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1729

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« Reply #18660 on: July 30, 2020, 07:32:01 pm »
+3
Hello,

Prove that if (a/n) is irreducible then (a/n)^2 will also be irreduicble for a =/= 1, 0, n and n =/= 1, 0, a.

Thanks.
Try a proof by contradiction; suppose (a/n) is irreducible and (a/n)^2 is reducible and see if you can arrive at the result that (a/n) must then be reducible.

So firstly, suppose (a/n)^2 is reducible; what does this say about a and n? Remember that the fraction a^2 / n^2 is reducible if a^2 and n^2 have a gcd > 1 but that does mean that a and n must have a gcd > 1. I'm not too sure how thorough your teacher wants you to be. So I'm just going to jump to the implication that a and n must have a gcd > 1. So if a and n have a gcd > 1 what does this say about the fraction a/n? (btw, gcd > 1 means that they have a common factor greater than 1) This just means that the fraction is "reducible". So we've assumed at the beginning that a/n is irreducible but we've just arrived at the conclusion that a/n must be reducible (the more formal way to do this would be to use contraposition, but i think this should be good enough) so since we've arrived at contradicting results, our initial assumption that (a/n)^2 is reducible must be false.

This is just to help you an idea. The more formal way to do this would be to use contraposition, but I think this should be good enough. So since we've arrived at contradicting results, our initial assumption that (a/n)^2 is reducible must be false.

However, this proof is technically a proof by contraposition; if we prove that (a/n)^2 is reducible implies (a/n) is reducible, this also proves that if (a/n) is irreducible, (a/n)^2 is irreducible.

But we can also frame it in a way where it's contradiction, but its a little bit iffy especially since we never actually used the assumption that (a/n) is irreducible. Framing it in a contradiction would be instead of saying that a/n must be reducible at the end, after we get to the point where gcd(a, n) > 1 we would get stuck. Because we've assumed that a/n is irreducible at the beginning, gcd(a, n) must be 1. But since we've just arrived at the conclusion that gcd(a, n) > 1, this contradicts our assumptions, meaning that (a/n)^2 must not be reducible.

Hope this helps! If you have any further questions don't hesitate to ask!

keltingmeith

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« Reply #18661 on: July 31, 2020, 01:15:48 am »
+6
Hello,

Prove that if (a/n) is irreducible then (a/n)^2 will also be irreduicble for a =/= 1, 0, n and n =/= 1, 0, a.

Thanks.

This feels so much like a specialist 1/2 question lol

could you explain why if you don't mind sorry if it sounds dumb

Don't forget that cos(v) is also a number. It's just like any pronumeral. Remember how when solving equations like:

$x^2-4x+2=7x$

You would move the 7x to the other side?

$x^2-4x-7x+2=0$

By this point, you might be on auto-pilot - but let's stop and think for a second. You COULD just work like you normally do, where you know that you can ignore the x and just do -4-7, and you would get:

$x^2-11x+2=0$

But, how does this step work? And why can we do -4x-7x but not -4x-7y? Well, you can think of this is as factorising out the x, like so:

$x^2-4x-7x+2=0\\
x^2-x(4+7)+2=0\\
x^2-x(11)+2=0\\
x^2-11x+2=0$

And this is why we can't do the same step with with -4x-7y - because there's nothing to factor out!

So, what does this have to do with cos(v)? Well, cos(v) is just a number - in fact, you can think of cos(v) AS x, and you could solve an equation like so:

$2\cos(v)=\cos(v)\\
\text{Set }\cos(v)=x\\
\therefore 2x=x\\
2x-x=0\\
x=0\\
\text{And since }x=\cos(v)\\
\therefore \cos(v)=0$

But, that's a lot of work! So, why don't we do this WITHOUT the substitution?

$2\cos(v)=\cos(v)\\
2\cos(v)-\cos(v)=0\\
\cos(v)(2-1)=0\\
\cos(v)(1)=0\\
\cos(v)=0$

In the third step, notice that all I'm doing is factoring out that cos(v) - just like I did earlier with x. Feel free to skip this step - but if doing it helps your understanding of WHY you can "cancel out" (which is a bad description btw - things aren't cancelling out) in the way that Cat did, then power to you
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rozmaaate

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« Reply #18662 on: July 31, 2020, 09:07:46 am »
+1
This feels so much like a specialist 1/2 question lol

Don't forget that cos(v) is also a number. It's just like any pronumeral. Remember how when solving equations like:

$x^2-4x+2=7x$

You would move the 7x to the other side?

