Proof relating to square root and floor function 
Prove that if $a>\left\lfloor\sqrt n\right\rfloor$, then $a>\sqrt n$ where $a,n\in\mathbb{Z}$.

I tried to prove with a contradiction:
Suppose there exists $a,n\in\mathbb{Z}$ such that $a>\left\lfloor\sqrt n\right\rfloor$ and $a\leq\sqrt n$. This means that $\left\lfloor\sqrt n\right\rfloor < a\leq\sqrt n$.
This can be rewritten as $\left\lfloor\sqrt n\right\rfloor < a<\sqrt n$ or $\left\lfloor\sqrt n\right\rfloor < a=\sqrt n$.
And I don't know how to proceed.
Is this the correct way to prove this statement?
 A: If $a,n\in\mathbb{Z}$, then $\sqrt{n}$ is either an integer, or an irrational. $\lfloor\sqrt{n}\rfloor$ however is always an integer.
Consider $\lfloor\sqrt{n}\rfloor\lt a\lt\sqrt{n}$: the case where the radical is an integer means it is equal to its floor, and therefore $a$ is somehow lesser than and greater than the same integer $\sqrt{n}$ - the first contradiction.
The second case where the radical is an irrational has $a$ strictly between an integer, and a non-integer that is within $1$ of the first integer; $\lfloor\sqrt{n}\rfloor\lt a\lt\lfloor\sqrt{n}\rfloor+ 1\implies a\notin\mathbb{Z}$, the second contradiction.
The case where the floor is equal to the radical is equally contradictory; when the radical is an integer, we have $a=x$ and simultaneously $a\gt x$, for some integer $x=\sqrt{n}$, a contradiction, and where the radical is irrational we have equality between $a$ and an irrational, giving $a\notin\mathbb{Z}$, another contradiction.
All cases are contradictory, completing the proof.
