8. Apply the definition to show that sqrt(2 + sqrt(2)) is an algebraic number.

Solution: Let x = sqrt(2 + sqrt(2)). Then x² = 2 + sqrt(2), so (x² - 2)² = 2. We can rewrite this as x⁴ - 4x² + 2 = 0. So, x is a root of a polynomial equation with integer coefficients, thus it is an algebraic number.

9. Is the cube root of 2 an algebraic number? Why?

Ans: Yes. Solution:It is the solution of the equation x³ - 2 = 0.

11. Prove that if a number of the form a + b sqrt(c) is a solution of x³ + dx² + ex + f = 0, then so is a - b sqrt(c).

Solution: Substitute a + b sqrt(c) in the equation, multiply out and collect terms to get
a³ + 3ab²c + a²d + b²cd + ea + f + (3a²b + b³c + 2abd + eb) sqrt(c) = 0,
i.e., an equation of the form P + Q sqrt(c) = 0, where P and Q are rational numbers. Now, assuming that the sqrt(c) is not rational, we must have P = Q = 0. We now substitute a - b sqrt(c) into the cubic polynomial and simplify as before to get, P - Q sqrt(c). But, since P = Q = 0, we see that a - b sqrt(c) is also a root of this polynomial.

12. Explain why the equation 8x³ - 6x - 1 = 0 has no rational solutions.

Solution: If the reduced rational number a/b was a solution to this equation, then after substituting this value for x and multiplying both sides of the resulting equation by b³, we would get the equation: 8a³ -6ab² - b³ = 0. This says that b is an integer solution of the equation -y³ -6ay² + 8a³ = 0. b would then have to divide the term 8a³, but since a and b have no common factor (a/b was reduced), this would mean that b must divide 8. In a similar manner, we see that a must divide -1. So, the only possible rational roots of the original equation are of the form, a/b where a is a divisor of -1 and b is a divisor of 8. This gives just 8 possibilities, and checking each of them shows that none is a root.

18.Show that the cube root of 7 is not constructible.

Solution:The equation x³ - 7 = 0 has one real root and two complex roots (divide the cubic by the linear factor x - 7^1/3 and note that the quadratic quotient has no real roots). If the cube root of 7 was constructible, then this equation would have a real root of the form a + b sqrt(c). By problem 11, the equation would also have to have a real root of the form a - b sqrt(c). Thus, unless b = 0, the equation would have to have two real roots, a contradiction. If b = 0, then the equation need only have the one real root, a, but a is a rational number and the cube root of 7 is not.