$x^2-4x-7x+2=0$

By this point, you might be on auto-pilot - but let's stop and think for a second. You COULD just work like you normally do, where you know that you can ignore the x and just do -4-7, and you would get:

$x^2-11x+2=0$

But, how does this step work? And why can we do -4x-7x but not -4x-7y? Well, you can think of this is as factorising out the x, like so:

$x^2-4x-7x+2=0\\
x^2-x(4+7)+2=0\\
x^2-x(11)+2=0\\
x^2-11x+2=0$

And this is why we can't do the same step with with -4x-7y - because there's nothing to factor out!

So, what does this have to do with cos(v)? Well, cos(v) is just a number - in fact, you can think of cos(v) AS x, and you could solve an equation like so:

$2\cos(v)=\cos(v)\\
\text{Set }\cos(v)=x\\
\therefore 2x=x\\
2x-x=0\\
x=0\\
\text{And since }x=\cos(v)\\
\therefore \cos(v)=0$

But, that's a lot of work! So, why don't we do this WITHOUT the substitution?

$2\cos(v)=\cos(v)\\
2\cos(v)-\cos(v)=0\\
\cos(v)(2-1)=0\\
\cos(v)(1)=0\\
\cos(v)=0$

In the third step, notice that all I'm doing is factoring out that cos(v) - just like I did earlier with x. Feel free to skip this step - but if doing it helps your understanding of WHY you can "cancel out" (which is a bad description btw - things aren't cancelling out) in the way that Cat did, then power to you

Thanks so much!!

keltingmeith

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« Reply #18663 on: July 31, 2020, 11:00:28 pm »
+4
wait how did you get the left hand side to equal 2 cos(v) ?? in regards to my question of  solving x for 4√2 cos(v) = √2 cos(v) + 3, -π≤v≤5π

2cos(v)=cos(v) is an equation I made up, it has nothing to do with your equation. But you should be able to apply the same principles to see how it works with your equation
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« Reply #18664 on: August 03, 2020, 06:53:44 pm »
+1
Hello.

find the general solution to the equation.

I did it but I found two solutions, one with (5pi/3) and one with (2pi/3). The textbook says only (2pi/3). Help
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james.lhr

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« Reply #18665 on: August 03, 2020, 07:18:40 pm »
+6
Hey a weaponized ikea chair!

Because the period of tan(x) is π, 2π/3 + nπ is actually the same as 5π/3 + nπ, where n ∈ Z

Hope this helps!
James
VCE Class of 2021
2020: Maths Methods [?]
2021: English [?] Biology [?] Chemistry [?] Physics [?] Specialist Maths [?]

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« Reply #18666 on: August 03, 2020, 07:20:53 pm »
0
Hey a weaponized ikea chair!

Because the period of tan(x) is π, 2π/3 + nπ is actually the same as 5π/3 + nπ, where n ∈ Z

Hope this helps!
James

Hello,

Is giving both solutions technically correct or would it be marked wrong in an exam? Thanks
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james.lhr

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« Reply #18667 on: August 03, 2020, 07:26:38 pm »
0
Although your solution is technically correct, I do not believe you will be awarded the mark if it was a VCAA exam. Either way, now that you know the correct method, you shouldn't have to worry about it
VCE Class of 2021
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2021: English [?] Biology [?] Chemistry [?] Physics [?] Specialist Maths [?]

schoolstudent115

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« Reply #18668 on: August 04, 2020, 04:11:46 pm »
+6
Hello,

Prove that if (a/n) is irreducible then (a/n)^2 will also be irreduicble for a =/= 1, 0, n and n =/= 1, 0, a.

Thanks.
I have a direct proof of this.

Suppose a/n is irreducible. Therefore a and n share no common factor.

Factors of a: $a_1,a_2,...a_n$
Factors of b: $b_1,b_2,...,b_n$

Now note that the quotient of any two factors of b and a $a_i/b_j$ will not be an integer.
*sub-proof:
*Suppose there does exist an i and j such that $\frac{a_i}{b_j} \in \mathbb{Z}$, then this would imply that $b_j=\frac{a_i}{n}, n\in \mathbb{Z}$. If some x is a factor of y, and y is a factor of z, then x is a factor of z. (Y=kx, z=ly = (lk)x). Therefore it follows that since b_j is a factor of a_i, then b_j is a factor of a, which is a contradiction since A and B share no factors.
— back to the proof
If there is no integer quotient $a_i/b_j$, then the square of this will also not be an integer, $(a_i)^2/(b_j)^2$ is not an integer. And more generally, no product of quotients is an integer. But every factor of a^2 is either a factor of a or a product of 2 factors. So if there is no integer result to $\frac{a_i}{b_j} \frac{a_k}{b_l}$, then there is no integer result to $\frac{(a^2)_i}{(b^2)_j}$, therefore $\frac{a^2}{b^2}=(\frac{a}{b})^2$ is irreducible**.

