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    (Original post by MathsNerd1)
    Okay I thought I got it but I couldn't make any sense of it. To express (K+1)! = K!(K+1) but do I now use the fact that K! > 2^k in my induction step, if so then how exactly as this inequality seems to be making me quite cautious


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    you're proving for n \geq 4
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    (Original post by Felix Felicis)
    Ok:

    i) How do you get from k! to (k+1)! ?

    ii) How do you get from 2^{k} to 2^{k+1} ?
    For the first bit you multiply by (K+1)?

    For the second bit you multiply by 2 so am I essentially trying to show K+1 > 2 for any number greater than 4? Which we can easily see anyway?


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    (Original post by MathsNerd1)
    For the first bit you multiply by (K+1)?

    For the second bit you multiply by 2 so am I essentially trying to show K+1 > 2 for any number greater than 4? Which we can easily see anyway?


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    Indeed, so the inequality still holds true for k+1 if true for k, thereby completing the inductive step.
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    (Original post by Felix Felicis)
    Indeed, so the inequality still holds true for k+1 if true for k, thereby completing the inductive step.
    Oh, I thought it would be more than that but I forgot about the fact it was for greater than or equal to 4, thanks for your time


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    (Original post by justinawe)
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    you're proving for n \geq 4
    Yeah, the part sort of left my mind at the time but it was the fact that I could see if n = 2 it would also be true, however I forgot to put it in the original expression to see that wouldn't work. Thanks for your help


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    Could someone please explain the last part of the C4 papers for OCR MEI (the comprehension part)? I haven't done any work on them in college, my teacher didn't even mention them - I didn't find out they were actually there until I looked at a past paper a few months ago

    I just tried to do one of them and found it really confusing...I'm worried enough about C4 as it is without having this extra part to confuse me as well. I just want to know what sort of things usually come up and whether there's any trick to understanding the questions apart from working from past papers.
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    (Original post by Felix Felicis)
    I was wondering...would you mind having a look at this problem? Proof by example's not permitted apparently
    Hey! Despite what the problem-poser seems to be saying, proof by example is incorrect. The problem is "for any x,y,z that satisfy [...] show that there exist a,b,c,d s.t. [...]" and not "show that there exists x,y,z,a,b,c,d s.t. [...]" (which is what an example proves)

    My initial impression is that the problem seems to be far above everything else on this thread in terms of difficulty - if you want me to have a go at it I can, just let me know.
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    (Original post by Lord of the Flies)
    Hey! Despite what the problem-poser seems to be saying, proof by example is incorrect. The problem is "for any x,y,z that satisfy [...] show that there exist a,b,c,d s.t. [...]" and not "show that there exists x,y,z,a,b,c,d s.t. [...]" (which is what an example proves)

    My initial impression is that the problem seems to be far above everything else on this thread in terms of difficulty - if you want me to have a go at it I can, just let me know.
    I see, that makes sense.

    Yes, please do! I've wasted all of today trying to do it and just want some peace of mind. :rofl:
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    (Original post by Imog)
    Could someone please explain the last part of the C4 papers for OCR MEI (the comprehension part)? I haven't done any work on them in college, my teacher didn't even mention them - I didn't find out they were actually there until I looked at a past paper a few months ago

    I just tried to do one of them and found it really confusing...I'm worried enough about C4 as it is without having this extra part to confuse me as well. I just want to know what sort of things usually come up and whether there's any trick to understanding the questions apart from working from past papers.
    I went through one of those papers... they are pretty weird, I had no idea MEI made such papers.

    I'm not sure if you can actually revise for it, seems more like a test of mathematical thinking/logic if you get me. Maybe do some past papers for familiarisation but other than that I'm not too sure what you can do.
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    Mods/whatever: this is not an A-level problem but some people on this thread have tried it and someone has asked for a solution, so I think it remains of interest, at least to some.

    By the way guys, I don't often come to this thread so I don't know quite what your rules are, but I think it's best if you post problems that you have done yourselves, so that you are aware of what you are posting. This question was not appropriate in my opinion.

