From Förberedande kurs i matematik 1

Revision as of 08:09, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 1.1:4c moved to Solution 1.1:4c: Robot: moved page) ← Previous diff		Current revision (13:54, 13 September 2008) (edit) (undo) Ian (Talk | contribs)
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-	<math>\sqrt{2}</math> är inte ett rationellt tal, dvs. går inte att skriva som en kvot mellan två heltal, och då går det inte heller att skriva <math>\sqrt{2}/3</math> som en kvot mellan två heltal. Talet <math>\sqrt{2}/3</math> är alltså ett irrationellt tal.	+	<math>\sqrt{2}</math> is not a rational number, i.e. it cannot be written as a ratio of two integers, and neither
		+	is it possible to write
		+	<math>\sqrt{2}/3</math> as a ratio of two integers. The number
		+	<math>\sqrt{2}/3</math> is thus irrational.
		+
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Current revision

\displaystyle \sqrt{2} is not a rational number, i.e. it cannot be written as a ratio of two integers, and neither is it possible to write \displaystyle \sqrt{2}/3 as a ratio of two integers. The number \displaystyle \sqrt{2}/3 is thus irrational.