Official TSR Mathematical Society

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  1. Blutooth's Avatar
    • Peer Of The TSR Realm
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    Re: Official TSR Mathematical Society
    (Original post by Dadeyemi)
    Question: If you choose an answer to this question at random, what is the probability that you will be correct?
    a. 25%
    b. 50%
    c. 0%
    d. 25%

    I'm guessing the law of the excluded middle features quite a bit in this paradox.

    Logical bit
    Spoiler:
    Show

    Assume a and d are correct. Then you have a 50% chance of answering the question right at random. SO a and d can't be right.

    Assume b is right. Then you have a 25% chance of getting the questions right. So b is wrong

    Assume c is right. Then you have a 0% chance iof getting the answer right so c is wrong.


    Assume all the options are wrong, then you have a 0% chance of getting the answer right, so the answer is c. But then all the answers can't be wrong.


    Philosophical bit
    Spoiler:
    Show

    So both the answers "all answers are wrong" and "there is at least one correct answer" lead to contradictions. Thus the answer can be neither of them- unless we wish to change the meaning of truth to allow for (P ∧ ¬P) i.e. paradoxes or we accept that some statements are neither true nor false, i.e. we drop the law of the excluded middle: ¬(P ∧ ¬P).

    I think this questions an example of one that can only be answered only with statements whose truth leads directly to a contradiction. Let's just say that such questions are undecidable and save a lot of time ;p

    Another example of an undecidable question would be:
    The sentence below is false
    The sentence above is true
    None of these three sentences are true.

    Which sentences are true?


    Nice problem though; a shame I had to ruin it with my logical analysis.: )
    Last edited by Blutooth; 14-04-2012 at 02:41.
  2. Zhen Lin's Avatar
    • Vengeful, Imperial Overlord of The Student Room
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    Re: Official TSR Mathematical Society
    (Original post by Blutooth)
    I'm guessing the law of the excluded middle features quite a bit in this paradox.
    The law of excluded middle is p \lor \lnot p. The formula you quote, \lnot (p \land \lnot p), is tautological even in weak fragments of intuitionistic logic. It is just an internalised modus ponens: (p \land (p \to \bot)) \to \bot.
  3. GreenLantern1's Avatar
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    Re: Official TSR Mathematical Society
    Alright fellow mathmos?!
  4. sheepstick's Avatar
    • Respected Member
    • Posts: 223
    Re: Official TSR Mathematical Society
    GCSE stats coursework help please:
    okay so i am doing coursework on track and field events in the Olympics i am focusing on the 2008 ones but i don't know if you can create a cumulative frequency diagram with the data that i have i have decathlon and individual data both with the events of 100m sprint and high jump help please its in for next week and i have no idea what i am doing please help!?!
  5. gff's Avatar
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    • Location: Milky way Posts: μ(Ø)
    Re: Official TSR Mathematical Society
    [*] During the weekends, Eli delivers milk in the complex plane.

    On Saturday, he begins at z and delivers milk to houses located at z^3, z^5, z^7, \cdots, z^{2013}, in that order.
    On Sunday, he begins at 1 and delivers milk to houses located at z^2, z^4, z^6, \cdots, z^{2012}, in that order.

    Eli always walks directly (in a straight line) between two houses.

    If the distance he must travel from his starting point to the last house is \sqrt{2012} on both days, find the real part of z^2.
  6. hasan123's Avatar
    • New Member
    • Posts: 18
    Re: Official TSR Mathematical Society
    I am doing some quantitative analysis on marginal costs and revenue, where I have to find the figure where marginal cost and marginal revenue are equal. I have
    revenue = 720x - 4x^2
    cost = 96-17x^2 + 560x

    Ive got
    marginal revenue = 720 - 8x
    marginal cost = 560 - 34x

    so when marginal cost = marginal revenue

    720 - 8x = 560 - 34x
    26x = -160
    x = -6.15....

    is it even possible to get a negative and what does this mean? at this point is it still maximum profit? which means you have to sell -6 units to make the maximum profit??