** That last statement is not true generally but is true in relation to factors. I could prove that but it is a bit complex.
« Last Edit: August 04, 2020, 09:06:36 pm by schoolstudent115 »
(2020) Year 11: Physics 3/4

keltingmeith

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« Reply #18669 on: August 04, 2020, 04:47:30 pm »
+4
Although your solution is technically correct, I do not believe you will be awarded the mark if it was a VCAA exam. Either way, now that you know the correct method, you shouldn't have to worry about it

Entirely untrue - there can very easily be multiple methods to solve a maths equation, with all of them being just as correct as each other. Hell, there's often multiple ways that any given mathematical fact can be proven - did you know that there are over 100 ways to prove Pythagoras' theorem?

Now, the study design does not specifically list general solutions to circular functions at all - and while I doubt this means they're NOT assessable (there are still points that can be interpreted to mean, you should know the general solution to circular functions), this DEFINITELY means VCAA does not have a prescribed method for answering these questions. If your answer is still correct, you'll still get the marks - and if you answered it a mathematically correct way, you'll still get the answers for that method. But also, to claim that there is a "correct" or an "incorrect" method in this case is kind of insulting to maths as a profession, so sorry if I got a little snarky

-snip-

I just want to add that I really like this proof, and discrete maths in general is just <3, but this further proves my point about this not being a methods question lol
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« Reply #18670 on: August 04, 2020, 05:41:14 pm »
0
Entirely untrue - there can very easily be multiple methods to solve a maths equation, with all of them being just as correct as each other. Hell, there's often multiple ways that any given mathematical fact can be proven - did you know that there are over 100 ways to prove Pythagoras' theorem?

Now, the study design does not specifically list general solutions to circular functions at all - and while I doubt this means they're NOT assessable (there are still points that can be interpreted to mean, you should know the general solution to circular functions), this DEFINITELY means VCAA does not have a prescribed method for answering these questions. If your answer is still correct, you'll still get the marks - and if you answered it a mathematically correct way, you'll still get the answers for that method. But also, to claim that there is a "correct" or an "incorrect" method in this case is kind of insulting to maths as a profession, so sorry if I got a little snarky

I just want to add that I really like this proof, and discrete maths in general is just <3, but this further proves my point about this not being a methods question lol

This is not a methods question. But it is also not a specialist question. Here's what happened:

I was doing methods homework when I reached 3/5 and I had to square it, getting 9/25.  Well, one thought led to another and I noticed that if (a/b) was irreducible so was (a/b)^2 (got completely sidetracked form the homework). I tried to prove it using my elementary understanding of proof by contradiction but could not. But because I was so interested in it, and I thought that there could be an exception to the rule, I had to find out. A google search came up with nothing, so, lo and behold, I came to my good old friend, AtarNotes.
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schoolstudent115

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« Reply #18671 on: August 04, 2020, 05:56:43 pm »
0
Lol true. Even in spec they wouldn't ask this most likely. If they did you would have to have first gone through the unit on number theory (euclidean algorithm etc.) to have a good base.
(2020) Year 11: Physics 3/4

keltingmeith

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« Reply #18672 on: August 04, 2020, 06:41:05 pm »
0

This is not a methods question. But it is also not a specialist question. Here's what happened:

I was doing methods homework when I reached 3/5 and I had to square it, getting 9/25.  Well, one thought led to another and I noticed that if (a/b) was irreducible so was (a/b)^2 (got completely sidetracked form the homework). I tried to prove it using my elementary understanding of proof by contradiction but could not. But because I was so interested in it, and I thought that there could be an exception to the rule, I had to find out. A google search came up with nothing, so, lo and behold, I came to my good old friend, AtarNotes.

Loooool

I'm not against asking random level questions on AN, but maybe discuss with the mods on a better place to post it so as to not frighten other students 😅

Lol true. Even in spec they wouldn't ask this most likely. If they did you would have to have first gone through the unit on number theory (euclidean algorithm etc.) to have a good base.

Funnily enough, there's an argument that it could fit specialist 1/2, but definitely not 3/4 because you're right - the relevant mathematics isn't taught there
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schoolstudent115

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« Reply #18673 on: August 04, 2020, 10:30:29 pm »
0

This is not a methods question. But it is also not a specialist question. Here's what happened:

I was doing methods homework when I reached 3/5 and I had to square it, getting 9/25.  Well, one thought led to another and I noticed that if (a/b) was irreducible so was (a/b)^2 (got completely sidetracked form the homework). I tried to prove it using my elementary understanding of proof by contradiction but could not. But because I was so interested in it, and I thought that there could be an exception to the rule, I had to find out. A google search came up with nothing, so, lo and behold, I came to my good old friend, AtarNotes.
Here is my complete proof (I left out one small part in the post I made before, but here is the full version).
« Last Edit: August 04, 2020, 10:33:32 pm by schoolstudent115 »
(2020) Year 11: Physics 3/4

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