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 $\mathbb{Q}$.
This is a contradiction.

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?

  • 7
    $\begingroup$ Your proof is quite correct. However, I would say it it is a proof by contrapositive, not by contradiction. $\endgroup$
    – Bernard
    Aug 29, 2019 at 15:40

1 Answer 1


Your proof is correct. Two minor nitpicks:

  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.


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