# Thread: Prove a product is rational

1. ## 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 $\prod_{(b_1,\cdots,b_n)\in B_n}(b_1 \sqrt{a_1}+\cdots + b_n \sqrt{a_n}) \in \mathbf{Q}$

2. ## Discriminant product

Originally Posted by elim
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.