# Proving Non-algebraic over $\mathbb{Q}$

Question:

Prove or disprove that $$(1+\sqrt{e})^{1/4}$$ is algebraic over $$\mathbb{Q}$$.

My Attempt:
We know that $$e$$ is transcendental over $$\mathbb{Q}$$.
Claim: $$\sqrt{e}$$ and $$1+\sqrt{e}$$ are also transcendental over $$\mathbb{Q}$$
Proof: We know that if $$a,b\in \mathbb{Q}$$, then $$a$$ and $$b$$ are algebraic over $$\mathbb{Q}$$ and $$a-b$$, $$ab$$ are also algebraic.
So, we assume that $$\sqrt{e}$$ and $$1+\sqrt{e}$$ are algebraic over $$\mathbb{Q}$$.
$$\implies(\sqrt{e})(\sqrt{e})=e$$ is algebraic over $$\mathbb{Q}$$.

Similarly, We assume that $$1+\sqrt{e}$$ is algebraic.
$$\implies$$ $$(1+\sqrt{e})-1=\sqrt{e}$$ is algebraic.
This is a contradiction by using above fact that $$\sqrt{e}$$ is transcendental.

Using The above arguments, I can show that $$(1+\sqrt{e})^{1/4}$$ is transcendental.

Is My proof Correct?
Is there any other method to disprove?

• Your proof is quite correct. However, I would say it it is a proof by contrapositive, not by contradiction. Aug 29, 2019 at 15:40

1. You arrive at a contradiction by assuming that $$\sqrt{e}$$ and $$1+\sqrt{e}$$ are algebraic. So then the conclusion should be that they are not both algebraic. In stead, you might want to first assume that $$\sqrt{e}$$ is algebraic, and reach a contradiction, and then assume that $$1+\sqrt{e}$$ is algebraic, and again reach a contradiction.
2. You might want to prove the subsequent claim that $$(1+\sqrt{e})^{1/4}$$ is transcendental. The proof is of course essentially the same as the proof for $$\sqrt{e}$$.
Alternatively, you could show more directly that if $$\alpha:=(1+\sqrt{e})^{1/4}$$ is algebraic, then so is $$\alpha^8-2\alpha^4+1=(\alpha^4-1)^2=e,$$ a contradiction.