$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{30}{60}+\frac{20}{60}+\frac{15}{60}+\frac{12}{60}+\frac{10}{60}=\frac{87}{60}=\frac{29}{20}$

Problem 3 Solution

Try to find the product of

$\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}$

without performing the actual multiplication.

$\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}=\frac{1}{6}$

The denominator of each fraction cancels with the numerator of the fraction following it.

The denominator $2$ in $\frac{1}{2}$ cancels the numerator $2$ in $\frac{2}{3}$ etc.

The only 2 numbers that don't cancel are the numerator $1$ in $\frac{1}{2}$ and the denominator $6$ in $\frac{5}{6}$ leaving $\frac{1}{6}$

We can also rearrange the numerators to make 4 of the fractions equivalent to $1$

$\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}= \frac{2}{2}\times\frac{3}{3}\times\frac{4}{4}\times\frac{5}{5}\times\frac{1}{6}=1\times1\times1\times1\times\frac{1}{6}=\frac{1}{6}$

Problem 4 Solution

Solve the equation

$\frac{1+2+3+4}{a}+\frac{1+2+3+4+5}{a}=5$

$\frac{1+2+3+4}{a}+\frac{1+2+3+4+5}{a}=\frac{10}{a}+\frac{15}{a}=\frac{25}{a}=5\implies a=5$

Problem 5 Solution

Solve the equation

$\frac{41}{7}-\frac{a}{14}=4$

$\frac{41}{7}-\frac{a}{14}=4\implies\frac{41}{7}-4=\frac{a}{14}$

$\frac{41}{7}-4=\frac{41}{7}-\frac{28}{7}=\frac{13}{7}\implies$

$\frac{13}{7}=\frac{a}{14}\implies a=(14)\frac{13}{7}=(2)(13)=26$

Problem 6 Solution

Find a Rational number in between

a. $\frac{1}{4}$ and $\frac{1}{3}$

$\frac{1}{4}=\frac{3}{12}$ and $\frac{1}{3}=\frac{4}{12}$

There is no fraction with a denominator of $12$ between $\frac{3}{12}$ and $\frac{4}{12}$ .

Converting denominators gives $\frac{3}{12}=\frac{6}{24}$ and $\frac{4}{12}=\frac{8}{24}$

There is a fraction in between $\frac{6}{24}$ and $\frac{8}{24}$ ,

namely $\frac{7}{24}$

$\frac{1}{4}=\frac{6}{24}<\frac{7}{24}<\frac{8}{24}=\frac{1}{3}$

This is just one of infinitely many fractions between $\frac{1}{4}$ and $\frac{1}{3}$ .

b. $\frac{5}{8}$ and $\frac{3}{4}$

$\frac{5}{8}$ and $\frac{3}{4}=\frac{6}{8}$

$\frac{5}{8}=\frac{10}{16}$ and $\frac{6}{8}=\frac{12}{16}$

There is a fraction between $\frac{10}{16}$ and $\frac{12}{16}$ ,

namely $\frac{11}{16}$ .

c. $-\frac{3}{5}$ and $-\frac{2}{5}$

$-\frac{3}{5}=-\frac{6}{10}$ and $-\frac{2}{5}=-\frac{4}{10}$

$-\frac{5}{10}=-\frac{1}{2}$

$-\frac{3}{5}<-\frac{1}{2}<-\frac{2}{5}$

Problem 7 Solution

Divide $21$ into $\frac{1}{5}$ 's and give each fractional value in improper and mixed fraction form.

Interval $0\to 21$ into $\frac{1}{5}'s$

1. $\frac{21}{5}=4\frac{1}{5}$

2. $\frac{42}{5}=8\frac{2}{5}$

3. $\frac{63}{5}=12\frac{3}{5}$

4. $\frac{84}{5}=16\frac{4}{5}$

5. $\frac{105}{5}=21$

Problem 8 Solution

Divide $24$ into $\frac{1}{10}'s$ and give each fractional value in both improper and mixed form.

