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Math Help - Prove a product is rational

  1. #1
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    Prove a product is rational

    Let B_n = \{-1,1\}^n = \{(b_1,\cdots,b_n) \mid b_k \in \{-1,1\}, k=1,\cdots,n\}, \; a_k \in \mathbf{Q},\; a_k > 0, \; \forall k
    Prove that \prod_{(b_1,\cdots,b_n)\in B_n}(b_1 \sqrt{a_1}+\cdots + b_n \sqrt{a_n}) \in \mathbf{Q}
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  2. #2
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    Discriminant product

    Quote Originally Posted by elim View Post
    Let B_n = \{-1,1\}^n = \{(b_1,\cdots,b_n) \mid b_k \in \{-1,1\}, k=1,\cdots,n\}, \; a_k \in \mathbf{Q},\; a_k > 0, \; \forall k
    Prove that \prod_{(b_1,\cdots,b_n)\in B_n}(b_1 \sqrt{a_1}+\cdots + b_n \sqrt{a_n}) \in \mathbf{Q}
    For example
    show that (b1 sqrt(a1) + b2(sqrt(a2)) ( b1 sqrt(a1) - b2 sqrt(a2)) (-b1 sqrt(a1) + b2 sqrt(a2)) (-b1 sqrt(a1) - b2 sqrt(a2)) is rational.

    A hand waving argument is as follows.

    For each b value, b1, b2, b3,....bk,
    then the b is multiplied by a different b,
    for example b1 * b2, it is , in another term,
    b1 is multiplied by another -b2.

    The basic idea is that all possible signs are assigned to the b values.
    so all the terms of b1 sqrt(a1) with a different b+j sqrt(a_j) will add to zero.

    this leaves the product to be

    b1^(2^k) a1^(2^(k-1)) b2^(2^k) a2^(2^(k-1)) ....
    which is rational.
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