square-root of a transcendental number

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?

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Yes. Any algebraic operation like roots done to a transcendental number will keep it transcendental. – Adam Hughes Jul 13 '14 at 10:12
What argument do you use to prove that the square root of an irrational number is irrational? Have you tried the exact same argument for transcendental numbers? – Amit Kumar Gupta Jul 13 '14 at 18:04

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.

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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.

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It is not true that adjoining an algebraic number to the integers always yields a finitely generated Z module. But just working over the rationals the problem goes away and as a bonus one only needs verctorspaces knowledge on which is wider spread. – quid Jul 13 '14 at 10:22
Oops, I was too eager to use algebraic integers. That's the number theorist in me. :-) – Adam Hughes Jul 13 '14 at 10:24

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.

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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.

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It's best to be a bit more precise here, exponentiation is a simple operation, yet $2^{\sqrt{2}}$ is transcendental, and one we can raise to an irrational power, namely $\sqrt{2}$ to get an algebraic number, in fact an integer. – Adam Hughes Jul 13 '14 at 10:21