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May 28, 2020, 09:38:42 pm

Author Topic: Class of 2021 Math Club  (Read 6359 times)

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AlphaZero

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Re: Class of 2021 Math Club
« Reply #30 on: December 18, 2018, 11:08:59 pm »
+5
Set 2 Solutions:

Question 1 Solution


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.

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, \(\sqrt{2}\notin\mathbb{Q}\).

Let me know guys when/if you want another set :)
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turtlesforeveryone

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Re: Class of 2021 Math Club
« Reply #31 on: December 20, 2018, 02:01:14 pm »
+2
Set 2 Solutions:

Question 1 Solution


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.

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, \(\sqrt{2}\notin\mathbb{Q}\).

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 :)

AlphaZero

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Re: Class of 2021 Math Club
« Reply #32 on: December 20, 2018, 05:25:46 pm »
+4
Set 3 Questions: 20 Dec 2018 to 2 Jan 2019

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!

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 :)
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turtlesforeveryone

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Re: Class of 2021 Math Club
« Reply #33 on: December 20, 2018, 11:04:41 pm »
+1
Set 3 Questions: 20 Dec 2018 to 2 Jan 2019

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!

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 :)

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 \(n\in\mathbb{N}.\)

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 \(n\in\mathbb{N}.\)

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.

Whew, this was my first time writing in LaTeX and it sure took a bit of time to complete! There may be typos.

turtlesforeveryone

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Re: Class of 2021 Math Club
« Reply #34 on: December 29, 2018, 04:51:22 pm »
+1
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).

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.

RuiAce

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Re: Class of 2021 Math Club
« Reply #35 on: December 29, 2018, 04:59:45 pm »
+7
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).

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.
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.
« Last Edit: December 29, 2018, 05:03:32 pm by RuiAce »

turtlesforeveryone

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Re: Class of 2021 Math Club
« Reply #36 on: December 29, 2018, 07:22:02 pm »
+1
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

RuiAce

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Re: Class of 2021 Math Club
« Reply #37 on: December 29, 2018, 07:48:41 pm »
+3
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
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.

(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.

aspiringantelope

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Re: Class of 2021 Math Club
« Reply #38 on: December 29, 2018, 09:09:38 pm »
0
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).

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.
Wow!!
Are you planning to do UMEP Maths?
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turtlesforeveryone

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Re: Class of 2021 Math Club
« Reply #39 on: January 23, 2019, 09:44:06 pm »
0
Wow!!
Are you planning to do UMEP Maths?

Most likely :p. Any sort of university extension would be great. You?

aspiringantelope

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Re: Class of 2021 Math Club
« Reply #40 on: February 11, 2019, 05:48:10 pm »
0
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 [>.<]
Spread kindness!

MB_

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Re: Class of 2021 Math Club
« Reply #41 on: February 11, 2019, 06:03:16 pm »
+2
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 [>.<]
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.


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aspiringantelope

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Re: Class of 2021 Math Club
« Reply #42 on: February 11, 2019, 09:14:05 pm »
0
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.

(Image removed from quote.)
OHHH WAITTT LOOOOL ONE OF THEM IS AT MY SCHOOOL LMFAOOOOOO
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 =(
Spread kindness!

w0lfqu33n89

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Re: Class of 2021 Math Club
« Reply #43 on: February 17, 2019, 05:19:26 pm »
0
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.

lacitam

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Re: Class of 2021 Math Club
« Reply #44 on: February 17, 2019, 05:39:47 pm »
0
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.
Monash BSci