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    (Original post by MathsNerd1)
    What do you mean?


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    2n is the sumation of all the coeffecients of a binomial expansion. So you can try to prove it by expanding anything which is raised to a power>4. I'm just suggesting a solution, so I'm not exactly sure wether I'm right or wrong
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    (Original post by MathsNerd1)
    What do you mean?


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    He means the question doesn't require induction, but in my defence, you asked for a proof by induction question
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    (Original post by Felix Felicis)
    He means the question doesn't require induction, but in my defence, you asked for a proof by induction question
    Oh okay and I did indeed, just not quite sure how to even attempt this one :-/


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    (Original post by Felix Felicis)
    Well yeah but how much effort do you need to put in to write a random number and then copy and paste it xD
    Ctrl+V doesn't really take more effort than typing '1'
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    (Original post by justinawe)
    Ctrl+V doesn't really take more effort than typing '1'
    Shush Fine fine I concede defeat :lol:
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    (Original post by MathsNerd1)
    Oh okay and I did indeed, just not quite sure how to even attempt this one :-/


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    What's another way of expressing (k+1)! ?

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    (Original post by MathsNerd1)
    What do you mean?
    Ah I should have read the preceding posts, if you're practising induction then carry on. I just meant that induction seems a waste of time on this one, because you can prove the inequality by simply re-writing it.
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    (Original post by justinawe)
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    What's another way of expressing (k+1)! ?

    Erm (K +1)K! ? I've never really understood how to manipulate factorials :-/


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    (Original post by Lord of the Flies)
    ...
    I was wondering...would you mind having a look at this problem? Proof by example's not permitted apparently
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    (Original post by MathsNerd1)
    Erm (K +1)K! ? I've never really understood how to manipulate factorials :-/


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    Ok, what've you got so far?
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    (Original post by Felix Felicis)
    question
    rational+irrational=irrational\rightarrow a+b=c
    where a=x/y and c=v/w as they are rational

    this means a+b=c\rightarrow \frac{x}{y}+b=\frac{v}{w}

    a+b=c\rightarrow \frac{x}{y}+b=\frac{v}{w} \rightarrow b=\frac{v}{w}-\frac{x}{y}

    b=\frac{vy}{wy}-\frac{xw}{wy}\therefore b=\frac{vy-xw}{wy}



Since the RHS is rational (as its a fraction with +ive intergers) B must be rational)
    However at the start of the eq we stated that b= irrational \therefore \frac{vy-xw}{wy} is irrational! This is contradicting so the sum must be irrational

    I think I did it right
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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
    I did not say that proof by example is not allowed. As my friend who gave me the question said that there is an another way of solving it. He did give a hint related to complex numbers so this must be the way

    Proof by example is all right, you just want to find an another solution for it.
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    (Original post by Felix Felicis)
    Ok, what've you got so far?
    I've got pretty much no where as I'm unsure how to prove on side is greater than another :-/ Could you work through a solution to show me how it's done and I'll see if I'm able to follow it?


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    (Original post by MAyman12)
    I did not say that proof by example is not allowed. As my friend who gave me the question said that there is an another way of solving it. He did give a hint related to complex numbers so this must be the way

    Proof by example is all right, you just want to find an another solution for it.
    That's what I've tried but I wasted most of today trying to and I just want to put my mind to rest.

    (Original post by MathsNerd1)
    I've got pretty much no where as I'm unsure how to prove on side is greater than another :-/ Could you work through a solution to show me how it's done and I'll see if I'm able to follow it?


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    I take it you've gotten to the inductive hypothesis that k! > 2^{k} ?

    Well, to complete the inductive step, you want to show that (k+1)! > 2^{k+1}

    Can you think of how to get to (k+1)! from k! ?
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    (Original post by MathsNerd1)
    Erm (K +1)K! ? I've never really understood how to manipulate factorials :-/


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    Yes.

    It's because,

    (k+1)!

    = 1 \times 2 \times 3 \times \dots \times k \times (k+1)

    = (1 \times 2 \times 3 \times \dots \times k) (k+1) = k!(k+1)
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    (Original post by Felix Felicis)
    That's what I've tried but I wasted most of today trying to and I just want to put my mind to rest.


    I take it you've gotten to the inductive hypothesis that k! > 2^{k} ?

    Well, to complete the inductive step, you want to show that (k+1)! > 2^{k+1}

    Can you think of how to get to (k+1)! from k! ?
    Isn't it
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    k!*(k+1)
    ?
    Edit: Already mentioned
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    (Original post by Felix Felicis)

    I take it you've gotten to the inductive hypothesis that k! > 2^{k} ?

    Well, to complete the inductive step, you want to show that (k+1)! > 2^{k+1}

    Can you think of how to get to (k+1)! from k! ?
    I keep forgetting that what I've assumed in n=K has to be used in the inductive step too, my mistake :-/


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    (Original post by justinawe)
    Yes.

    It's because,

    (k+1)!

    = 1 \times 2 \times 3 \times \dots \times k \times (k+1)

    = (1 \times 2 \times 3 \times \dots \times k) (k+1) = k!(k+1)
    Okay, thanks I think I can see how to prove it all now


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    (Original post by Felix Felicis)

    I take it you've gotten to the inductive hypothesis that k! > 2^{k} ?

    Well, to complete the inductive step, you want to show that (k+1)! > 2^{k+1}

    Can you think of how to get to (k+1)! from k! ?
    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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    (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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    Ok:

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

    ii) How do you get from 2^{k} to 2^{k+1} ?
 
 
 
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