A complex number, c, is said to be algebraic if it satisfies a polynomial with i

Question

A complex number, c, is said to be algebraic if it satisfies a polynomial with integer coefficients (i.e. there exists a polynomial. P(x) with integer coefficients, such that P(c) = 0). Otherwise it is called transcendental. Prove that every rational number is algebraic. Prove that each of the following is algebraic: - Squareroot 3 3 Squareroot 2 - 3 Squareroot 2 1/3 Squareroot 2 Squareroot 2 + Squareroot 3 Suppose a is algebraic and satisfies the polynomial 2x^3 + 3x^2 + 4x + 5. Prove that - a and 1/a are algebraic by finding polynomials that they satisfy.

Explanation / Answer

a.

Let, p/q be a rational number , so p,q are integers

So, qx=p has this rational number as a root

Hence every rational number is algebraic

b.

x=-sqrt{3}

Squaring gives

x^2=3

hence it is a root of polynomial

x^2-3

x=(2)^{1/3}

Taking cube of both sides gives

x^3=2

So it is a root of

x^3-2

x=-(2)^{1/3}

Taking cube of both sides gives

x^3=-2

So it is a root of

x^3+2

x=1/(2)^{1/3}

Taking cube of both sides gives

x^3=1/2

So it is a root of

2x^3-1=0

x=sqrt{2}+sqrt{3}

x-sqrt{2}=sqrt{3}

Squaring gives

x^2+2-2xsqrt{2}=3

x^2-1=2xsqrt{2}

Squaring both sides gives

(x^2-1)^2=8x

x^4+1-4x^2=8x

Hence it is a root of

x^4-4x^2-8x+1

c.

Substitute -x in place of x we get

2(-x)^3+3(-x)^2+4(-x)+5=

-2x^3+3x^2-4x+5

x=a is root of original polynomial

So, -x=a ie x=-a is root of : -2x^3+3x^2-4x+5

Now substitute:x=1/x we get

2/x^3+3/x^2+4/x+5=

(2+3x+4x^2+5x^3)/x^3

x=a is root of this

Hence, 1/x=a is root of this

x =1/a is root of (2+3x+4x^2+5x^3)/x^3

Hence, x=1/a is root of

2+3x+4x^2+5x^3

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