Definition 34.8.5 (Positive Square Root).label Let $A$ be a $C^{*}$-algebra and $x \in A$ be positive, then there exists a unique positive element $y \in A$ such that $y^{2} = x$. The element $y$ is the positive square root of $x$, denoted $\sqrt{x}$.
Proof. Since $x$ is positive, $\sigma_{A}(x) \subset [0, \infty)$ by Proposition 34.8.3. Therefore the square root function $f(t) = \sqrt{t}$ is defined and continuous on $\sigma_{A}(x)$. Using the continuous functional calculus, $f(x)$ is a positive element of $A$ such that $f(x)^{2} = x$.
Let $y \in A$ such that $y^{2} = x$, then by the Spectral Mapping Theorem, $y = f(y^{2}) = f(x)$, so the square root is unique.$\square$
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