Prove expressions rational - Printable Version +- OMath! (http://math.elinkage.net) +-- Forum: Math Forums (/forumdisplay.php?fid=4) +--- Forum: Algebra (/forumdisplay.php?fid=5) +--- Thread: Prove expressions rational (/showthread.php?tid=82) Prove expressions rational - elim - 07-26-2010 03:30 PM Let $$B_n = \{-1,1\}^n, \; a_k \in \mathbb{Q}, \; k=1,\cdots,n$$ Prove that (1) $$\sum_{(b_1,\cdots,b_n)\in B_n} (b_1 \sqrt{a_1}+ \cdots +b_n \sqrt{a_n})^{2m} \in \mathbb{Q}, \; m \in \mathbf{N}$$ (2) $$\prod_{(b_1,\cdots,b_n)\in B_n} (b_1 \sqrt{a_1}+ \cdots +b_n \sqrt{a_n}) \in \mathbb{Q}$$ RE: Prove expressions rational - elim - 07-27-2010 09:17 AM (1) Let $L_k(\bf{v}) = b_1\sqrt{a_1}+\cdots+b_k \sqrt{a_k}, \; \bf{v} =(b_1,\cdots,b_k)\in B_k = \{-1,1\}^k$, then by induction we have $\sum_{\bf{v} \in B_n} L_k(\bf{v})^{2m} = \sum_{\bf{v}\in B_{n-1}} ((L_{n-1}(\bf{v})+\sqrt{a_n})^{2m}+(L_{n-1}(\bf{v})-\sqrt{a_n})^{2m})$ $=\sum_{\bf{v}\in B_{n-1}}\sum_{j=0}^{2m}C_{2m}^j L_{n-1}^{2m-j}(\bf{v})(\sqrt{a_n})^j (1+(-1)^j)=\sum_{\bf{v}\in B_{n-1}}\sum_{j=0}^m 2C_{2m}^{2j} L_{n-1}^{2(m-j)}(\bf{v})a_n^j \in \mathbb{Q}$ (2) For $L_k(\bf{x,v}) = b_1 x_1+\cdots + b_k x_k\;$ then $\prod_{v \in B_{n}}L_n(\bf{v}) = \prod_{v \in B_{n-1}} (L_{n-1}(\bf{x,v}) + x_n)(L_{n-1}(\bf{x,v})- x_n)= \prod_{v \in B_{n-1}} (L_{n-1}^2(\bf{x,v}) - x_n^2)$ is a integer-coefficient, symmetric polynomial $P(x_1,\cdots , x_n)$ and been even with respect to $x_n$ hence all $x_k$. Therefore $P(x_1,\cdots , x_n) = G(x_1^2, \cdots, x_n^2)$ for some symmetric polynomial $G(t_1,\cdots, t_n)$ Consequently, $P(\sqrt{a_1},\cdots , \sqrt{a_n}) = G(a_1,\cdots, a_n) \in \mathbb{Q}$