From Förberedande kurs i matematik 1

Revision as of 11:20, 20 March 2008 (edit) Lina (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:1_1_5c-1(2).gif </center> <center> Bild:1_1_5c-2(2).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (09:21, 22 September 2008) (edit) (undo) Ian (Talk | contribs)
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-	{{NAVCONTENT_START}}	+	It is quite easy to see that
-	<center> [[Bild:1_1_5c-1(2).gif]] </center>	+
-	<center> [[Bild:1_1_5c-2(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	<math>\begin{align}
		+	& \frac{1}{2}=0.5 \\
		+	& \\
		+	& \frac{2}{3}=2\centerdot \frac{1}{3}=0.666... \\
		+	& \\
		+	& \frac{3}{5}=3\centerdot \frac{1}{5}=3\centerdot 0.2=0.6 \\
		+	\end{align}</math>
		+
		+	<math></math>
		+
		+
		+	which means that
		+
		+
		+	<math>{1}/{2<{3}/{5<{2}/{3}\;}\;}\;</math> .
		+
		+	Because it is more difficult to evaluate the decimal expansion of
		+	<math>{5}/{8}\;</math>
		+	and
		+	<math>{21}/{34}\;</math>, we compare instead
		+	<math>{5}/{8}\;</math>
		+	and
		+	<math>{21}/{34}\;</math>
		+	with
		+	<math>{1}/{2,\ \ {3}/{5}\;}\;</math> and
		+	<math>{2}/{3}\;</math>
		+	by rewriting the fractions so that they have a common denominator. We start by comparing
		+	<math>{5}/{8}\;</math>
		+	with
		+	<math>{1}/{2,\ \ {3}/{5}\;}\;</math>
		+	and
		+	<math>{2}/{3}\;</math>
		+
		+
		+	* We have
		+	<math>\frac{1}{2}=\frac{1\centerdot 4}{2\centerdot 4}=\frac{4}{8}</math>
		+	and thus
		+	<math>\frac{1}{2}<\frac{5}{8}</math>
		+	.
		+
		+	* Then, we have
		+	<math>\frac{3}{5}=\frac{3\centerdot 8}{5\centerdot 8}=\frac{24}{40}</math>
		+	and
		+	<math>\frac{5}{8}=\frac{5\centerdot 5}{8\centerdot 5}=\frac{25}{40}</math>
		+	, which gives that
		+	<math>\frac{3}{5}<\frac{5}{8}</math>.
		+
		+	* Finally,
		+	<math>\frac{2}{3}=\frac{2\centerdot 8}{3\centerdot 8}=\frac{16}{24}</math>
		+	and
		+	<math>\frac{5}{8}=\frac{5\centerdot 3}{8\centerdot 3}=\frac{15}{24}</math>, and this gives that
		+	<math>\frac{5}{8}<\frac{2}{3}</math>.
		+
		+	Thus,
		+	<math>{1}/{2}\;<{3}/{5}\;<{5}/{8}\;<{2}/{3}\;</math>.
		+
		+	When we compare
		+	<math>{21}/{34}\;</math>
		+	with
		+	<math>{1}/{2,\ \ {3}/{5}\;}\;</math>
		+	and
		+	<math>{2}/{3}\;</math>, we obtain:
		+
		+	* because
		+	<math>\frac{1}{2}=\frac{1\centerdot 17}{2\centerdot 17}=\frac{17}{34}</math>, so
		+	<math>\frac{1}{2}<\frac{21}{34}</math>
		+
		+
		+	* furthermore,
		+	<math>\frac{3}{5}=\frac{3\centerdot 34}{5\centerdot 34}=\frac{102}{170}</math>
		+	and
		+	<math>\frac{21}{34}=\frac{21\centerdot 5}{34\centerdot 5}=\frac{105}{170}</math>, i.e.
		+	<math>\frac{3}{5}<\frac{21}{34}</math>.
		+
		+	* we have
		+	<math>\frac{5}{8}=\frac{5\centerdot 17}{8\centerdot 17}=\frac{85}{136}</math>
		+	and
		+	<math>\frac{21}{34}=\frac{21\centerdot 4}{34\centerdot 4}=\frac{84}{136}</math>, which gives that
		+	<math>\frac{21}{34}<\frac{5}{8}</math>.
		+
		+	The answer is
		+	<math>{1}/{2}\;<{3}/{5}\;<{21}/{34}\;<{5}/{8}\;<{2}/{3}\;</math>.

