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# STEP Prep Thread 2016 (Mark. II)

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Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016
1. I did the same thing as you and I can't see how anyone can deduct a mark for that; drawing 6 sketches doesn't make it any more rigorous.

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2. (Original post by physicsmaths)
If someone can check Q6 2015 III for me
Attachment 552436

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3. (Original post by physicsmaths)
Bump

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sorry for drowning out ur post Ill just do the q first quickly
4. (Original post by EnglishMuon)
IrrationalRoot Quick q about 4 i on that paper if u wouldnt mind correcting me
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for part i I thought it would be pretty clear that after noting what happens as z->+- infinity and that f(z) is continuous there must be one root, but the ms has 6 sketches as well for a separate mark. Should I draw these in future for a rule of thumb?
Precisely what I did. No clue why the **** they did 6 sketches. Nobs

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5. (Original post by physicsmaths)
If someone can check Q6 2015 III for me
Attachment 552436
Seems fine. There are a lot of ways to do that first bit.
6. (Original post by EnglishMuon)
IrrationalRoot Quick q about 4 i on that paper if u wouldnt mind correcting me
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for part i I thought it would be pretty clear that after noting what happens as z->+- infinity and that f(z) is continuous there must be one root, but the ms has 6 sketches as well for a separate mark. Should I draw these in future for a rule of thumb?
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Nah the ms approach seems really excessive. Your argument was right (maybe mention the IVT). I use the conjugate pair argument .
7. (Original post by to4ka)
STEP III 2015, Q5, part (ii):
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Isn't proving that the cube root of 2 is irrational enough to do the second part of (ii)? Letting S = {n | n is natural and n*2^1/3 is an integer} and knowing that 2^1/3 < 2 gives the same contradiction. Now, if 2^2/3 is rational, then 2^1/3 is rational (first part of (ii)), which we just proved false, so 2^2/3 is also irrational.

I'm asking because the set proposed in the mark scheme is different from mine.
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Q5 needs a different method as 2^(1/3)*(2^(1/3)-1)*k isn't necessarily an integer. If that's what you did.
8. (Original post by Hauss)
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Q5 needs a different method as 2^(1/3)*(2^(1/3)-1)*k isn't necessarily an integer. If that's what you did.
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That's what I did:
Let S = { n | n is natural and n * 2^1/3 is an integer}, assuming 2^1/3 is rational (=p/q). Obviously, q is in S, so it's non-empty. Now, if k is the smallest member of S, (2^1/3 - 1)k is also in S, but 2^1/3 - 1 < 1, so (2^1/3-1)k < k, which is a contradiction. The irrationality of 2^2/3 follows from the irrationality of 2^1/3.
9. (Original post by to4ka)
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That's what I did:
Let S = { n | n is natural and n * 2^1/3 is an integer}, assuming 2^1/3 is rational (=p/q). Obviously, q is in S, so it's non-empty. Now, if k is the smallest member of S, (2^1/3 - 1)k is also in S, but 2^1/3 - 1 < 1, so (2^1/3-1)k < k, which is a contradiction. The irrationality of 2^2/3 follows from the irrationality of 2^1/3.
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But since 2^(1/3)*(2^(1/3)-1)*k isn't necessarily an integer you can't say (2^1/3 - 1)k is also in S.
10. (Original post by to4ka)
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That's what I did:
Let S = { n | n is natural and n * 2^1/3 is an integer}, assuming 2^1/3 is rational (=p/q). Obviously, q is in S, so it's non-empty. Now, if k is the smallest member of S, (2^1/3 - 1)k is also in S, but 2^1/3 - 1 < 1, so (2^1/3-1)k < k, which is a contradiction. The irrationality of 2^2/3 follows from the irrationality of 2^1/3.
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I don't think the step saying that (2^1/3 - 1)k is also in S follows, as in (i) you used the fact that 2^1/2 * 2^1/2 = 2 is an integer to prove it, whereas here you have 2^1/3 * 2^1/3 != 2. So what you have to do (at least what I did) is consider (2^2/3 - 1)k, as 2^2/3 * 2^1/3 = 2 like before.
11. (Original post by Hauss)
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But since 2^(1/3)*(2^(1/3)-1)*k isn't necessarily an integer you can't say (2^1/3 - 1)k is also in S.
(Original post by sweeneyrod)
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I don't think the step saying that (2^1/3 - 1)k is also in S follows, as in (i) you used the fact that 2^1/2 * 2^1/2 = 2 is an integer to prove it, whereas here you have 2^1/3 * 2^1/3 != 2. So what you have to do (at least what I did) is consider (2^2/3 - 1)k, as 2^2/3 * 2^1/3 = 2 like before.
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Yeah, I see where I made a mistake - assuming that, since (2^1/3 - 1)k is an integer, it must be in S... I guess I have to deduct a lot of marks from this question lol.
12. (Original post by to4ka)
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Yeah, I see where I made a mistake - assuming that, since (2^1/3 - 1)k is an integer, it must be in S... I guess I have to deduct a lot of marks from this question lol.
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Yeah, I almost did that in the first part (forgetting to multiply by root 2).
14. (Original post by Insight314)
Its not answering your question, but from a FoI request, a certain mathmo (not me, nor anyone on TSR AFAIK) extracted this from Cambridge:

https://www.whatdotheyknow.com/reque...and%20data.pdf
15. STEP III 2015 Q 6:
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Not sure about my proof for the first part:

u^2 <= 2v and u and v real
hence (w + z)^2 <= 2(w^2 + z^2)
hence (w - z)^2 >= 0
hence w - z is real (is this right?)
hence w and z are real or have an equal imaginary part
If the latter, u = w + z has an imaginary part, so isn't real.
So w and z must be real.
16. (Original post by Mathemagicien)
Its not answering your question, but from a FoI request, a certain mathmo (not me) extracted this from Cambridge:

https://www.whatdotheyknow.com/reque...and%20data.pdf
Yeah, I am looking for a more specific statistics.
18. (Original post by gasfxekl)
That's a serious allegation.
19. (Original post by Insight314)
[redacted] also made a request (much better one) a while ago, which is what I meant when mentioning "statistics" in my previous post.

But yeah, I am looking for a more specific statistics.

Btw, I know you are not [redacted].
They're the same request

I know you know
Although I could have been using that as a pseudonym
20. (Original post by Insight314)
That's a serious allegation.

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