Prove $\sqrt{5}$ is both irrational and algebraic. I've got the irrational part down but how do I prove algebraic? Is there a way to prove both simultaneously? 
Algebraic number: a number that is a root of a non-zero polynomial in one variable with rational coefficients (or equivalently — by clearing denominators — with integer coefficients). Numbers such as π that are not algebraic are said to be transcendental.
 A: $x^2 - 5 = 0$ for the algebraic part. If $\sqrt{5} = \frac{p}{q}$ ($p$ and $q$ - co-prime), then $5q^2 = p^2$, so $5$ divides $p^2 \Rightarrow 5$ divides $p$, because $5$ is prime. So $25$ divides $5q^2$. From where you can conclude that $5$ divides $q^2$, which means that $5$ divides $q$ (again because $5$ is prime). So $p$ and $q$ are not co-prime since $5$ divides both. Contradiction! So $\sqrt{5}$ is irrational.  
A: To prove both simultaneously, notice that $\sqrt{5}$ is a root of the polynomial
$$x^2-5$$
and that, by Eisenstein's Irreducibility Criterion, we can conclude that $x^2-5$ is irreducible. Therefore, there is no linear polynomial of which $\sqrt{5}$ is a root, implying it is irrational.
A: One can prove both simultaneously by using the rational roots theorem that $\sqrt{5}$ satisfies the "algebraic equation" $x^2-5=0$, which if it has rational roots of the form $\frac{p}{q}$ (assuming $p,q$ to have GCD $1$) then $p$ should divide $-5$ and $q$ should divide $1$ (the lowest and highest degree coefficients respectively). Now this means $q=\pm1,p=\pm 5, \pm1$ and none of the combinations of such $p,q$ satisfy that $x^2-5=0$ and thus a contradiction arises.