Interval $0\to 24$ into $\frac{1}{10}'s$

1. $\frac{24}{10}=\frac{12}{5}=2\frac{2}{5}$

2. $\frac{48}{10}=\frac{24}{5}=4\frac{4}{5}$

3. $\frac{72}{10}=\frac{36}{5}=7\frac{1}{5}$

4. $\frac{96}{10}=\frac{48}{5}=9\frac{3}{5}$

5. $\frac{120}{10}=\frac{60}{5}=12$

6. $\frac{144}{10}=\frac{72}{5}=14\frac{2}{5}$

7. $\frac{168}{10}=\frac{84}{5}=16\frac{4}{5}$

8. $\frac{192}{10}=\frac{96}{5}=19\frac{1}{5}$

9. $\frac{216}{10}=\frac{108}{5}=21\frac{3}{5}$

10. $\frac{240}{10}=\frac{120}{5}=24$

Problem 9 Solution

Divide the number line interval of $\frac{1}{6}$ to $\frac{5}{6}$ into $\frac{1}{4}'s$ and give each value.

Number Line Distance: $\frac{5}{6}-\frac{1}{6}=\frac{4}{6}$

Spaces: $\frac{\frac{4}{6}}{4}=\frac{1}{6}$

Interval 1: $\frac{1}{6}\to\frac{2}{6}\implies\frac{1}{6}\to\frac{1}{3}$

Interval 2: $\frac{1}{3}\to\frac{3}{6}\implies\frac{1}{3}\to\frac{1}{2}$

Interval 3: $\frac{1}{2}\to\frac{4}{6}\implies\frac{1}{2}\to\frac{2}{3}$

Interval 4: $\frac{2}{3}\to\frac{5}{6}$

Problem 10 Solution

Divide the number line interval of $\frac{6}{5}$ to $\frac{8}{5}$ into $\frac{1}{10}'s$ and give each value.

Number Line Distance: $\frac{8}{5}-\frac{6}{5}=\frac{2}{5}$

Spaces: $\frac{\frac{2}{5}}{10}=\frac{1}{25}$

Number Line Start and Finish: $\frac{6}{5}=\frac{30}{25}$ and $\frac{8}{5}=\frac{40}{25}$

Interval 1: $\frac{30}{25}\to\frac{31}{25}\implies\frac{6}{5}\to\frac{31}{25}$

Interval 2: $\frac{31}{25}\to\frac{32}{25}$

Interval 3: $\frac{32}{25}\to\frac{33}{25}$

Interval 4: $\frac{33}{25}\to\frac{34}{25}$

Interval 5: $\frac{34}{25}\to\frac{35}{25}\implies\frac{34}{25}\to\frac{7}{5}$

Interval 6: $\frac{7}{5}\to\frac{36}{25}$

Interval 7: $\frac{36}{25}\to\frac{37}{25}$

Interval 8: $\frac{37}{25}\to\frac{38}{25}$

Interval 9: $\frac{38}{25}\to\frac{39}{25}$

Interval 10: $\frac{39}{25}\to\frac{40}{25}\implies\frac{39}{25}\to\frac{8}{5}$

Problem 11 Solution

Name at least 4 Rational numbers that are between

a. $\frac{1}{2}$ and $1\frac{1}{2}$

Here are quite a few Rational numbers between $\frac{1}{2}$ and $1\frac{1}{2}$

$\frac{3}{5}, \frac{3}{4}, \frac{4}{7}, \frac{5}{8}, \frac{2}{3},\frac{7}{8}, \frac{9}{10}, \frac{11}{10}, \frac{5}{4}, \frac{4}{3}, \frac{11}{8}$

