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    I have only just realised that this is on the spec. (and according to the spec it is guaranteed that 1 question will involve proof).

    I have looked it up on Google and I follow the idea:

    Take something that you want to disprove and show that it leads to a contradiction.

    Many of the proofs involve rational/irrational numbers and showing that such and such is irrational.

    Could someone go through the steps to show that log(2)5 is irrational?

    Thank you.
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    (Original post by samdavyson)
    I have only just realised that this is on the spec. (and according to the spec it is guaranteed that 1 question will involve proof).
    Any links?
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    For that on the spec or for information on proof by contradiction?

    The spec is at: http://www.edexcel.org.uk/VirtualCon...cification.pdf

    And the bit of interest is page 33, but I have attached a screen shot of the paragraph of importance.

    As for proof by contradiction try: http://zimmer.csufresno.edu/~larryc/...ontradict.html
    http://www.delphiforfun.org/Programs...tradiction.htm
    http://www.math.uncc.edu/~droyster/m...om/node24.html

    and loads more here
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    Could someone go through the steps to show that log(2)5 is irrational?
    Assume that it's rational, i.e.
    log(2) 5 = p/q, where both p & q are non-zero integers.
    5 = 2^(p/q)
    5^q = 2^p

    The LHS is odd for all integral q, while the RHS is even for all integral p. Contradiction.
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    (Original post by samdavyson)
    For that on the spec or for information on proof by contradiction?

    The spec is at: http://www.edexcel.org.uk/VirtualCon...cification.pdf

    And the bit of interest is page 33, but I have attached a screen shot of the paragraph of importance.

    As for proof by contradiction try: http://zimmer.csufresno.edu/~larryc/...ontradict.html
    http://www.delphiforfun.org/Programs...tradiction.htm
    http://www.math.uncc.edu/~droyster/m...om/node24.html

    and loads more here
    thanx didn tknow that =o abit sneaky the way its put in the spec tho
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    (Original post by dvs)
    Assume that it's rational, i.e.
    log(2) 5 = p/q, where both p & q are non-zero integers.
    5 = 2^(p/q)
    5^q = 2^p

    The LHS is odd for all integral q, while the RHS is even for all integral p. Contradiction.
    That is good.

    How do you go from the secon dfrom bottom line to the bottom line so fast?

    I would go:
    5 = 2^(p/q)
    ln5 = (p/q)ln2
    qln5 = pln2
    ln5^q = ln2^p
    => 5^q = 2^p

    This may be what you did but you just did it in one step, or is there a good quick way?
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    (Original post by TheWolf)
    thanx didn tknow that =o abit sneaky the way its put in the spec tho
    It also says tan(θ/2) is not required!

    See attachment.
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    Raise both sides to the power of q, since [2^(p/q)]^q = 2^p.
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    Thanks.
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    (Original post by samdavyson)
    It also says tan(θ/2) is not required!

    See attachment.
    Looks like they changed the spec recently? If so, that's v.gay. :mad:
 
 
 
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