Solution 1.1:5c
From Förberedande kurs i matematik 1
m  (Lösning 1.1:5c moved to Solution 1.1:5c: Robot: moved page)  | 
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| - | {{  | + | It is quite easy to see that  | 
| - | <  | + | |
| - | <  | + | |
| - | {{  | + | <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
532=2
31=0
666

53=3
51=3
0
2=0
6
which means that
   
2
3
5
2
3
Because it is more difficult to evaluate the decimal expansion of 
8
34
8
34
2
  3
5
3
8
2
  3
5
3
- We have
 
41
4=84
85
- Then, we have
 
83
8=4024
55
5=4025
85
- Finally,
 
82
8=2416
35
3=2415
32
Thus,  
2
3
5
5
8
2
3
When we compare 
34
2
  3
5
3
- because
 
171
17=3417
3421
- furthermore,
 
343
34=170102
521
5=170105
3421
- we have
 
175
17=85136
421
4=84136
85
The answer is \displaystyle {1}/{2}\;<{3}/{5}\;<{21}/{34}\;<{5}/{8}\;<{2}/{3}\;.
