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Suppose that $A$ is a unital C*-algebra and $a\in A$ is positive, that is, $a$ is normal and $\sigma(a)\subset[0,\infty)$.

Then we can define the element $a^{1/2}\in A$ to be the unique element satisfying $(a^{1/2})^{2}=a$.

On the other hand, we can consider the continuous map $f\colon\sigma(a)\to\mathbb{C}$ defined by $f(x):=\sqrt{x}$ and apply the (continuous) functional calculus. Then we get an element $f(a)\in A$.

I don't understand why the definitions of $f(a)$ and $a^{1/2}$ coincide. Can someone explain what's going on? Any help will be greatly appreciated!

By the way, I use the definitions in Murphy's book on C*-algebras and Operator Theory.

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    $\begingroup$ Note that $a^{1/2}$ is the unique positive element such that $(a^{1/2})^2=a$. indeed, one can cook up many different $b$ such that $b^2=a$, but only one of them can be positive. $\endgroup$ – Aweygan Mar 5 at 19:31
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Suppose that $b $ is positive and $b^2=a $. Let $g(t)=t^2$. The functional calculus respects composition: that is, since $t=f (g (t))$ on $\sigma (b) $, then $b=f (g (b))$.

Then $$ b=f (g (b))=f (b^2)=f (a). $$

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