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 An example of infinite dimensional linear space
04-24-2010, 09:15 PM
Post: #1
 elim Moderator Posts: 581 Joined: Feb 2010 Reputation: 0
An example of infinite dimensional linear space
Let $B=\left\{b\; | \;b=1 \text{ or }b=\sqrt{p}, \quad p \text{ is prime}\right\}$
Prove that $\displaystyle{\left\{ \sum_{k=1}^n a_k b_k \; |\; a_k \in Q, \; b_k \in B, \; n \in N\right\}}$ is a linear space over rationals $Q$ with basis $B$
04-26-2010, 03:43 PM
Post: #2
 elim Moderator Posts: 581 Joined: Feb 2010 Reputation: 0
RE: An example of infinite dimensional linear space
We prove this $$1,\sqrt{2},\sqrt{3},\cdots$$ are linear independent over $$\mathbb{Q}$$
Notations and Definitions
$$p_1,\cdots,p_n,\cdots$$ are sequence of all primes, $$p_0 = 1$$
We assume that $$J \subset N^+, \; |J| \in N, \; J_i = J\setminus \left\{i\right \}\quad\; (i \in J)$$
$$A=\left\{\sqrt{p_n} \mid n \in N^+\right \},\quad\quad A(J)=\left\{a \mid a = \sqrt{p_k}, \quad k\in J\right \}$$
$$a(J)=\prod_{u \in A(J)} u, \quad a_i = (\left\{i\right \}), \quad a(\emptyset )=1$$,
$$Q(A(J))=\left\{\displaystyle{\frac{\sum_{B\subset J} c_B a(B)}{\sum_{B \subset J} d_B a(B)}} \mid J \neq \emptyset, \quad c_B, d_B \in Q, \sum_B d_B a_B \ne 0, \quad Q(\emptyset)\right\}, \quad Q(\emptyset) = Q$$

Lemma $$Q(A(J))=\left\{b+ca_i \mid b,c \in Q(A(J_i))\right\}$$
Proof. It is enough to show that LHS $$\subset$$ RHS. Consider the general element $$\displaystyle{\frac{\sum_B c_B a(B)}{\sum_B d_B a(B)}}$$.
Since $$a_i^2 \in Q \subset Q(A(J_i)), \quad \displaystyle{\frac{\sum_B c_B a(B)}{\sum_B d_B a(B)} = \frac{s+ta_i}{u+va_i}}$$, where $$s,t,u,v \in Q(A(J_i))$$
If $$v \ne 0, u-va_i = 0$$, then $$\displaystyle{\frac{s+ta_i}{u+va_i} = \frac{s}{2u}+\frac{t}{2u} a_i, \quad \quad\frac{s}{2u},\frac{t}{2u}} \in Q(A(J_i))$$
Otherwise $$\displaystyle{\frac{s+ta_i}{u+va_i} = \frac{(u-va_i)(s+ta_i)}{u^2-v^2 a_i^2} = b+ca_i} \quad \quad (b,c \in Q(A(J_i))) \quad \square$$

Propersition $$(I,J \subset N^+, \; I \bigcap J = \emptyset \ne I, \; |I|,|J| \in N) \Rightarrow (a(I) \notin Q(A(J)))$$
Proof. If not, $$a(I) \in Q(A(J))$$ for some $$I, \; J$$. We then take $$J$$ the smallest possible set that make this happen for some $$I$$.
Clearly $$J \ne \emptyset$$ and so we can pick an $$i \in J$$, that give us $$a(I)=b+ca_i$$ with $$b,c \in Q(A(J_i))$$
Square this to get $$2bca_i = a(I)^2-b^2-c^2a_i^2 \in Q(A(J_i))$$
If $$bc \ne 0$$, then $$a_i \in Q(A(J_i))$$ and so $$a(I) = b+ca_i \in Q(A(J_i))$$
If $$c = 0$$, then $$a(I)=b+ca_i = b \in Q(A(J_i))$$
If $$b = 0$$, since $$i \notin I$$, $$a(I\bigcup \left\{i\right \} ) = a(I)a_i=ca_i^2 \in Q(A(J_i))$$
thus any cases contradict the minimal property of $$J$$. $$\square$$

Corollary $$\sqrt{p_{n+1}} \notin Q(A(\left\{1,\cdots,n\right \})) = Q(\sqrt{p_1},\cdots,\sqrt{p_n}) \supset \left\{\sum_{k=0}^n c_k\sqrt{p_k} \mid c_k \in Q\right \}$$

Corollary $$1,\sqrt{2},\sqrt{3},\sqrt{5},\cdots,\sqrt{p_n},\cdots$$ are linear independent over $$Q$$.

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