How do you prove that vectors are linearly independent in $ \mathcal{C}[0,1]$?

$C[0,1]$ usually denotes the collection of continuous functions $f: [0,1]\to \mathbb{R}$. This is a vector space over $\mathbb{R}$, with multiplication defined pointwise: $(af)(x) = af(x)$.

What would it mean if two functions, $f,g\in C[0,1]$, were linearly dependent? It would meant that there were two scalars, $a,b\in\mathbb{R}$, not both zero, such that $af + bg = 0$ as functions. In other words, for all $x$, $af(x) + bg(x) = 0$.

To show that $f=x^{3/2}$ and $g=x^{5/2}$ are linearly independent, let's assume that such $a$ and $b$ exist, and try to show that they're both zero.. If such $a,b\in\mathbb{R}$ exist, then $ax^{3/2} + bx^{5/2} = 0$ for all $x$. Let's try plugging in $x=1$—we get $a+b=0$. Plugging in $x=1/4$, we get $a/8 + b/32=0$ (note: you could also plug in different values of $x$, but I chose these to keep the math as simple as possible).

Now, we can solve the two equations $a+b=0$ and $a/8+b/32=0$, to get $a=b=0$. Ah, but this is exactly what we wanted to show: $x^{3/2}$ and $x^{5/2}$ cannot have a linear sum which equal zero unless it is the trivial sum, i.e. $0\cdot f + 0\cdot g = 0$.

$C[0,1]$ is the vector space of continuous functions from $[0,1]$ into $\mathbb{R}$. You can add functions, subtract them, there is a zero function, multiply by elements in $\mathbb{R}$, etc. Thus, it makes sense to ask about linearly independent elements of this vector space. Take the elements $x^{3/2}$ and $x^{5/2}$. We want to show that these are linearly independent, so that one is not the multiple of another by an element in $\mathbb{R}$. Can you show this?