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    (Original post by bananarama2)
    I like to live dangerously.

     (\box + \mu ^2 ) \psi =0

    Look at that beauty.

    Edit: It won't let me put put the D'alembert symbol.
    I was wondering for a while why on Earth you'd put a floating index in a bracket and times it by the wavefunction... and why it was then zero. :lol:

    (Original post by bananarama2)
    Solution 63

    \forall x,y,z,w, \ x^2+y^2+z^2+w^2+1 \geq x+y+z+w

     \iff \displaystyle x^2-x+y^2-y+z^2-z+w^2-w \geq -1

     \iff (x-\frac{1}{2})^2 -\frac{1}{4} + (y-\frac{1}{2})^2 -\frac{1}{4} + (z-\frac{1}{2})^2 -\frac{1}{4} + (w-\frac{1}{2})^2 -\frac{1}{4} \geq -1

    \iff (x-\frac{1}{2})^2 + (y-\frac{1}{2})^2  + (z-\frac{1}{2})^2 + (w-\frac{1}{2})^2 \geq 0

    Which is true since a square is equal greater than one.
    Indeed. I thought that after some of the previous questions, I would post some gentle questions.
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    (Original post by bananarama2)
    :pierre: One does no simply use square.

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    That'll do, thanks
    Tell that to Frodo...
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    (Original post by metaltron)
    I have another solution that I'm quite happy with so I thought I'd post it on here:

     2^{60} - 1 = \frac{1}{2}(2^{61} - 2) = \frac{1}{2}(\displaystyle \binom{61}{1} + \displaystyle \binom{61}{2} + \displaystyle \binom{61}{3} + .. + \displaystyle \binom{61}{60} )

    The bracket is divisible by 61 and divisible by 2 since only 61C1 and 61C60 are odd, due to the otherwise everpresent factor of 60. Hence:

     2^{60} -1 is divisible by 61.
    Very nice also.
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    (Original post by bananarama2)
    Why does the first one seem so familiar? :pierre:
    I may have reused it from a thread long ago... shhhh... :ninjagirl:

    However, I posted it in that other thread, so it's fine.
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    Problem 64 */**(?)

    Prove that the following are irrational:

    (a)  \sqrt{2} + \sqrt{3};
    (b) the real root of x^3-4x+7=0; and
    (c)  \log_2{3}

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    This was on a first year Cambridge paper. Standards these days
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    (Original post by shamika)
    Problem 64 */**(?)

    Prove that the following are irrational:

    (a)  \sqrt{2} + \sqrt{3};
    (b) the real root of x^3-4x+7=0; and
    (c)  \log_2{3}

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    This was on a first year Cambridge paper. Standards these days
    Really? Short questions on N+S, or analysis or something?
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    (Original post by Slumpy)
    Really? Short questions on N+S, or analysis or something?
    Long question in N+S (I missed out the bookwork bit)
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    (Original post by shamika)
    Long question in N+S (I missed out the bookwork bit)
    I vaguely recall the proof that e is irrational taking a while, but the log one takes 2 lines, and the first one is very standard. To my shame, I'm not immediately seeing the cubic one!
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    Solution 64

    log_{2}3=\frac{p}{q} leads to 3^{q}=2^{p} which is a contradiction.
    If \sqrt{2}+\sqrt{3} is rational, then \sqrt{6} is rational - contradiction.
    If x^{3}-4x+7 has rational root x_{0}, then x_{0}=7 or x_{0}=-7, or x= \pm 1. None of those satisfies the equation. Hence the conclusion.
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    (Original post by shamika)
    Problem 64 */**(?)

