Existence of Real Polynomial with no Real Root

Theorem

There exist polynomials in real numbers $\R$ which have no roots in $\R$.

Proof

Proof by Counterexample

Take the quadratic equation:

$(1): \quad x^2 + 1 = 0$

From the Quadratic Formula, the solution to $(1)$ is:

\(\ds x\)

\(=\)

\(\ds \dfrac {-0 \pm \sqrt {0^2 - 4 \times 1 \times 1} } {2 \times 1}\)

\(\ds \)

\(=\)

\(\ds \pm \sqrt {-1}\)

But there is no real number $x$ such that:

$x^2 = -1$

Hence the result.

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: The Complex Number System