    (Original post by Felix Felicis)
    Yes, please do! I've wasted all of today trying to do it and just want some peace of mind.
    Let \mathcal{P}(x,y,z) be the statement we wish to prove. Note that \mathcal{P}(2,1,1) is clearly true. Now assume that \mathcal{P}(x,y,z) is true for all z\leq k and all x,y satisfying xy=z^2+1 for these z. Additionally, let m,n be any pair satisfying mn=(k+1)^2+1 with m\geq n

    Observe that adding n\big(n-2(k+1)\big) yields n\big(m+n-2(k+1)\big)=(k+1-n)^2+1. Note that the RHS is positive hence 0<m+n-2(k+1) and also that n^2\leq mn=(k+1)^2+1\Rightarrow 0< k+1-n.
    Hence \mathcal{P}\big(m+n-2(k+1),n,k+1-n\big) is true and we have m+n-2(k+1)=a^2+b^2, n=c^2+d^2, k+1-n=ac+bd for some integers a,b,c,d. This yields m=(a+c)^2+(b+d)^2.
    (a^2+b^2)(c^2+d^2)=(ac+bd)^2+1 gives (ad-bc)^2=1 and hence \big((a+c)d-(b+d)c\big)^2=1 which yields \big((a+c)^2+(b+d)^2\big)(c^2+d^  2)=\big((a+c)c+(b+d)d\big)^2+1 \Rightarrow k=(a+c)c+(b+d)d\Rightarrow \mathcal{P}(m,n,k+1) true \Rightarrow \mathcal{P}(x,y,z) true \forall x,y,z satisfying xy=z^2+1.



    For the inequality, split the 4 in the factorial: n!=( 2\cdot 3 \cdot 2)(2\cdot 5\cdot 6\cdots n)>(2\cdot 2\cdot 2)(2\cdot 2\cdot 2\cdots 2)=2^n
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    Hi can someone help me with 6(iii) please?
    http://www.ocr.org.uk/Images/136141-...echanics-1.pdf


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    (Original post by Lord of the Flies)
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    Mods/whatever: this is not an A-level problem but some people on this thread have tried it and someone has asked for a solution, so I think it remains of interest, at least to some.

    By the way guys, I don't often come to this thread so I don't know quite what your rules are, but I think it's best if you post problems that you have done yourselves, so that you are aware of what you are posting. This question was not appropriate in my opinion.



    Let \mathcal{P}(x,y,z) be the statement we wish to prove. Note that \mathcal{P}(2,1,1) is clearly true. Now assume that \mathcal{P}(x,y,z) is true for all z\leq k and all x,y satisfying xy=z^2+1 with for these z. Additionally, let m,n be any pair satisfying mn=(k+1)^2+1 with m\geq n

    Observe that adding n\big(n-2(k+1)\big) yields n\big(m+n-2(k+1)\big)=(k+1-n)^2+1. Note that the RHS is positive hence 0<m+n-2(k+1) and also that n^2\leq mn=(k+1)^2+1\Rightarrow 0< k+1-n.
    Hence \mathcal{P}\big(m+n-2(k+1),n,k+1-n\big) is true and we have m+n-2(k+1)=a^2+b^2, n=c^2+d^2, k+1-n=ac+bd for some integers a,b,c,d. This yields m=(a+c)^2+(b+d)^2.
    (a^2+b^2)(c^2+d^2)=(ac+bd)^2+1 gives (ad-bc)^2=1 and hence \big((a+c)d-(b+d)c\big)^2=1 which yields \big((a+c)^2+(b+d)^2\big)(c^2+d^  2)=\big((a+c)c+(b+d)d\big)^2+1 \Rightarrow k=(a+c)c+(b+d)d\Rightarrow \mathcal{P}(m,n,k+1) true \Rightarrow \mathcal{P}(x,y,z) true \forall x,y,z satisfying xy=z^2+1.



    For the inequality, split the 4 in the factorial: n!=( 2\cdot 3 \cdot 2)(2\cdot 5\cdot 6\cdots n)>(2\cdot 2\cdot 2)(2\cdot 2\cdot 2\cdots 2)=2^n
    Woah! You piece of insanity :zomg: That is brilliant, thanks very much PRSOM

    (Original post by Lord of the Flies)
    it's best if you post problems that you have done yourselves
    True that, I wouldn't post a problem I couldn't do myself unless I was asking for help. :L
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    (Original post by Lord of the Flies)

    By the way guys, I don't often come to this thread so I don't know quite what your rules are, but I think it's best if you post problems that you have done yourselves, so that you are aware of what you are posting. This question was not appropriate in my opinion.

