Answer to Question #234770 in Discrete Mathematics for Varshi

Question

Use a proof by contradiction to show that there is no rational number r for which r+r+1=0.

(Assume that r=a/h is a root, where a and b are integers and a/b is in lowest terms.obtain an equation involving integers by multiplying by then look at whether a and bare each odd or

even.

Answer 1

Use a proof by contradiction to show that there is no rational number $r$ for which $r^3 + r + 1 = 0.$

Proof

Assume that $r = a/b$ is a root, where $a$ and $b$ are integers and $a/b$ is in lowest terms.

We see that

if $a$ is odd, then $b$ is even,

if $a$ is even, then $b$ is odd.

Substitute

$$\dfrac{a^3}{b^3}+\dfrac{a}{b}+1=0$$

Multiply by $b^3\not=0$

$$a^3+ab^2+b^3=0$$

If $a$ is odd, $b$ is even, then

$a^3$is odd, $ab^2$ is even, $b^3$ is even. Hence the sum $a^3+ab^2+b^3$ is odd. But zero is an even number.

We have a contradiction in this case.

If $a$ is even, $b$ is odd, then

$a^3$is even, $ab^2$ is even, $b^3$ is odd. Hence the sum $a^3+ab^2+b^3$ is odd. But zero is an even number.

We have a contradiction in this case.

Therefore our assumption leads to the contradiction.

Therefore there is no rational number $r$ for which $r^3 + r + 1 = 0.$