b. $\frac{1}{4}$ and $\frac{1}{2}$

$\frac{1}{4}\to\frac{2}{4}$

$\frac{2}{8}\to\frac{4}{8}$

$\frac{2}{8}<\frac{3}{8}<\frac{4}{8}$

$\frac{4}{16}\to\frac{8}{16}$

$\frac{4}{16}<\frac{5}{16}<\frac{7}{16}<\frac{8}{16}$

$\frac{8}{32}<\frac{9}{32}<\frac{11}{32}<\frac{13}{32}<\frac{15}{32}<\frac{16}{32}$

c. $0$ and $\frac{1}{10}$

$0\to\frac{1}{10}$

$0 ... <\frac{1}{17}<\frac{1}{16}<\frac{1}{15}<\frac{1}{14}<\frac{1}{13}<\frac{1}{12}<\frac{1}{11}<\frac{1}{10}$

Problem 12 Solution

Solve:

$1-\frac{1}{1-\frac{1}{1-\frac{1}{2}}}$

$1-\frac{1}{1-\frac{1}{\frac{1}{2}}}$

$=1-\frac{1}{1-2}=1-\frac{1}{-1}=1+1=2$

Problem 13 Solution

How many $\frac{1}{5}$ 's are in

a. $\frac{16}{5}$

$\frac{\frac{16}{5}}{\frac{1}{5}}=\frac{16}{5}\times\frac{5}{1}=16$

b. $41\frac{4}{5}$

$41\frac{4}{5}=\frac{209}{5}$

$\frac{\frac{209}{5}}{\frac{1}{5}}=\frac{209}{5}\times\frac{5}{1}=209$

c. $2$

$\frac{\frac{2}{1}}{\frac{1}{5}}=\frac{2}{1}\times\frac{5}{1}=10$

d. $\frac{3}{10}$

$\frac{\frac{3}{10}}{\frac{1}{5}}=\frac{3}{10}\times\frac{5}{1}=\frac{15}{10}=\frac{3}{2}=1\frac{1}{2}$

Problem 14 Solution

What is the fractional distance $d$ between these sets of points $[a,b]$ on the number line?

Interval 1: $a=\frac{14}{3}$ and $b=\frac{21}{3}$

$d=b-a$

$[a,b]=[\frac{7}{3},\frac{14}{3}]$

$d=\frac{14}{3}-\frac{7}{3}=\frac{7}{3}=2\frac{1}{3}$

Interval 2: $a=4\frac{3}{7}$ and $b=8\frac{2}{7}$

$4\frac{3}{7}=\frac{31}{7}$ and $8\frac{2}{7}=\frac{58}{7}$

$d=b-a$

$[a,b]=[\frac{31}{7},\frac{58}{7}]$

$d=\frac{58}{7}-\frac{31}{7}=\frac{27}{7}=3\frac{1}{7}$

Interval 3: $a=-\frac{4}{3}$ and $b=4\frac{1}{3}$

$4\frac{1}{3}=\frac{13}{3}$

$d=b-a$

$[a,b]=[-\frac{4}{3},\frac{13}{3}]$

$d=\frac{13}{3}-(-\frac{4}{3})=\frac{13}{3}+\frac{4}{3}=\frac{17}{3}=5\frac{2}{3}$

Interval 4: $=2\frac{1}{3}$ and $b=5\frac{1}{6}$

$2\frac{1}{3}=2\frac{2}{6}=\frac{14}{6}$ and $5\frac{1}{6}=\frac{31}{6}$

$d=b-a$

$[a,b]=[\frac{14}{6},\frac{31}{6}]$

$d=\frac{31}{6}-\frac{14}{6}=\frac{17}{6}=2\frac{5}{6}$

Problem 15 Solution

What is the midpoint $m$ of each of these intervals on the number line?

