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    (Original post by tigerz)
    \log_{3}2=\frac{a}{b} \therefore 3^\frac{a}{b}=2 this means 2^b=3^a
    We know that 2^b will be even so can be represented by 2w
    We also know that 3^a will be odd so is  2x+1
    This means 2^b=3^a\equiv 2w=2x+1\therefore w=x+1

    URM i'm lost now, I need to link it back to the original question but confused :confused:
    You have that 2w = 2x+1
    Where w and x are positive integers. Can you spot a problem in this?
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    (Original post by joostan)
    You have that 2w = 2x+1
    Where w and x are positive integers. Can you spot a problem in this?
    OHH basically, if 2w and 2x are positive integers they can be simplified, and the initial statement said that a/b is in its simplest form is untrue!
    I dunno really sorry, I'll come back in a bit gotta sort something out
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    (Original post by joostan)
    Basis case.
    Inductive step. (n=k)
    n=k+1

    e.g.
    Prove that:
    \displaystyle\sum_{r=1}^n r = \frac{n}{2}(n+1)
    Okay, I'm stuck on the inductive step, would the this assumption be correct or not?
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    This seems completely wrong though :-/


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    (Original post by DJMayes)
    The question relies on one specific result for generating sets, although there are probably other ways you can obtain the result to the ones we've been through. However, you cannot really go further with it due to the infinite number of potential solutions. By the way, I'm not entirely convinced your use of modular arithmetic is correct:

    Spoiler:
    Show


    You have:
     5q^3 = p^3
     \Rightarrow p^3 \equiv 0 (mod 5)
     \Rightarrow p \equiv 0 (mod 5)
     \Rightarrow q^3 \equiv 0 (mod 5)
     \Rightarrow q \equiv 0 (mod 5)

    And then you get the simplification contradiction.  p^3mod5 is absolutely meaningless as a statement, and I think you'd have been better writing up your solution without the modular arithmetic.



    (Also, you think you've been watching too much? Started watching it a month or two ago, now halfway through the fifth season. )
    I know, but in essence it gives you the same equation to generate sets of solutions which is what joostan and justinawe proposed earlier - Mayman said it was unsatisfactory so presumably there's a more general solution that he knows of and it's bugging me :dontknow:

    And my apologies but I don't really see the problem I've written exactly what you have (albeit skipped a step) and nowhere have I written p^{3} mod 5 But yeah, I suppose it would've been easier to write 5 | p^{3} or something :lol:

    It's good isn't it? I started last week and I'm on season 2 :ninja: Hugh Laurie is a god
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    (Original post by MathsNerd1)
    Okay, I'm stuck on the inductive step, would the this assumption be correct or not?
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    This seems completely wrong though :-/


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    Not quite:
    \displaystyle\sum_{r=1}^{k+1} r = \displaystyle\sum_{r=1}^{k} r + (k+1)
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    (Original post by tigerz)
    OHH basically, if 2w and 2x are positive integers they can be simplified, and the initial statement said that a/b is in its simplest form is untrue!
    I dunno really sorry, I'll come back in a bit gotta sort something out
    Spoiler:
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    One is always even, the other is always odd. So . . .
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    (Original post by joostan)
    Not quite:
    \displaystyle\sum_{r=1}^{k+1} r = \displaystyle\sum_{r=1}^{k} r + (k+1)
    Of course! I have no idea what was making me believe what I wrote down as it was complete and utter nonsense!


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    (Original post by MathsNerd1)
    Can anyone run through the basics of induction and then give me some relatively easy questions so I can work at it, then slowly bump up the difficulty so I'm able to try a few STEP questions on the topic, I just feel like I've forgotten too much of it to do anything that would be any good :-/


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    Induction is a method of proof used when you want to prove a statement concerning integers, and this is usually restricted to positive integers. The logic is essentially this:

    - Prove that it is true for one specific case
    - Prove that, if it is true for one case, it must be true for the next.

