1.2.3 Exercises

Exercise 1.2.618

Reduce the following fractions to the lowest terms:

$\frac{216}{240}$ $=$
.
Since $\gcd (216,240) = 24$ we have $\frac{216}{240} = \frac{216 : 24}{240 : 24} = \frac{9}{10}$ .
$\frac{36}{72}$ $=$
.
$36$ divides $72$ , hence $\frac{36}{72} = \frac{1}{2} .$
$\frac{48}{144}$ $=$
.
$48$ divides $144$ , hence $\frac{48}{144} = \frac{1}{3} .$
$\frac{- a + 2 b}{- 4 b + 2 a}$ $=$
if $a$ not equals
.
Reducing gives $\frac{- a + 2 b}{- 4 b + 2 a} = \frac{(- 1) \cdot (- 2 b + a)}{2 \cdot (- 2 b + a)} = - \frac{1}{2}$ . The fraction is only defined for $a \neq 2 b$ .

Enter the fraction as numerator/denominator fully reduced and with positive denominator. Always place the signs in front of the fractions, and do not use brackets.

Exercise 1.2.619

Calculate and fully reduce the following expressions for appropriate numbers

a, b, x, y

$\frac{1}{2} - \frac{2}{7} + \frac{3}{8} + \frac{3}{4}$ $=$
.
Adding up the numerators over the least common denominator gives $\frac{1}{2} - \frac{2}{7} + \frac{3}{8} + \frac{3}{4} = \frac{28}{56} - \frac{16}{56} + \frac{21}{56} + \frac{42}{56} = \frac{75}{56}$ , since $\gcd (2,7, 8,4) = 56$ .
$\frac{3}{13} : \frac{7}{26}$ $=$
.
Dividing by a fraction is the same as multiplying by its reciprocal: $\frac{3}{13} : \frac{7}{26} = \frac{3}{13} \cdot \frac{26}{7} = \frac{3 \cdot 26}{13 \cdot 7} = \frac{3 \cdot 2}{1 \cdot 7} = \frac{6}{7}$ .
$(1. \overline{4} \cdot 3 - \frac{1}{2}) \cdot \frac{6}{7}$ $=$
.
Converting the decimal expression to a fraction gives $(1. \overline{4} \cdot 3 - \frac{1}{2}) \cdot \frac{6}{7} = (1. \overline{4} \cdot 18 - 3) \cdot \frac{1}{7} = (26 - 3) \cdot \frac{1}{7} = \frac{23}{7}$ .

Enter the fractions fully reduced and with positive numerator.

Exercise 1.2.620

Convert the following infinite repeating decimal fractions to fractions and fully reduce them:

$0. \overline{4}$ $=$ .
$0. \overline{23}$ $=$ .
$0.12 \overline{34}$ $=$ .
$0. \overline{9}$ $=$ .

Enter the fractions fully reduced and with positive numerator.

Using the trick for converting infinite repeating decimal numbers one gets the following solutions:

$x = 0. \overline{4}$ , hence $10 x - x = 4 \Rightarrow 9 x = 4 \Rightarrow x = \frac{4}{9}$ ,
$x = 0. \overline{23}$ , hence $100 x - x = 23 \Rightarrow 99 x = 23 \Rightarrow x = \frac{23}{99}$ ,
$x = 0.12 \overline{34}$ , hence $100 x - x = 12.22 \Rightarrow 99 x = \frac{1222}{100} \Rightarrow x = \frac{1222}{9900} = \frac{611}{4950}$ ,
$x = 0. \overline{9}$ , hence $10 x - x = 9 \Rightarrow 9 x = 9 \Rightarrow x = 1$ .

Note for the last part of this exercise, that

1 = 1.000 \dots

and

1 = 0.999 \dots = 0. \overline{9}

are two decimal representations of the same number.

Onlinebrückenkurs Mathematik

1. Elementary Arithmetic

2. Equations in one Variable

3. Inequalities in one Variable

4. System of Linear Equations

5. Geometry

6. Elementary Functions

7. Differential Calculus

8. Integral Calculus

9. Objects in the Two-Dimensional Coordinate System

10. Basic Concepts of Descriptive Vector Geometry

11. Language of Descriptive Statistics

Chapter 1 Elementary Arithmetic