Evaluating and Estimating Square Roots
1. What Is a Square Root?
The square root of a number $n$ is a value $x$ such that $x^2 = n$. We write it as $\sqrt{n}$.
$\sqrt{1} = 1$, $\sqrt{4} = 2$, $\sqrt{9} = 3$
$\sqrt{16} = 4$, $\sqrt{25} = 5$, $\sqrt{36} = 6$
$\sqrt{49} = 7$, $\sqrt{64} = 8$, $\sqrt{81} = 9$
$\sqrt{100} = 10$, $\sqrt{121} = 11$, $\sqrt{144} = 12$
2. Theorem: If $a > b > 0$, then $\sqrt{a} > \sqrt{b}$
This theorem tells us that the square root function preserves order for positive numbers. Bigger inputs give bigger outputs.
Proof (by contradiction):
Assume $a > b > 0$. We want to show $\sqrt{a} > \sqrt{b}$.
Step 1: Since $a > 0$ and $b > 0$, both $\sqrt{a}$ and $\sqrt{b}$ exist and are positive.
Step 2: We know that $a - b > 0$. We can factor:
$$a - b = (\sqrt{a})^2 - (\sqrt{b})^2 = (\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b})$$
Step 3: Since $\sqrt{a} > 0$ and $\sqrt{b} > 0$, we know $(\sqrt{a} + \sqrt{b}) > 0$.
Step 4: Since $a - b > 0$ and $(\sqrt{a} + \sqrt{b}) > 0$, we must have:
$$(\sqrt{a} - \sqrt{b}) = \frac{a - b}{\sqrt{a} + \sqrt{b}} > 0$$
Step 5: Therefore $\sqrt{a} - \sqrt{b} > 0$, which means $\sqrt{a} > \sqrt{b}$. $\blacksquare$