    You usually perform the 2nd step by assuming that it is true for some value of n (Denoted k) and showing that this implies it is true when n=k+1. First of all, do the question joostan set you. Then, look at this question:

    Prove that  \frac{d^n}{dx^n} (xe^x) = (x+n)e^x

    Induction hint:

    Spoiler:
    Show


    - We can very easily link one case to another - we just need to differentiate the case!



    Full solution:

    Spoiler:
    Show


     \frac{d^n}{dx^n}( xe^x) = (x+n)e^x

    We can differentiate using the product rule, giving  \frac{d}{dx} xe^x = xe^x+e^x = (x+1)e^x therefore it is true when n = 1. Assume it is true when n = k:

     \frac{d^k}{dx^k}xe^x = (x+k)e^x

    Now, differentiate both sides with respect to x:

     \frac{d^{k+1}}{dx^{k+1}}xe^x = (x+k)e^x+e^x = (x+k+1)e^x therefore true when n = k+1.

    We have shown that, if the statement is true for some value of n, it must be true for the next. It is true when n=1 therefore it is true for all positive integers n by mathematical induction.

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    (Original post by joostan)
    Not quite:
    \displaystyle\sum_{r=1}^{k+1} r = \displaystyle\sum_{r=1}^{k} r + (k+1)
    Okay I've been able to prove that one, here's my solution, if you can read it.

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    (Original post by DJMayes)
    Induction is a method of proof used when you want to prove a statement concerning integers, and this is usually restricted to positive integers. The logic is essentially this:

    - Prove that it is true for one specific case
    - Prove that, if it is true for one case, it must be true for the next.

    You usually perform the 2nd step by assuming that it is true for some value of n (Denoted k) and showing that this implies it is true when n=k+1. First of all, do the question joostan set you. Then, look at this question:

    Prove that  \frac{d^n}{dx^n} (xe^x) = (x+n)e^x

    Induction hint:

    Spoiler:
    Show


    - We can very easily link one case to another - we just need to differentiate the case!



    Full solution:

    Spoiler:
    Show


     \frac{d^n}{dx^n}( xe^x) = (x+n)e^x

    We can differentiate using the product rule, giving  \frac{d}{dx} xe^x = xe^x+e^x = (x+1)e^x therefore it is true when n = 1. Assume it is true when n = k:

     \frac{d^k}{dx^k}xe^x = (x+k)e^x

    Now, differentiate both sides with respect to x:

     \frac{d^{k+1}}{dx^{k+1}}xe^x = (x+k)e^x+e^x = (x+k+1)e^x therefore true when n = k+1.

    We have shown that, if the statement is true for some value of n, it must be true for the next. It is true when n=1 therefore it is true for all positive integers n by mathematical induction.

    Onto your question now, thanks for your time by the way and that's a great way of explaining it too


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    (Original post by MathsNerd1)
    Okay I've been able to prove that one, here's my solution, if you can read it.

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    Looks good to me
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    (Original post by DJMayes)
    Induction is a method of proof used when you want to prove a statement concerning integers, and this is usually restricted to positive integers. The logic is essentially this:

    - Prove that it is true for one specific case
    - Prove that, if it is true for one case, it must be true for the next.

    You usually perform the 2nd step by assuming that it is true for some value of n (Denoted k) and showing that this implies it is true when n=k+1. First of all, do the question joostan set you. Then, look at this question:

    Prove that  \frac{d^n}{dx^n} (xe^x) = (x+n)e^x

    Induction hint:

    Spoiler:
    Show


    - We can very easily link one case to another - we just need to differentiate the case!



    Full solution:

    Spoiler:
    Show


     \frac{d^n}{dx^n}( xe^x) = (x+n)e^x

    We can differentiate using the product rule, giving  \frac{d}{dx} xe^x = xe^x+e^x = (x+1)e^x therefore it is true when n = 1. Assume it is true when n = k:

     \frac{d^k}{dx^k}xe^x = (x+k)e^x

    Now, differentiate both sides with respect to x:

     \frac{d^{k+1}}{dx^{k+1}}xe^x = (x+k)e^x+e^x = (x+k+1)e^x therefore true when n = k+1.

