It is proved that there exists an irreducible polynomial f (x) over the field of rational numbers, such that f (a) = 0
Prove: because (√ 2 + √ 3) (√ 2 - √ 3) = - 1, (√ 2 + √ 3) + (√ 2 - √ 3) = 2 √ 2, so √ 2 + √ 3 is the root of the equation x ^ 2-2 √ 2x-1 = 0, x ^ 2-2 √ 2x-1 = 0, multiplied by x ^ 2 + 2 √ 2x-1 to get: (x ^ 2-1) ^ 2 - (2 √ 2x) ^ 2 = 0, that is: x ^ 4-10x ^ 2 + 1 = 0, take F (x) = x ^ 4-10x ^ 2 + 1, then f (x) is a rational number field
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