    If someone could help today please
  7. gff's Avatar
    • Exalted and Worshipped Member
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    Re: Official TSR Mathematical Society
    Somebody who doesn't like milk has negged me. :rolleyes:
    This crappy hand pointing downwards should be removed!
  8. nohomo's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by gff)
    Milk Delivery Problem
    Here's a solution (I hope):

    Spoiler:
    Show


    From the problem statement, we have

    |z^3-z|+|z^5-z^3|+|z^7-z^5|+\cdots + |z^{2013}-z^{2011}| = \sqrt{2012} \ \ \ \ \ (1)
    |z^2-1|+|z^4-z^2|+|z^6-z^4|+\cdots + |z^{2012}-z^{2010}| = \sqrt{2012} \ \ \ \ \ (2)

    From (1),

    |z|(|z^2-1|+|z^4-z^2|+|z^6-z^4|+\cdots + |z^{2012}-z^{2010}|) = \sqrt{2012}

    so from (2),

    |z|\sqrt{2012} = \sqrt{2012} \implies |z| = 1

    Then, from (2),

    |z^2-1|+|z|^2|z^2-1|+|z|^4|z^2-1|+\cdots + |z|^{2010}|z^2-1| = \sqrt{2012}
    1006|z^2-1| = \sqrt{2012}
    |z^2-1| = \frac{\sqrt{2}}{\sqrt{1006}} = \frac{1}{\sqrt{503}}

    Let z = a+bi, where a,b \in \mathbb{R}. Then |z| = 1 implies a^2+b^2 = 1 and |z^2-1| = \frac{1}{\sqrt{503}} implies |(a+bi)^2-1| = \frac{1}{\sqrt{503}}, so |(a^2-b^2-1)+2abi| = \frac{1}{\sqrt{503}} and (a^2-b^2-1)^2+4a^2b^2 = \frac{1}{503}, so a^4+b^4+2a^2b^2-2(a^2-b^2)+1 = (a^2+b^2)^2-2(a^2-b^2)+1 = 1-2(a^2-b^2)+1 = \frac{1}{503}, so a^2-b^2 = \frac{1005}{1006}.

    We have z^2=(a+bi)^2 = (a^2-b^2)+2abi, so \Re (z^2) = a^2-b^2 = \frac{1005}{1006}.

    Last edited by nohomo; 23-04-2012 at 16:35.
  9. gff's Avatar
    • Exalted and Worshipped Member
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    Re: Official TSR Mathematical Society
    (Original post by nohomo)
    Here's a solution (I hope):
    It is a solution.
  10. Farhan.Hanif93's Avatar
    • Section Moderator
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    Re: Official TSR Mathematical Society
    (Original post by nohomo)
    Here's a solution (I hope):

    Spoiler:
    Show


    From the problem statement, we have

    |z^3-z|+|z^5-z^3|+|z^7-z^5|+\cdots + |z^{2013}-z^{2011}| = \sqrt{2012} \ \ \ \ \ (1)
    |z^2-1|+|z^4-z^2|+|z^6-z^4|+\cdots + |z^{2012}-z^{2010}| = \sqrt{2012} \ \ \ \ \ (2)

    From (1),

    |z|(|z^2-1|+|z^4-z^2|+|z^6-z^4|+\cdots + |z^{2012}-z^{2010}|) = \sqrt{2012}

    so from (2),

    |z|\sqrt{2012} = \sqrt{2012} \implies |z| = 1

    Then, from (2),

    |z^2-1|+|z|^2|z^2-1|+|z|^4|z^2-1|+\cdots + |z|^{2010}|z^2-1| = \sqrt{2012}
    1006|z^2-1| = \sqrt{2012}
    |z^2-1| = \frac{\sqrt{2}}{\sqrt{1006}} = \frac{1}{\sqrt{503}}
    [/latex].

    Slightly neater from this point to...

    Spoiler:
    Show
    Note that |z^2-1| = \sqrt{(z^2-1)(\bar{z}^2-1)} = \sqrt{(z\bar{z})^2 - (z^2 + \bar{z}^2) + 1} = \sqrt{2-(z^2+\bar{z}^2)} so 2 - (z^2 + \bar{z}^2) = \dfrac{1}{503} \implies \dfrac{1005}{1006} = \dfrac{1}{2}(z^2+\bar{z}^2) = \mathrm{Re}(z^2)
  11. user's Avatar
    • Exalted Member
    • Posts: 278
    Re: Official TSR Mathematical Society
    A fair die is thrown four times and the number of times it falls with a 6 on the top, Y, is noted. Write down all the possible values of y.


