Study whether $\succsim$ represented by $u(x)=\lfloor x \rfloor$ is continuous Using the following definition of continuity: $\succsim$ is continuous if for any bundles $x,y,z$ such that x$\succ$y$\succ$z, there exists $\alpha \in (0,1)$ such that $\alpha x + (1-\alpha)z \sim y$.

I am unable to show continuity/not continuity using this definition.
Are you sure that the definition you want? 

With your definition then yes its true.

You should show that $x\succ y\succ z\implies x&gt;y&gt;z$. Hence, you can always find a $\alpha$ such that $\alpha x+(1-\alpha)y=z$, so $u(\alpha x+(1-\alpha)y)=u(z)$ which then implies $\alpha x+(1-\alpha)y\sim z$.

Perhaps the definition you want to use is for all $x^n$ and $y^n$ are two sequences with $x^n\to x$ and $y^n\to y$ and $x^n\succeq y^n$ for all $n$ then $x\succeq y$? But maybe not, I dont want to presume.