Square Root of 2 is Algebraic of Degree 2

Theorem

The square root of $2$ is an algebraic number of degree $2$.

Proof

Suppose $\sqrt 2$ could be expressed as the root of the linear polynomial:

$a_1 x + a_0 = 0$

for some $a_0, a_1 \in \Q$.

Then:

$\sqrt 2 = -\dfrac {a_0} {a_1}$

and would be rational.

But as Square Root of 2 is Irrational, this is not the case.

However, $\sqrt 2$ is a root of the polynomial of degree $2$:

$x^2 - 2 = 0$

Hence the result by definition of degree of algebraic number.

$\blacksquare$

Sources

• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental