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    These are my own problems, so they may not make sense! I made them up after some exploration in areas of factorials and algebra stuff.

    If you have the answer, please post it in white colour. This way, the problems are not spoilt for people who read it after it has been solved. People looking for solutions can highlight the text to read it:

    \/-\/-\/
    Like This
    /\-/\-/\

    Thanks.

    1. (a-b)/(b-a) = 5; What does [(a²-a²-a²+b²)/(b²-b²-b²+a²)]³ equal?
    2. Simplify: [(c^d-1)!(c^d+1)]/[(c^2d-1)(c^d-2)!]
    3. If m and n are integers, and p = -(3m²) and q = (2n³), prove that p^q > [p^(q-1)](p-1)

    More may come later... They are quite hard to make up. If anyone else has original problems feel free to post them here, if you want.
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    1. EDIT: No answer because the first equation is meaningless
    2. 1
    3. EDIT: No proof because original statement is false.
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    1)
    (a²-a²-a²+b²) = -a²+b²
    (b²-b²-b²+a²) = -b²+a²
    so,
    (a²-a²-a²+b²)/(b²-b²-b²+a²) = (-a²+b²)/(-b²+a²) = -1
    Ans is (-1)³ = -1
    =============

    2)
    [(c^d-1)!(c^d+1)]/[(c^2d-1)(c^d-2)!]
    [(c^d-1)(c^d-2)!(c^d+1)]/[(c^2d-1)(c^d-2)!]
    [(c^d-1)(c^d+1)]/[(c^2d-1)] - dividing top and bottom by (c^d-2)!
    [(c^2d)]/[(c^2d-1)]
    c
    =

    3)
    let m=1, n=1
    p = -3, q = 2

    lhs = p^q = (-3)^2 = 9
    rhs = [p^(q-1)](p-1) = [(-3)^1](-4) = (-3)(-4) = 12

    lhs !> rhs
    fault in question!!
    =============
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    (Original post by AntiMagicMan)

    Lemma 4 p^q-1 is always positive
    Proof:
    p is negative
    a negative number raised to the power of a positive odd integer is always positive
    ......
    Sorry, but not so.
    p= -3 is a negative number
    q-1 = 1 is a positive odd integer

    p^(q-1) is a negative number raised to the power of a positive odd integer
    p^(q-1) = (-3)^1 = -3
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    (Original post by mik1a)
    ...
    If you have the answer, please post it in white colour. This way, the problems are not spoilt for people who read it after it has been solved. People looking for solutions can highlight the text to read it:

    \/-\/-\/
    Like This
    /\-/\-/\

    ...
    Good suggestion. Perhaps that should/could be done in the UKL maths soc?
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    Meh, of course. I was thinking of negative numbers raised to the power of positive even integers.
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    (Original post by Fermat)
    1)
    (a²-a²-a²+b²) = -a²+b²
    (b²-b²-b²+a²) = -b²+a²
    so,
    (a²-a²-a²+b²)/(b²-b²-b²+a²) = (-a²+b²)/(-b²+a²) = -1
    Ans is (-1)³ = -1
    =============

    2)
    [(c^d-1)!(c^d+1)]/[(c^2d-1)(c^d-2)!]
    [(c^d-1)(c^d-2)!(c^d+1)]/[(c^2d-1)(c^d-2)!]
    [(c^d-1)(c^d+1)]/[(c^2d-1)] - dividing top and bottom by (c^d-2)!
    [(c^2d)]/[(c^2d-1)]
    c

    For 1)

    The reason we got different answers is because the equation (a-b)/(b-a) = 5 has no solution, so it is meaningless to use it anyway.

    For 2)
    Actually (c^d-1)(c^d+1) = c^2d - 1
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    (Original post by AntiMagicMan)
    For 1)

    The reason we got different answers is because the equation (a-b)/(b-a) = 5 has no solution, so it is meaningless to use it in anyway.

    For 2)
    Actually (c^d-1)(c^d+1) = c^2d - 1
    (c^d-1)(c^d+1) = [c^(d-1)][c^(d+1)] = c^(d-1+d+1) = c^(2d)
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    (Original post by AntiMagicMan)
    For 1)

    The reason we got different answers is because the equation (a-b)/(b-a) = 5 has no solution, so it is meaningless to use it anyway.
    Of course! That eqn is saying,

    -1 = 5!!!
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    (Original post by Fermat)
    (c^d-1)(c^d+1) = [c^(d-1)][c^(d+1)] = c^(d-1+d+1) = c^(2d)
    Indeed, but ((c^d)-1)((c^d)+1) = c^(2d) + c^d - c^d -1 = (c^2d)-1.

    I agree the question is ambiguous, but really if it was meant to be c^(d+/-1) then that would have been made clear.
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    (Original post by Fermat)
    Of course! That eqn is saying,

    -1 = 5!!!
    Yeah, I think the poster's disclaimer was very wise "These are my own problems, so they may not make sense!"
    • Thread Starter
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    For the first, my method was:

    1. (a-b)/(b-a) = 5; What does [(a²-a²-a²+b²)/(b²-b²-b²+a²)]³ equal?

    (a²-a²-a²+b²)/(b²-b²-b²+a²) = (b²-a²)/(a²-b²)
    = (b+a)(b-a)/(a+b)(a-b)
    = (b-a)/(a-b)

    Now as (a-b)/(b-a) = 5, (b-a)/(a-b) = 1/5

    [(a²-a²-a²+b²)/(b²-b²-b²+a²)]³ = (1/5)³ = 1/125
    • Thread Starter
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    New original problem. Spot the fallacy.!!!
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    (Original post by mik1a)
    For the first, my method was:

    1. (a-b)/(b-a) = 5; What does [(a²-a²-a²+b²)/(b²-b²-b²+a²)]³ equal?

    (a²-a²-a²+b²)/(b²-b²-b²+a²) = (b²-a²)/(a²-b²)
    = (b+a)(b-a)/(a+b)(a-b)
    = (b-a)/(a-b)

    Now as (a-b)/(b-a) = 5, (b-a)/(a-b) = 1/5

    [(a²-a²-a²+b²)/(b²-b²-b²+a²)]³ = (1/5)³ = 1/125
    Yes, that was my solution, but it's still wrong .
 
 
 
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