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    I'm reading a book which gives a proof of the Cauchy-Schwarz inequality. It starts by proving the following theorem (F is a general field with characteristic not 2):

    If m=2^n and c_1,...,c_m, d_1,...,d_m \in F then there exists a_2,...,a_m \in F such that

    \begin{equation}(c_1^2+...+ c_m^2)\cdot (d_1^2+...+d_m^2) = (c_1 d_1+...+c_m d_m)^2+a_2^2+...+a_m^2.\end{equa  tion}

    After the proof the book reads, "For F being the field of real numbers, the above theorem gives a hilarious proof of the Cauchy-Schwarz inequality!".
    I can see how the theorem proves the C/S inequality for m=2^n but does it prove it for all n? Also, does anyone else find it hilarious?
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    Say we call the vectors c=(c_1, \dots, c_m) and d=(d_1, \dots, d_m) but m \ne 2^n. Then suppose 2^{n-1}<m<2^n, and extend the vectors to be of length 2^n, with c_{m+1}=d_{m+1}=\cdots=c_{2^n}=d  _{2^n}=0.

    I'm not sure how it's particularly hilarious, but it's quite cool I suppose.
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    (Original post by 0-))
    After the proof the book reads, "For F being the field of real numbers, the above theorem gives a hilarious proof of the Cauchy-Schwarz inequality!"
    :eyebrow:
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    (Original post by nuodai)
    Say we call the vectors c=(c_1, \dots, c_m) and d=(d_1, \dots, d_m) but m \ne 2^n. Then suppose 2^{n-1}<m<2^n, and extend the vectors to be of length 2^n, with c_{m+1}=d_{m+1}=\cdots=c_{2^n}=d  _{2^n}=0.

    I'm not sure how it's particularly hilarious, but it's quite cool I suppose.
    Thank you!

    I now realise why it's hilarious but it's one of those jokes where you have to be there .
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    (Original post by 0-))
    Thank you!

    I now realise why it's hilarious but it's one of those jokes where you have to be there .
    Clearly :p:
 
 
 

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