    Let \mathcal{P}(x,y,z) be the statement we wish to prove. Note that \mathcal{P}(2,1,1) is clearly true. Now assume that \mathcal{P}(x,y,z) is true for all z\leq k and all x,y satisfying xy=z^2+1 with for these z. Additionally, let m,n be any pair satisfying mn=(k+1)^2+1 with m\geq n

    Observe that adding n\big(n-2(k+1)\big) yields n\big(m+n-2(k+1)\big)=(k+1-n)^2+1. Note that the RHS is positive hence 0<m+n-2(k+1) and also that n^2\leq mn=(k+1)^2+1\Rightarrow 0< k+1-n.
    Hence \mathcal{P}\big(m+n-2(k+1),n,k+1-n\big) is true and we have m+n-2(k+1)=a^2+b^2, n=c^2+d^2, k+1-n=ac+bd for some integers a,b,c,d. This yields m=(a+c)^2+(b+d)^2.
    (a^2+b^2)(c^2+d^2)=(ac+bd)^2+1 gives (ad-bc)^2=1 and hence \big((a+c)d-(b+d)c\big)^2=1 which yields \big((a+c)^2+(b+d)^2\big)(c^2+d^  2)=\big((a+c)c+(b+d)d\big)^2+1 \Rightarrow k=(a+c)c+(b+d)d\Rightarrow \mathcal{P}(m,n,k+1) true \Rightarrow \mathcal{P}(x,y,z) true \forall x,y,z satisfying xy=z^2+1.



    For the inequality, split the 4 in the factorial: n!=( 2\cdot 3 \cdot 2)(2\cdot 5\cdot 6\cdots n)>(2\cdot 2\cdot 2)(2\cdot 2\cdot 2\cdots 2)=2^n
    Darn, nice proof. I'd been having a go at solving it last night but wasn't making much headway.

    Also, I agree that people should do their problems first (And potentially think through whether the difficulty is really suitable, as this is an A level thread after all) as there have been other things come up that aren't suitable, like "Prove that  \pi + e is irrational.

    Now if you'll excuse me, I'm off to sacrifice a goat to various gods in the hope that one day I'll be able to do a proof to a question like that.
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    Hello. !
    For mechanics..collisions...there's always a problem of giving a negative value (as an expression) when stating the speed of P & Q or whatever. Do anyone know whether we could just avoid this by adding a modulus | | sign ?:cool:
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    (Original post by TheKingOfTSR)
    Hello. !
    For mechanics..collisions...there's always a problem of giving a negative value (as an expression) when stating the speed of P & Q or whatever. Do anyone know whether we could just avoid this by adding a modulus | | sign ?:cool:
    Is this for M2? I normally find that the algebra sorts out whether or not you've got the correct direction, so it shouldn't really make a difference, although I could be wrong.


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    (Original post by MathsNerd1)
    Is this for M2? I normally find that the algebra sorts out whether or not you've got the correct direction, so it shouldn't really make a difference, although I could be wrong.


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    Yes it is for next weeks M2.
    But what I mean is this:
    They frequently ask us to find the speed of object and as it comes in constants and in terms of e , its a frequent mistake to give a -ve answer like 3u(e-2) which is wrong. So can we just state |3u(e-2)| specially when it gets complicate?.
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    Anyone got a nice question? I'm in the mood for some maths.
    Preferably some calculus, but anything goes.
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    (Original post by TheKingOfTSR)
    Yes it is for next weeks M2.
    But what I mean is this:
    They frequently ask us to find the speed of object and as it comes in constants and in terms of e , its a frequent mistake to give a -ve answer like 3u(e-2) which is wrong. So can we just state |3u(e-2)| specially when it gets complicate?.
    Well because you know e has to lie between 0 and 1 you should be able to tell when you've got it going in the wrong direction and just correct it from there? I've never used the modulus sign for it as I wouldn't feel very confident with it unfortunately :-/


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    (Original post by joostan)
    Anyone got a nice question? I'm in the mood for some maths.
    Preferably some calculus, but anything goes.
    LOOL, I shall try Googling one if you want?
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    (Original post by tigerz)
    LOOL, I shall try Googling one if you want?
    Lol, I tried doing that but they never seem interesting enough, but go for it
 
 
 
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