]
(https://i.imgur.com/dYohbPG.png)
Post by: aspiringantelope on February 11, 2019, 09:14:05 pm
UMEP maths isn't actually taken at the university, its conducted at various schools. I've attached an image with all the locations, maybe there is something closer for you.OHHH WAITTT LOOOOL ONE OF THEM IS AT MY SCHOOOL LMFAOOOOOO
(https://i.imgur.com/dYohbPG.png)
but I think i'm too dumb for it cause look at turtlesforeveryone!!! He/she's solving these insanely hard questions that I've never learnt/heard of at the same level at me -_____-
I wouldn't call myself THAT excelled in maths anyways XP
Do you think there would be a way to excel at maths and possibly do UMEP? I only got a Distinction in the AMC which is not really that impressing and I'm not even in the Top 5 Math students in my cohort =(
Post by: w0lfqu33n89 on February 17, 2019, 05:19:26 pm
Post by: lacitam on February 17, 2019, 05:39:47 pm
Hey guys! assuming this the right place to post this. I am in year 10 this year and doing general mathematics and feel I have made a mistake and should of done methods. Math isn't my best subject, I struggle quite a bit, I am finding the content easy as and feel like I am wasting what I am capable of. I have been told its a good choice as its one less subject I don't have to worry about but I want to be prepared and knowledgable for my future endeavours. Does General Mathematics year 10 get harder? should I try to move to methods? I am really torn and I have missed the deadline but I feel I can try to move if I am finding it easy as.General maths? As in you're doing year 11 general maths in year 10?
Methods and General/Further aren't that related besides graphs and relations. I personally, back in year 11, found general to be easy, but methods was difficult.
'I have been told its a good choice as its one less subject' isn't really a good thought to have. I spent the most time in general fine-tuning and understanding the concept to perfection because of all the silly mistakes I made.
I suggest you talk to your subject counsellor for more details, but for my two cents, you should choose methods if you're actually capable of doing it (take into consideration that you also did mention that you struggle with maths). Look into the methods 1/2 study design for synopsis.
What do you want to do in uni? perhaps look into the prerequisites as many need at least 25 in methods.
Post by: w0lfqu33n89 on February 17, 2019, 11:45:39 pm
General maths? As in you're doing year 11 general maths in year 10?
Methods and General/Further aren't that related besides graphs and relations. I personally, back in year 11, found general to be easy, but methods was difficult.
'I have been told its a good choice as its one less subject' isn't really a good thought to have. I spent the most time in general fine-tuning and understanding the concept to perfection because of all the silly mistakes I made.
I suggest you talk to your subject counsellor for more details, but for my two cents, you should choose methods if you're actually capable of doing it (take into consideration that you also did mention that you struggle with maths). Look into the methods 1/2 study design for synopsis.
What do you want to do in uni? perhaps look into the prerequisites as many need at least 25 in methods.
my school is quite weird so I am doing year 10 general. the year 10 math are split into 3 classes Foundation; basically you are planning to ditch math and this is the last compulsory year and you basically budge.
general; you go over previously learnt content but more in depth
Methods; new content and quite tricky. I know a few people struggling in math methods and a few who find it great.
I am looking towards healthcare, as in I am aiming high (doctor, or even higher, and if I don't make it I can always fall back to nursing)
Post by: AngelWings on February 18, 2019, 02:10:21 pm
Hey guys! assuming this the right place to post this. I am in year 10 this year and doing general mathematics and feel I have made a mistake and should of done methods. Math isn't my best subject, I struggle quite a bit, I am finding the content easy as and feel like I am wasting what I am capable of. I have been told its a good choice as its one less subject I don't have to worry about but I want to be prepared and knowledgable for my future endeavours. Does General Mathematics year 10 get harder? should I try to move to methods? I am really torn and I have missed the deadline but I feel I can try to move if I am finding it easy as.
my school is quite weird so I am doing year 10 general. the year 10 math are split into 3 classes Foundation; basically you are planning to ditch math and this is the last compulsory year and you basically budge.Seeing as Year 10 doesn't technically count towards VCE, I'd still probably talk to your maths teacher to see if you would be capable of doing the Year 10 Methods stream and can switch into the Methods stream next semester for a good foundation* and to test the waters. If you struggle too much, either switch back to General Year 10 or, if you miss the cut off date, then you'll have to stick it out for one semester and take General Maths U1/2 next year. Just be aware that many nursing, health care and common back up plans for people who miss out on undergrad med usually need some form of maths at the Year 11/12 stages, sometimes being Methods, sometimes not, so be wary of any prereqs your back up plans need. (I talked about how to check prereqs on another of your posts.)
general; you go over previously learnt content but more in depth
Methods; new content and quite tricky. I know a few people struggling in math methods and a few who find it great.
I am looking towards healthcare, as in I am aiming high (doctor, or even higher, and if I don't make it I can always fall back to nursing)
*This isn't 100% a must have for VCE Methods, but would be preferable to take if you're sure you want to study Methods later on.
Post by: lacitam on February 18, 2019, 03:55:49 pm
my school is quite weird so I am doing year 10 general. the year 10 math are split into 3 classes Foundation; basically you are planning to ditch math and this is the last compulsory year and you basically budge.What uni do you have in mind? I'm pretty sure only monash or melb uni need a prereq of methods for their science courses (you might need to double check on this).
general; you go over previously learnt content but more in depth
Methods; new content and quite tricky. I know a few people struggling in math methods and a few who find it great.
I am looking towards healthcare, as in I am aiming high (doctor, or even higher, and if I don't make it I can always fall back to nursing)
If methods doesn't work out for you, you can always go to other uni without methods. Deakin biomed only needs an english prerequisite for example