What is the relationship between span and dimension?

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Homework Statement



Hi, I have to prove that there's no exist a generating set for "x" with less of "n" vectors when "n" is the dimension of the basis of "x"

Homework Equations


is there a span(x) whith dimension m? when m<n and n is the dimension of the basis

The Attempt at a Solution



I know that all the basis of a vectorial space must have the same dimension, and the basis have linearly independient vectors, so I can't remove any of them, but I don't know how sow that its not possible that existence of that span
 
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to start this proof, I think you need to start with a mathematical definition of what a basis actually is.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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