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An example of infinite dimensional linear space
04-24-2010, 09:15 PM
Post: #1
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 $B$
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04-26-2010, 03:43 PM
Post: #2
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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