Since the OP has not come back to this, I will go ahead and answer it for other who might be interested.
To say that an infinite set of vectors is "independent" means that if, for any finite subset, \(\displaystyle v_1, v_2, ..., v_n\) and corresponding scalars, \(\displaystyle a_1, a_2, ..., a_n\), \(\displaystyle a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0\), then we must have \(\displaystyle a_1= a_2= \cdot\cdot\cdot= a_n= 0\). That is, all the scalars must be 0.
To show that "for \(\displaystyle \alpha\) a transcendental number the set \(\displaystyle \{1, \alpha, \alpha^2, \alpha^3, ...\}\) is linearly independent, use "indirect proof". That is, assume the set is NOT linearly independent and derive a contradiction.
Suppose the set \(\displaystyle \{1, \alpha, \alpha^2, \alpha^3...\}\) is NOT linearly independent. Then it is linearly dependent and there exist a finite set of rational numbers, \(\displaystyle \{a_1, a_2, ..., a_n\}\), not all 0, such that \(\displaystyle a_1+ a_2\alpha+ a_2\alpha^2+ \cdot\cdot\cdot+ a_n\alpha^n= 0\). Since \(\displaystyle \{a_1, a_2, ..., a_n\}\) is a finite set of rational numbers, they have a "least common denominator". Multiplying the equation by that least common denominator, we have \(\displaystyle b_1+ b_2\alpha+ b_2\alpha^2+ \cdot\cdot\cdot+ b_n\alpha^n= 0\) where the coefficients are now integers. But that means that \(\displaystyle \alpha\) satisfies a polynomial equation with integer coefficients, contradicting the hypothesis that \(\displaystyle \alpha\) is a transcendental number.
as a linear space over
and let
be a transcendental number
