Solution 1.1:5a
From Förberedande kurs i matematik 1
(Difference between revisions)
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+ | 2, 3/5, 5/3 and 7/3 . | ||
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It is easier to see the mutual order of the numbers if we write them as decimals. | It is easier to see the mutual order of the numbers if we write them as decimals. | ||
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& \\ | & \\ | ||
& \frac{7}{3}=\frac{6+1}{3}=2+\frac{1}{3}=2.333... \\ | & \frac{7}{3}=\frac{6+1}{3}=2+\frac{1}{3}=2.333... \\ | ||
+ | \end{align}</math> | ||
+ | |||
+ | and then we see that. | ||
+ | |||
+ | |||
+ | <math>\begin{align} | ||
+ | & \frac{3}{5}<\frac{5}{3}<2<\frac{7}{3} \\ | ||
+ | & \\ | ||
\end{align}</math> | \end{align}</math> |
Current revision
2, 3/5, 5/3 and 7/3 .
It is easier to see the mutual order of the numbers if we write them as decimals.
Because we know that \displaystyle {1}/{5}\;=0.2 and \displaystyle {1}/{3}\;=0.333..., we obtain
\displaystyle \begin{align}
& \frac{3}{5}=3\centerdot \frac{3}{5}=3.02=0.6 \\
& \\
& \frac{5}{3}=\frac{3+2}{3}=1+\frac{2}{3}=1.666... \\
& \\
& \frac{7}{3}=\frac{6+1}{3}=2+\frac{1}{3}=2.333... \\
\end{align}
and then we see that.
\displaystyle \begin{align}
& \frac{3}{5}<\frac{5}{3}<2<\frac{7}{3} \\
& \\
\end{align}