    We have shown that, if the statement is true for some value of n, it must be true for the next. It is true when n=1 therefore it is true for all positive integers n by mathematical induction.

    This is what I got, which now looking at your solution seems the same really

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    (Original post by joostan)
    Looks good to me
    Thanks, do you have a more challenging one like DJ provided me I feel the hardest part is finding what else to add when proving n= K+1 but once I have it's very simple, or at least these cases were


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    (Original post by Felix Felicis)
    I know, but in essence it gives you the same equation to generate sets of solutions which is what joostan and justinawe proposed earlier - Mayman said it was unsatisfactory so presumably there's a more general solution that he knows of and it's bugging me :dontknow:

    And my apologies but I don't really see the problem I've written exactly what you have (albeit skipped a step) and nowhere have I written p^{3} mod 5 But yeah, I suppose it would've been easier to write 5 | p^{3} or something :lol:

    It's good isn't it? I started last week and I'm on season 2 :ninja: Hugh Laurie is a god
    Protip: You can edit your own post to try save face but you can't edit joostan's post quoting you on page 194, and so trying to do so carries the risk that someone might remember said quote, and you end up looking a lot worse than if you'd just said "Fair enough, fixed."

    http://www.thestudentroom.co.uk/show...e=194&page=194
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    (Original post by MathsNerd1)
    Thanks, do you have a more challenging one like DJ provided me I feel the hardest part is finding what else to add when proving n= K+1 but once I have it's very simple, or at least these cases were
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    DJ Set a slightly harder version of this on the old thread.
    Prove by induction that:
    \dfrac{d}{dx}(x^n)=nx^{n-1}
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    (Original post by DJMayes)
    Protip: You can edit your own post to try save face but you can't edit joostan's post quoting you on page 194, and so trying to do so carries the risk that someone might remember said quote, and you end up looking a lot worse than if you'd just said "Fair enough, fixed."

    http://www.thestudentroom.co.uk/show...e=194&page=194
    That was a LaTeX error which I edited after I saw it...if you look, the last edit to my solution was at 20:04 and your post was at 20:30... -.-
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    (Original post by Felix Felicis)
    That was a LaTeX error which I edited after I saw it...if you look, the last edit to my solution was at 20:04 and your post was at 20:30... -.-
    It's not actually possible to see what time my post was at, it just says "1 hour ago" which means anything up to 1 : 59 : 999 and doesn't help us.

    (And even if that's the case, why would you not simply say "that was an error, I've edited it out?" rather than stating that you never wrote it, which is clearly false...)
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    (Original post by joostan)
    DJ Set a slightly harder version of this on the old thread.
    Prove by induction that:
    \dfrac{d}{dx}(x^n)=nx^{n-1}
    I've gotten stuck on the inductive step :-/

    Hmm let me take another look at it


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    (Original post by DJMayes)
    It's not actually possible to see what time my post was at, it just says "1 hour ago" which means anything up to 1:59:999 and doesn't help us.

    (And even if that's the case, why would you not simply say "that was an error, I've edited it out?" rather than stating that you never wrote it, which is clearly false...)
    Hover above the "1 hour ago" and it says 20:30....

    Ok, I did write it but I didn't realise that you saw the solution with the LaTeX error and thought you were scrutinising my solution based on the updated one. My bad. And tbf, I didn't say "I never wrote it" I said "Nowhere have I written it" which is true as I made that post after updating my solution.
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    (Original post by Felix Felicis)
    Hover above the "1 hour ago" and it says 20:30....

    Ok, I did write it but I didn't realise that you saw the solution with the LaTeX error and thought you were scrutinising my solution based on the updated one. My bad. And tbf, I didn't say "I never wrote it" I said "Nowhere have I written it" which is true as I made that post after updating my solution.
    Well, I learned something new today. :lol:

    And fair enough - this did come across very fishy from my perspective though.
 
 
 
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