February 20, 2019, 01:35:56 am

### AuthorTopic: Class of 2021 Math Club  (Read 2778 times)

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#### miniturtle

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##### Re: Class of 2021 Math Club
« Reply #15 on: December 06, 2018, 10:45:15 am »
+1
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?
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#### AlphaZero

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##### Re: Class of 2021 Math Club
« Reply #16 on: December 06, 2018, 12:03:36 pm »
+2
The plus/minus symbol is written as \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.
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#### aspiringantelope

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##### Re: Class of 2021 Math Club
« Reply #17 on: December 06, 2018, 07:31:50 pm »
+2
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?
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?

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#### miniturtle

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##### Re: Class of 2021 Math Club
« Reply #18 on: December 06, 2018, 08:12:32 pm »
+5
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?

No worries

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.
How are 1/5 and 1/10 similar?

Hope this helps
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#### fun_jirachi

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##### Re: Class of 2021 Math Club
« Reply #19 on: December 06, 2018, 08:16:00 pm »
+5
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
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#### aspiringantelope

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##### Re: Class of 2021 Math Club
« Reply #20 on: December 07, 2018, 09:04:23 pm »
+3
$\log_2\left(\dfrac{1}{10}\right)=-3.322$
$\log_2\left(\frac{2}{5}\right)=\log_2\left(\frac{4}{10}\right)$
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
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#### Aaron

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##### Re: Class of 2021 Math Club
« Reply #21 on: December 07, 2018, 09:06:53 pm »
+3

Remember everyone - to encourage users to demonstrate understanding with their questions.... it will help them in the long run!  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.
« Last Edit: December 07, 2018, 09:45:14 pm by Aaron »
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#### miniturtle

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##### Re: Class of 2021 Math Club
« Reply #22 on: December 07, 2018, 09:44:54 pm »
+4
$\log_2\left(\dfrac{1}{10}\right)=-3.322$
$\log_2\left(\frac{2}{5}\right)=\log_2\left(\frac{4}{10}\right)$
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

Remember everyone - to encourage users to demonstrate understanding with their questions.... it will help them in the long run!  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).
Thanks Aaron

Definitely agree that it's more beneficial for users to show their thoughts & what they've tried, than only getting a suggested solution
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#### turtlesforeveryone

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##### Re: Class of 2021 Math Club
« Reply #23 on: December 07, 2018, 10:05:12 pm »
+3
Hello! This thread suddenly became active again :p

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.

#### AlphaZero

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##### Re: Class of 2021 Math Club
« Reply #24 on: December 07, 2018, 11:09:50 pm »
+6
Set 1 Solutions

Question 1 Solution
\begin{align*} ax^2+bx+c&=0\\
\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}{a}&=0\\
\left(x+\frac{b}{2a}\right)^2&=\frac{b^2-4ac}{4a^2}\\
x+\frac{b}{2a}&=\frac{\pm\sqrt{b^2-4ac}}{2a}\\
x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a},\ \ \text{as required.}
\end{align*}

Question 2 Solution
\begin{align*}\log_2\left(\frac{2}{5}\right)&=\log_2\left(4\cdot\frac{1}{10}\right)\\
&=\log_2(4)+\log_2\left(\frac{1}{10}\right)\\
&=2-3.322\\
&=-1.322,\ \ \text{correct to three decimal places.}\end{align*}

Question 3 Solution
\begin{align*}A&=(\text{area of equilateral triangle with side length 12 cm})-(\text{area of 3 times 1/6 circles of diameter 12 cm})\\
&=\frac{1}{2}\cdot 12\cdot 12\cdot \sin(60^\circ)-3\cdot\frac{1}{6}\pi\cdot 6^2\\
&=36\sqrt{3}-18\pi\ \text{cm}^2.
\end{align*}

Well done guys. I'll post 3 more questions soon
« Last Edit: December 08, 2018, 12:04:57 am by dantraicos »
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#### GMT. -_-

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##### Re: Class of 2021 Math Club
« Reply #25 on: December 07, 2018, 11:36:40 pm »
+2
Question 2
log2(2/5)= log2(4)-log2(10)
= 2- log2(10)
= 2 + log2(10^-1)
= 2 - 3.322
= -1.322

#### AlphaZero

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##### Re: Class of 2021 Math Club
« Reply #26 on: December 08, 2018, 01:28:26 am »
+7
Set 2 Questions: 8 Dec 2018 to 13 Dec 2018

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

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.

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.
« Last Edit: December 08, 2018, 05:16:22 pm by dantraicos »
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#### turtlesforeveryone

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##### Re: Class of 2021 Math Club
« Reply #27 on: December 08, 2018, 12:06:05 pm »
+5
I will try
(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.

Spoiler
Q2) Le'ts assume x is an odd number (proof by contradiction).
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.
∴ 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.
√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.
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.
« Last Edit: December 08, 2018, 01:28:58 pm by turtlesforeveryone »

#### AlphaZero

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##### Re: Class of 2021 Math Club
« Reply #28 on: December 08, 2018, 12:38:26 pm »
+6

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.

« Last Edit: December 20, 2018, 04:20:37 pm by dantraicos »
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#### fun_jirachi

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##### Re: Class of 2021 Math Club
« Reply #29 on: December 08, 2018, 06:27:45 pm »
+3
Hey there!
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
Failing everything, but I'm still Flareon up.

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HSC 2019: Physics | Chemistry | English Advanced | Maths Extension 1 | Maths Extension 2