From Förberedande kurs i matematik 1

Revision as of 06:50, 21 August 2008 (edit) Tekbot (Talk | contribs) m (Robot: Automated text replacement (-[[Bild: +[[Image:)) ← Previous diff		Current revision (07:52, 30 September 2008) (edit) (undo) Tek (Talk | contribs) m
(2 intermediate revisions not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	That which is under the root sign is the same as <math>(-3)^{2} = 9</math> and because
-	<center> [[Image:3_1_2b.gif]] </center>	+	<math>9 = 3\cdot 3 = 3^{2}</math>, hence
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>\sqrt{(-3)^{2}} = \sqrt{9} = 9^{1/2} = \bigl(3^{2}\bigr)^{1/2} = 3^{2\cdot\frac{1}{2}} = 3^{1} = 3</math>.}}
		+
		+
		+	Note:
		+	The calculation <math>\sqrt{(-3)^{2}} = \bigl((-3)^{2}\bigr)^{1/2} = (-3)^{2\cdot \frac{1}{2}} = (-3)^1 = -3</math> is wrong at the second equals sign. Remember that the power rules apply when the base is positive.

---

Current revision

That which is under the root sign is the same as \displaystyle (-3)^{2} = 9 and because \displaystyle 9 = 3\cdot 3 = 3^{2}, hence

\displaystyle \sqrt{(-3)^{2}} = \sqrt{9} = 9^{1/2} = \bigl(3^{2}\bigr)^{1/2} = 3^{2\cdot\frac{1}{2}} = 3^{1} = 3.

Note: The calculation \displaystyle \sqrt{(-3)^{2}} = \bigl((-3)^{2}\bigr)^{1/2} = (-3)^{2\cdot \frac{1}{2}} = (-3)^1 = -3 is wrong at the second equals sign. Remember that the power rules apply when the base is positive.