---

Current revision

It is quite easy to see that

\displaystyle \begin{align} & \frac{1}{2}=0.5 \\ & \\ & \frac{2}{3}=2\centerdot \frac{1}{3}=0.666... \\ & \\ & \frac{3}{5}=3\centerdot \frac{1}{5}=3\centerdot 0.2=0.6 \\ \end{align}

\displaystyle

which means that

\displaystyle {1}/{2<{3}/{5<{2}/{3}\;}\;}\; .

Because it is more difficult to evaluate the decimal expansion of \displaystyle {5}/{8}\; and \displaystyle {21}/{34}\;, we compare instead \displaystyle {5}/{8}\; and \displaystyle {21}/{34}\; with \displaystyle {1}/{2,\ \ {3}/{5}\;}\; and \displaystyle {2}/{3}\; by rewriting the fractions so that they have a common denominator. We start by comparing \displaystyle {5}/{8}\; with \displaystyle {1}/{2,\ \ {3}/{5}\;}\; and \displaystyle {2}/{3}\;

• We have

\displaystyle \frac{1}{2}=\frac{1\centerdot 4}{2\centerdot 4}=\frac{4}{8} and thus \displaystyle \frac{1}{2}<\frac{5}{8} .

• Then, we have

\displaystyle \frac{3}{5}=\frac{3\centerdot 8}{5\centerdot 8}=\frac{24}{40} and \displaystyle \frac{5}{8}=\frac{5\centerdot 5}{8\centerdot 5}=\frac{25}{40} , which gives that \displaystyle \frac{3}{5}<\frac{5}{8}.

• Finally,

\displaystyle \frac{2}{3}=\frac{2\centerdot 8}{3\centerdot 8}=\frac{16}{24} and \displaystyle \frac{5}{8}=\frac{5\centerdot 3}{8\centerdot 3}=\frac{15}{24}, and this gives that \displaystyle \frac{5}{8}<\frac{2}{3}.

Thus, \displaystyle {1}/{2}\;<{3}/{5}\;<{5}/{8}\;<{2}/{3}\;.

When we compare \displaystyle {21}/{34}\; with \displaystyle {1}/{2,\ \ {3}/{5}\;}\; and \displaystyle {2}/{3}\;, we obtain:

• because

\displaystyle \frac{1}{2}=\frac{1\centerdot 17}{2\centerdot 17}=\frac{17}{34}, so \displaystyle \frac{1}{2}<\frac{21}{34}

• furthermore,

\displaystyle \frac{3}{5}=\frac{3\centerdot 34}{5\centerdot 34}=\frac{102}{170} and \displaystyle \frac{21}{34}=\frac{21\centerdot 5}{34\centerdot 5}=\frac{105}{170}, i.e. \displaystyle \frac{3}{5}<\frac{21}{34}.

• we have

\displaystyle \frac{5}{8}=\frac{5\centerdot 17}{8\centerdot 17}=\frac{85}{136} and \displaystyle \frac{21}{34}=\frac{21\centerdot 4}{34\centerdot 4}=\frac{84}{136}, which gives that \displaystyle \frac{21}{34}<\frac{5}{8}.

The answer is \displaystyle {1}/{2}\;<{3}/{5}\;<{21}/{34}\;<{5}/{8}\;<{2}/{3}\;.