    Why is 0 the first answer?
  12. nohomo's Avatar
    • Benevolent Member
    • Posts: 662
    Re: Official TSR Mathematical Society
    (Original post by Farhan.Hanif93)
    Slightly neater from this point to...

    Spoiler:
    Show
    Note that |z^2-1| = \sqrt{(z^2-1)(\bar{z}^2-1)} = \sqrt{(z\bar{z})^2 - (z^2 + \bar{z}^2) + 1} = \sqrt{2-(z^2+\bar{z}^2)} so 2 - (z^2 + \bar{z}^2) = \dfrac{1}{503} \implies \dfrac{1005}{1006} = \dfrac{1}{2}(z^2+\bar{z}^2) = \mathrm{Re}(z^2)
    That's a good improvement to the solution.

    Does anyone have any more problems for people to solve?
  13. sputum's Avatar
    • Adored and Respected Member
    • Posts: 425
    Re: Official TSR Mathematical Society
    (Original post by user)
    A fair die is thrown four times and the number of times it falls with a 6 on the top, Y, is noted. Write down all the possible values of y.


    Why is 0 the first answer?
    What result do you get if you roll 5 4 times?
  14. gff's Avatar
    • Exalted and Worshipped Member
    • Location: Milky way Posts: μ(Ø)
    Re: Official TSR Mathematical Society
    (Original post by nohomo)
    Does anyone have any more problems for people to solve?
    I have tons of fun problems to solve, but, unfortunately, not enough time. Help yourself.
  15. Intriguing Alias's Avatar
    • TSR Idol
    • Location: Yorkshire
    Re: Official TSR Mathematical Society
    I know this is old but I just read the solutions by nohomo (and then the bit farhan posted); thought they were both a bit convoluted (algebra-wise) nearing the end (for me anyway)

    Where nohomo says let z = a+bi

    It's a lot easier to let z^2 = a+bi

    and we have |z^2-1| = \frac{1}{\sqrt{503}}
    or (a-1)^2 + b^2 = 1/503

    and since a^2 + b^2 = 1 then 2a-1 = 1 - 1/503 and then finally a = 1 - 1/1006

    Sorry for the lack of latex; couldn't really be bothered
    Last edited by Intriguing Alias; 25-04-2012 at 12:23.
  16. HypErTwisT's Avatar
    • Full Member
    • Posts: 77
    Re: Official TSR Mathematical Society
    Is this just random questions?

    5x0 = 0
    Re-arrange.
    5 = 0/0

    seems legit
  17. GreenLantern1's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by HypErTwisT)
    Is this just random questions?

    5x0 = 0
    Re-arrange.
    5 = 0/0

    seems legit
    0/0 is an indeterminate. And the answer proves that 0/0 is infinity as it provides no solution, so there can be an arbitrary constant equal to this equation, as in this example:5.
  18. GreenLantern1's Avatar
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    Re: Official TSR Mathematical Society
    Posted for Usycool:

    When perpendicular lines meet, the product of their gradient is -1, right?

    However, say if we have the lines y = 3 and x = 5.

    The gradient of the y=3 in undefined, and the gradient of the x = 5 is 1.

    So is this not an exception for the rule that perpendicular lines's gradients have a product of -1? Or is their another explanation for this?
  19. Slumpy's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by GreenLantern1)
    Posted for Usycool:

    When perpendicular lines meet, the product of their gradient is -1, right?

    However, say if we have the lines y = 3 and x = 5.

    The gradient of the y=3 in undefined, and the gradient of the x = 5 is 1.

    So is this not an exception for the rule that perpendicular lines's gradients have a product of -1? Or is their another explanation for this?
    The gradient of the line x=5 is not 1.
  20. GreenLantern1's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by Slumpy)
    The gradient of the line x=5 is not 1.
    I only posted this for my friend not myself (so it has just been copied from what he has written to me). Can you please tell us what all the gradients are and if the prodyct of the gradients doesn't equal -1, then why!!
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