Is this an acceptable definition for $\sqrt{a}$, where $a \in \mathbb{R}$?
If $a\geq 0, \sqrt{a} = b \in \mathbb{R}$ s.t. $b\geq 0, b^2 = a$.
I'm proving some theorems involving $\sqrt{a}$, in the context of introductory real analysis.
Is this an acceptable definition for $\sqrt{a}$, where $a \in \mathbb{R}$?
If $a\geq 0, \sqrt{a} = b \in \mathbb{R}$ s.t. $b\geq 0, b^2 = a$.
I'm proving some theorems involving $\sqrt{a}$, in the context of introductory real analysis.
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Let $a, b \in \mathbb{R}$. $\sqrt{a}$ is the square root of $a$ and is defined as $f(a) = b$ such that $f:a \to a\times a = b$. Notice the product then can never be negative, because $\mathbb{R}$ is reflexive.