1231+3112 is irrational?

Question

Elois Puryear · Accepted Answer

I would assume that 1231+3112 is rational and try to find a contradiction However, I don&#039;t know where to start. Can someone give me a tip on how to approach this problem?

Jillian Edgerton · Accepted Answer

Let Q(α) denote the smallest field containing Q and α The theory of field extensions tells us that Q(1231) has degree 31 over Q, Q(3112) has degree 12 over Q , and, because (31,12)=1, we have Q(1231)∩Q(3112)=Q If 1231+3112 were a rational number, we would have 1231∈Q(1231)∩Q(3112)=Q But 1231 is not rational, contradiction.

Vasquez · Accepted Answer

It is known that algebraic integers are closed under addition, subtraction, product and taking roots.
Since 12 and 31 are algebraic integers, so does their roots $\sqrt[31]{12}, \sqrt[12]{31}$ . Being the sum of two such roots, $\sqrt[31]{12} + \sqrt[12]{31}$ is an algebraic integer.
It is also known that if an algebraic integer is a rational number, it will be an ordinary integer. Notice
$2 < \sqrt[31]{12} + \sqrt[12]{31} < \sqrt[31]{2^{4}} + \sqrt[12]{2^{5}} = 2^{\frac{4}{31}} + 2^{\frac{5}{12}} < 2 \sqrt{2} < 3$
$\sqrt[31]{12} + \sqrt[12]{31}$ isn't an integer and hence is an irrational number.

\sqrt[31]{12}+\sqrt[12]{31} is irrational?

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