a. $[a,b]=[\frac{1}{10},\frac{7}{10}]$

$m=\frac{a+b}{2}$

$m=\frac{\frac{7}{10}+\frac{1}{10}}{2}=\frac{\frac{8}{10}}{2}=\frac{8}{20}=\frac{4}{10}=\frac{2}{5}$

b. $[a,b]=[\frac{17}{15},\frac{20}{15}]$

$m=\frac{a+b}{2}$

$a=\frac{17}{15}=\frac{34}{30}$ and $b=\frac{20}{15}=\frac{40}{30}$

$m=\frac{\frac{34}{30}+\frac{40}{30}}{2}=\frac{74}{60}=\frac{37}{30}$

c. $[a,b]=[\frac{3}{5},\frac{2}{3}]$

$m=\frac{a+b}{2}$

$a=\frac{3}{5}=\frac{18}{30}$ and $b=\frac{2}{3}=\frac{20}{30}$

$m=\frac{\frac{18}{30}+\frac{20}{30}}{2}=\frac{38}{60}=\frac{19}{30}$

Problem 16 Solution

How many Rational numbers $x$ are there between $-1\leq x\leq1$ ?

There are an infinite amount of Rational numbers in between any 2 other Rational numbers.

Problem 17 Solution

What happens to the denominator and overall value of the Rational number $\frac{1}{x}$ when

a. $x$ is less than $1$ but greater than $0$ and $x$ is approaching $0$

$(x: 0<x<1)$ with $x\to0$ or $x\to0^{+}$

The Rational number $\frac{1}{x}$ gets bigger and bigger as $x\to0$ .

$x=\frac{1}{2}$ and $(0<\frac{1}{2}<1)$ gives $\frac{1}{x}=\frac{1}{\frac{1}{2}}=2$

$x=\frac{1}{4}$ and $(0<\frac{1}{4}<1)$ gives $\frac{1}{x}=\frac{1}{\frac{1}{4}}=4$

$x=\frac{1}{10}$ and $(0<\frac{1}{10}<1)$ gives $\frac{1}{x}=\frac{1}{\frac{1}{10}}=10$

$x=\frac{1}{100}$ and $(0<\frac{1}{100}<1)$ gives $\frac{1}{x}=\frac{1}{\frac{1}{100}}=100$

b. $x$ is greater than $-1$ but less than $0$ and $x$ is approaching $0$ (approaches $0$ from the negative side)

$(x:-1<x<0)$ with $x\to0$ or $x\to0^{-}$

The Rational number $\frac{1}{x}$ gets bigger and bigger negatively as $x\to0$ .

$x=-\frac{1}{2}$ and $(-1<-\frac{1}{2}<0)$ gives $\frac{1}{x}=\frac{1}{-\frac{1}{2}}=-2$

$x=-\frac{1}{4}$ and $(-1<-\frac{1}{4}<0)$ gives $\frac{1}{x}=\frac{1}{-\frac{1}{4}}=-4$

$x=-\frac{1}{10}$ and $(-1<-\frac{1}{10}<0)$ gives $\frac{1}{x}=\frac{1}{-\frac{1}{10}}=-10$

$x=-\frac{1}{100}$ and $(-1<-\frac{1}{100}<0)$ gives $\frac{1}{x}=\frac{1}{-\frac{1}{100}}=-100$

c. $x$ is less than $-1$ and approaches $-1000$

$(x:x<-1)$ and $x\to -1000$

The Rational number $\frac{1}{x}$ gets smaller and smaller negatively as $x\to-1000$ .

$x=-2$ and $(-2<-1)$ gives $\frac{1}{x}=\frac{1}{-2}=-\frac{1}{2}$

$x=-4$ and $(-4<-1)$ gives $\frac{1}{x}=\frac{1}{-4}=-\frac{1}{4}$

$x=-10$ and $(-10<-1)$ gives $\frac{1}{x}=\frac{1}{-10}=-\frac{1}{10}$

$x=-100$ and $(-100<-1)$ gives $\frac{1}{x}=\frac{1}{-100}=-\frac{1}{100}$

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