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

Proof by Counterexample

Take the quadratic equation:

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

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

\(\displaystyle x\) \(=\) \(\displaystyle \dfrac {-0 \pm \sqrt {0^2 - 4 \times 1 \times 1} } {2 \times 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \pm \sqrt {-1}\)

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

$x^2 = -1$

Hence the result.

$\blacksquare$


Sources