Question

Use a proof by contradiction to show that there is no rational number \(r\) for which\({r^3} + r + 1 = 0\). (Hint: Assume that \(r = \frac{a}{b}\)is a root, where \(a\) and \(b\) are integers and \(\frac{a}{b}\) is in lowest terms. Obtain an equation involving integers by multiplying by\({b^3}\). Then look at whether \(a\) and \(b\) are each odd or even.)

Short answer

There is no rational number\(r = \frac{a}{b}\)such that\({r^3} + r + 1 = 0\)

Step-by-step solution

Introduction

The purpose is to show that there is no rational number \(r\) for which \({r^3} + r + 1 = 0\) by using a proof of contradiction.

Converse assumption for proof

For proving that there is no rational number \(r\) for which\({r^3} + r + 1 = 0\),

Conversely assume that there is a rational number\(r = \frac{a}{b}\)is a root of\({r^3} + r + 1 = 0\).

Where, \(a\) and \(b\) are integers and this function is in lowest terms (that is \(a\) and \(b\) have no common divisor greater than 1).

Since \(r = \frac{a}{b}\)is a root of the equation.

Substitute \(r = \frac{a}{b}\)in the equation.

\(\frac{{{a^3}}}{{{b^3}}} + \frac{a}{b} + 1 = 0\) (As\(r = \frac{a}{b}\))

Multiply through by\({b^3}\) to obtain,

\({a^3} + a{b^2} + {b^3} = 0\) …………………. (1)

Case (1) \(a\) and \(b\) are both odd

Let \(a\) is odd and \(b\)is odd

Therefore, for \(a\) and\(b\)are both odd, the left hand side of equation (1) is the sum of three odd numbers must be odd and right hand side of equation (1) is 0.

Thus, this is contradiction to the assumption “the rational number \(r = \frac{a}{b}\)is a root of \({r^3} + r + 1 = 0\)”

Hence, there is no rational number\(r = \frac{a}{b}\) where both \(a\)and \(b\)odd numbers for which\({r^3} + r + 1 = 0\).

Case (2) \(a\)is odd and \(b\)is even

Let \(a\) is odd and \(b\)is even.

Therefore, for \(a\)is odd and \(b\)is even, the left hand side of equation (1) is odd +even +even, must be odd and right hand side of equation (1) is 0.

Thus, this is contradiction to the assumption “the rational number \(r = \frac{a}{b}\)is a root of\({r^3} + r + 1 = 0\)”

Hence, there is no rational number\(r = \frac{a}{b}\) where\(a\)is odd and \(b\)is even for which\({r^3} + r + 1 = 0\).

Case (3) \(a\) is even and \(b\)is odd.

Let \(a\) is even and \(b\)is odd.

Therefore, for \(a\)is even and \(b\)is odd, the left hand side of equation (1) is even +even+ odd, which is odd and right hand side of equation (1) is 0.

Thus, this is contradiction to the assumption “the rational number \(r = \frac{a}{b}\)is a root of\({r^3} + r + 1 = 0\)”

Hence, there is no rational number\(r = \frac{a}{b}\) where \(a\)is even and \(b\)is odd for which\({r^3} + r + 1 = 0\).

Case (1) \(a\) and \(b\) are both even

Because the fraction \(\frac{a}{b}\)is in simplest terms, it cannot happen that both\(a\) and \(b\) are even.

Thus, in all cases the left hand side is odd; right hand side is zero and this is a contradiction.

Hence, there is no rational number\(r = \frac{a}{b}\)such that\({r^3} + r + 1 = 0\)