Existence of Real Polynomial with no Real Root
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Theorem
There exist polynomials in real numbers $\R$ which have no roots in $\R$.
Proof
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