Show that $f(y)=0$ for every $y \in Y$. 
If $Y$ is a subspace of a vector space $X$ and $f$ is a linear functional on $X$ such that $f(Y)$ is not the whole scalar field of $X$, show that $f(y)=0$ for all $y \in Y$.

Suppose to the contrary, that there exists some $y_0 \in Y$ such that $f(y_0) = \alpha \neq 0$. Now notice that for any scalar $\beta$ is the underlying scalar field of $X$, we have that $$\beta = \frac{\beta}{\alpha}f(y_0) = f(\frac{\beta}{\alpha}y_0).$$ I am having trouble forming or seeing the contradiction though? Can someone please help point it out to me? 

You wrote "for ANY scalar $\beta$".
That means NO MATTER WHICH scalar $\beta$ is, the conclusion holds.
The conclusion is $\beta=f\text{ of something}$.
Thus EVERY scalar is $f$ of something.
Thus $f(Y)$ is the whole scalar field.
This contradicts the assumption that $f(Y)$ is not the whole scalar field. 
