Exercise 1.1

Question

Prove that $7 + \sqrt[3]{2}$ and $\sqrt{3} + \sqrt{-5}$ are algebraic numbers.

Abdullah Zubair · Accepted Answer

First, let $\alpha = 7 + \sqrt[3]{2}$. Then  
\begin{align*}
	\sqrt[3]{2} &= \alpha - 7\
	2 &= \left( \alpha - 7 ight)^{3}\
	0 &= \alpha^{3} - 21\alpha^2 + 147\alpha - 345
.\end{align*}
From here, we see that $\alpha$ is a root to $f(x) = x^{3} - 21x^2 + 147x - 345 \in \mathbb Z[x]$, so that $\alpha$ is algebraic.

As for the latter, let $\beta  = \sqrt{3} + \sqrt{5}i$. Then 
\begin{align*}
	\beta &= \sqrt{3} + \sqrt{5}i\
	\beta^2 &= 3 - 5 + 2\sqrt{15}i\
	\beta^2 + 2 &= 2\sqrt{15}i\
	\beta^{4} + 2\beta^2 + 4 &= -60\
	\beta^{4} + 2\beta^2 + 64 &= 0
.\end{align*}

We see that $\beta$ is a root to $f(x) = 	x^{4} + 2x^2 + 64 \in \mathbb Z[x]$, so that $\beta$ is algebraic.

Exercise 1.1

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Exercise 1.1

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