square-root of a transcendental number

Question

I know that a square-root of an irrational number is also irrational. Is it also true that the square root of a transcendental number is transcendental?

Answer 1

By definition a transcendental number, $t$, satisfies no polynomial with rational coefficients. If we have a number $\sqrt{t}$ which is algebraic, then--by definition--it satisfies some polynomial $p(x)$ with rational coefficients, and so the $\Bbb Q$-vector space $\Bbb Q[\sqrt{t}]$ is finitely generated. But every subspace of a $\Bbb Q$ vector space of finite dimension is finite dimensional, hence $\Bbb Q[t]\subseteq\Bbb Q[\sqrt{t}]$ must be finite dimensional, hence if $\sqrt{t}$ is algebraic, so is $t$. In fact, this is how one generally proves that the sum/difference and product of algebraic numbers is algebraic.

In simpler terms: By definition since $p(\sqrt{t})=0$ we can write this as:

$$\sum_{i=0}^n a_i\sqrt{t}^i=0$$ with $a_n\ne 0$

$$\iff \sqrt{t}^n=-{1\over a_n}\sum_{i=0}^{n-1}a_ix^i$$

so that we can always write it in terms of lower powers, but then the subset of things of the form $$\sum_{i=0}^Nb_it^i=\sum_{i=0}^N b_i\sqrt{t}^{2i}$$ is write-able in terms of only lower powers of $\sqrt{t}$ (up to $N$) so that there is a largest power, $n_0$ of $t$ this is independent of the lower powers. This gives rise to a relationship:

$$\sum_{i=0}^{n_0}c_it^i=0$$

i.e. $t$ is algebraic.

Answer 2

I'm sure someone else can give you a more elaborate answer, but the idea is this:

Hint(-ish) If you, by simple algebraic operations, could turn a transcendental number into an algebraic number, you could to the "reverse" and turn an algebraic number into a transcendental number. That would mean you'd have a transcendental number expressed in terms of algebraic operations and an algebraic number.

Answer 3

Since the product of algebraic numbers is algebraic, if $\sqrt{a}$ is algebraic, then $a = \sqrt{a} \sqrt{a}$ is algebraic. So $a$ trascendental implies $\sqrt{a}$ trascendental.

Answer 4

Following is a more elementary proof.

Given any transcendental number $t$. If $\sqrt{t}$ is algebraic, then it is a root of some polynomial $P(x) \in \mathbb{Z}[x]$. We can split $P(x)$ into two polynomials, one for those power of $x$ which are even, the other one for odd. i.e there exists two polynomials $Q(x)$ and $R(x)$ such that

$$P(x) = Q(x^2) + xR(x^2)$$

It is clear $t$ is a root of the polynomial $$Q(x)^2 - xR(x)^2 = \underbrace{\left(Q(\sqrt{x}^2) + \sqrt{x}R(\sqrt{x}^2)\right)}_{\Large= P\left(\sqrt{x}\right)} \left(Q(\sqrt{x}^2) - \sqrt{x}R(\sqrt{x}^2)\right) $$ contradict with the given condition that $t$ is transcendental.

Answer 5

Lots of good answers, but here is the way I managed to understand it:

Suppose $\sqrt{\alpha}$ is algebraic. Then it is a root of a polynomial $f \in \mathbb{Q}[x]$ of degree $n$. But then {$1, \sqrt{\alpha}, \ldots , (\sqrt{\alpha})^{n-1}$} is a basis for $\mathbb{Q}(\alpha)$ (see theorem 30.23 in Fraleigh's book), and the degree of the extension is $n$. But since every finite finite extension is algebraic, every element in $\mathbb{Q}(\alpha)$ is algebraic. But then, note that $\alpha \in \mathbb{Q}(\alpha)$ (since you can always multiply two members of $\mathbb{Q}(\alpha)$, then you just multiply $\sqrt{\alpha} * \sqrt{\alpha} = \alpha \in \mathbb{Q}(\alpha)$. So, $\alpha$ is algebraic, a contradiction, since $\alpha$ is transcendental, by hypostesis. Therefore, $\sqrt{\alpha}$ is transcendental.