Show that every positive number \(a\) has exactly one positive square root, as follows:
- Show that if \(x > 0\) and \(0 \leq h < 1\), then
\[ (x + h) ^ {2} \leq x ^ {2} + h (2 x + 1), \]
\[ (x - h) ^ {2} \geq x ^ {2} - h (2 x). \]
Let \(x > 0\) . Show that if \(x^2 < a\), then \((x + h)^2 < a\) for some \(h > 0\) ; and if \(x^2 > a\), then \((x - h)^2 > a\) for some \(h > 0\) .
Given \(a > 0\), let \(B\) be the set of all real numbers \(x\) such that \(x^2 < a\) . Show that \(B\) is bounded above and contains at least one positive number. Let \(b = \sup B\) ; show that \(b^2 = a\) .
Show that if \(b\) and \(c\) are positive and \(b^2 = c^2\), then \(b = c\) .
Solution
1. For \(x > 0\) and \(0 \le h < 1\), \((x + h)^2 \le x^2 + h(2x + 1)\) and \((x - h)^2 \ge x^2 - h(2x)\).
Since \(0 \le h < 1\), \(h^2 \le h\), so \[(x + h)^2 = x^2 + 2xh + h^2 \le x^2 + 2xh + h = x^2 + h(2x + 1).\] Since \(h^2 \ge 0\), \[(x - h)^2 = x^2 - 2xh + h^2 \ge x^2 - h(2x).\]
2. Let \(x > 0\). If \(x^2 < a\), then \((x + h)^2 < a\) for some \(h > 0\); if \(x^2 > a\), then \((x - h)^2 > a\) for some \(h\) with \(0 < h < x\).
Suppose \(x^2 < a\). Put \(h = \min\left(\frac{a - x^2}{2x + 1}, \frac12\right) > 0\) and \(h' = h/2\). Then \(0 < h' < h < 1\), and step 1 gives \[(x + h')^2 \le x^2 + h'(2x + 1) < x^2 + h(2x + 1) \le x^2 + (a - x^2) = a.\]
Suppose \(x^2 > a\). Put \(h = \min\left(\frac{x^2 - a}{2x}, \frac{x}{2}, \frac12\right) > 0\) and \(h' = h/2\). Then \(0 < h' < x\) and \(h' < 1\), and step 1 gives \[(x - h')^2 \ge x^2 - h'(2x) > x^2 - h(2x) \ge x^2 - (x^2 - a) = a.\]
3. The set \(B = \{x \in \RR \mid x^2 < a\}\) contains a positive number and is bounded above by \(1 + a\).
Put \(x_0 = \min(1, a/2) > 0\). Then \(x_0^2 \le x_0 < a\), so \(x_0 \in B\). If \(x > 1 + a\), then \(x^2 > (1 + a)^2 = 1 + 2a + a^2 > a\), so \(x \notin B\).
4. \(b = \sup B\) exists, \(b > 0\), and \(b^2 = a\).
By step 3 and the least upper bound property of \(\RR\), \(b = \sup B\) exists and \(b \ge x_0 > 0\).
If \(b^2 < a\), step 2 gives \(h > 0\) with \((b + h)^2 < a\). Then \(b + h \in B\) and \(b + h > b\), which contradicts that \(b\) is an upper bound of \(B\).
If \(b^2 > a\), step 2 gives \(h\) with \(0 < h < b\) and \((b - h)^2 > a\). Let \(x \in B\). If \(x \le 0\), then \(x < b - h\). If \(x > 0\), then \(x^2 < a < (b - h)^2\) with \(x, b - h > 0\), so \(x < b - h\). Hence \(b - h\) is an upper bound of \(B\) smaller than \(b\), which contradicts that \(b\) is the least upper bound.
By trichotomy, \(b^2 = a\).
5. If \(b, c > 0\) and \(b^2 = c^2\), then \(b = c\).
\((b - c)(b + c) = b^2 - c^2 = 0\) and \(b + c > 0\), so \(b - c = 0\).
6. Q.E.D.
Steps 1, 2, 3 with 4, and 5 prove parts (a), (b), (c), and (d). By step 4, \(\sup\{x \in \RR \mid x^2 < a\}\) is a positive square root of \(a\), and by step 5 it is the only one.