    Prove that the following are irrational:

    (a)  \sqrt{2} + \sqrt{3};
    (b) the real root of x^3-4x+7=0; and
    (c)  \log_2{3}

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    This was on a first year Cambridge paper. Standards these days
    a) and c)

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    a)
    asume they are rational, square it to get 5+2sqrt(6) is rational so sqrt(6) is ration, and sqrt(6) can be easily proven irrational in the same way as sqrt(2)

    c)
    assume rational and equal to a/b so 2^(a/b)=3

    2^a=3^b

    this is clearly a contradiction
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    Solution 14 (EDIT)


    2^{60}-1\equiv ((61+3)^{5})^{2}-1

    which in turn (diff of 2 squares) is:

     ((61+3)^{5}-1) ((61+3)^{5}+1)

    expand the right hand bracket by the binom theorem and you get:

    61^{5}+5(61^{4})(3)+10(61^{3})(9  )+10(61^{2})(27)+5(61)(81)+3^{5}  +1

    and 3^{5}+1=244=4 \times 61

    giving (for the right hand bracket):

    61(61^{4}+5(61^{3})(3)+10(61^{2}  )(9)+10(61)(27)+409)
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    (Original post by Mladenov)
    Solution 64

    If x^{3}-4x+7 has rational root x_{0}, then x_{0}=\frac{1}{7} or x_{0}=-\frac{1}{7}, or x= \pm 1. None of those satisfies the equation. Hence to conclusion.
    I think you mean x_0=\pm 7 or x_0=\pm 1
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    (Original post by Slumpy)
    I vaguely recall the proof that e is irrational taking a while, but the log one takes 2 lines, and the first one is very standard. To my shame, I'm not immediately seeing the cubic one!
    Come to think of it, I can't remember the proof for e being irrational either
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    (Original post by Lord of the Flies)
    I think you mean x_0=\pm 7 or x_0=\pm 1
    That is exactly what I mean.
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    It is high time for me to go to bed.
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    Another physicsy maths one :awesome:

    Problem 65 **/***

    A cone of base radius R, height H, mass M and homogeneous mass density ϱ is orientated as shown below.

    Name:  Untitled.jpg
Views: 256
Size:  11.5 KB

    The angle the cone makes to the xy plane is α. The cone is then attached to an axis along its outer mantle in the direction of  \hat{\mathbf{u}}.

    Calculate the moment of inertia of the cone, with respect to this axis.
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    (Original post by Mladenov)
    Solution 64

    If x^{3}-4x+7 has rational root x_{0}, then x_{0}=7 or x_{0}=-7, or x= \pm 1. None of those satisfies the equation. Hence the conclusion.
    You should write out the proof for the last one properly - if I hadn't set the question I might not follow the solution (but maybe tomorrow )
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    (Original post by Hasufel)
    Problem 14


    2^{60}-1\equiv ((61+3)^{5})^{2}-1)

    which in turn (diff of 2 squares) is:

     ((61+3)^{5})-1) ((61+3)^{5})+1)

    expand the right hand bracket by the binom theorem and you get:

    61^{5}+5(61^{4})(3)+10(61)^{3}(9  )+10(61^{2})(27)+5(61)(81)+3^{5}  +1

    and 3^{5}+1=244=4 \times 61

    giving:

    61(61^{4}+5(61^{3})(3)+10(61)^{2  }(9)+10(61)(27)+409)
    You mean solution 14.

    The alternative methods are just rolling in...
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    (Original post by Star-girl)
    You mean solution 14.

    The alternative methods are just rolling in...
    I think Boy Wonder's solution should get credited in the OP too :teehee:
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    (Original post by shamika)
    Come to think of it, I can't remember the proof for e being irrational either
    I won't type it out in LaTeX as it will take too long but I believe that the standard proof goes something like this:

    1. Assume the converse that e is rational and thus express it as a/b where a and b are coprime integers.
    2. Express e in its series expansion and equate this to a/b.
    3. Multiply through by b factorial.
    4. (b-1)!a is clearly an integer, so the other side must also be an integer and so the series on the other side must sum to an integer.
    5. Deduce that the terms in the series on the other side beyond b!/b! add to an integer bounded by 0 and 1.
    6. No such integer exists - contradiction.
    7. It follows that e must be irrational.
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    (Original post by shamika)
    You should write out the proof for the last one properly - if I hadn't set the question I might not follow the solution (but maybe tomorrow )
    Can you explain it? I don't get the sudden jump.
 
 
 
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