There is no real number that is the square root of -1

by RoRi

June 30, 2015

Prove that $x^2+1 = 0$ has no solution $x \in \mathbb{R}$ .

Proof. From Theorem I.20 we know that if $x \neq 0$ , then $x^2 > 0$ . From Theorem I.21 we know $1 > 0$ . Thus, if $x \neq 0$ , then $x^2 + 1 > 0$ . On the other hand, if $x = 0$ , then $x^2+1=1 > 0$ . Hence, there is no $x \in \mathbb{R}$ such that $x^2 + 1 = 0. \qquad \blacksquare$

Stumbling Robot

There is no real number that is the square root of -1

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