Exercise 3.4 Practice

First, convert all numbers to a common form to compare their positions:

• $\dfrac{2}{3} \approx 0.667$ — lies between 0 and 1, closer to 1.
• $-\dfrac{5}{4} = -1.25$ — lies between $-2$ and $-1$, closer to $-1$.
• $1\dfrac{1}{2} = \dfrac{3}{2} = 1.5$ — lies between 1 and 2.

Number line representation:

To place $\dfrac{2}{3}$ precisely: divide the segment from 0 to 1 into 3 equal parts. Mark the 2nd division.
To place $-\dfrac{5}{4}$: divide the segment from $-1$ to $-2$ into 4 equal parts. Mark the 1st division (i.e., 1 part to the left of $-1$).
To place $1\dfrac{1}{2}$: the midpoint between 1 and 2.

Convert the two bounds to a common denominator to make finding values easier. $$-\frac{1}{2} = -\frac{4}{8} \qquad \frac{1}{4} = \frac{2}{8}$$ We need rational numbers strictly between $-\dfrac{4}{8}$ and $\dfrac{2}{8}$.

Method — using the common denominator 8:
Integers between $-4$ and $2$ (exclusive): $-3,\,-2,\,-1,\,0,\,1$.
This gives the fractions: $-\dfrac{3}{8},\,-\dfrac{2}{8},\,-\dfrac{1}{8},\,\dfrac{0}{8},\,\dfrac{1}{8}$.

Picking any three distinct values from this list (simplified): $$-\frac{3}{8}, \quad -\frac{1}{8}, \quad \frac{1}{8}$$ Verification: $-0.5 < -0.375 < -0.125 < 0.125 < 0.25$ ✓

Find the LCD of 4 and 12.
$4 = 2^2$, $\quad 12 = 2^2 \times 3$, so LCD $= 12$.

Convert $-\dfrac{1}{4}$ to twelfths: $$-\frac{1}{4} = -\frac{3}{12}$$

Add the fractions: $$-\frac{3}{12} + \frac{5}{12} = \frac{-3 + 5}{12} = \frac{2}{12} = \frac{1}{6}$$

Convert the mixed numbers to improper fractions: $$15\frac{3}{4} = \frac{15 \times 4 + 3}{4} = \frac{63}{4}$$ $$2\frac{1}{4} = \frac{2 \times 4 + 1}{4} = \frac{9}{4}$$

Divide the total silk by the silk needed per kurta: $$\text{Number of kurtas} = \frac{63}{4} \div \frac{9}{4} = \frac{63}{4} \times \frac{4}{9} = \frac{63}{9} = 7$$

Since 7 is a whole number, there is no leftover silk and the division is exact.

Between any two distinct real numbers, infinitely many rational numbers exist. We extend the decimal places to find values strictly between the two bounds.

$3.1415 \lt \,?\, \lt 3.1416$

Using five decimal places:

• $3.14151$ — lies between $3.1415$ and $3.1416$ ✓
• $3.14153$ — lies between $3.1415$ and $3.1416$ ✓
• $3.14159$ — lies between $3.1415$ and $3.1416$ ✓   (Note: $\pi \approx 3.14159\ldots$ — a famous irrational number also in this interval!)

As fractions, these can also be written as: $$\frac{314151}{100000}, \quad \frac{314153}{100000}, \quad \frac{314159}{100000}$$ All three are rational numbers (ratio of integers with non-zero denominator).

Method 1 – The Average (Midpoint) Method:
Given two rational numbers $p$ and $q$ with $p \lt q$, their arithmetic mean always lies strictly between them: $$\text{Mean} = \frac{p + q}{2}$$ Since the sum and the division of rational numbers is always rational, $\dfrac{p+q}{2}$ is a rational number between $p$ and $q$.

Example: Between $\dfrac{1}{3}$ and $\dfrac{1}{2}$: $$\frac{\frac{1}{3} + \frac{1}{2}}{2} = \frac{\frac{5}{6}}{2} = \frac{5}{12}$$ Indeed $\dfrac{1}{3} \lt \dfrac{5}{12} \lt \dfrac{1}{2}$.

Method 2 – Scaling the Denominator:
Convert both rationals to a common denominator $d$. If $\dfrac{a}{d}$ and $\dfrac{b}{d}$ (with $a \lt b$) are the two numbers, and if $b - a \gt 1$, then any $\dfrac{k}{d}$ for $a \lt k \lt b$ is a rational between them. If $b - a = 1$ (i.e., consecutive integers in the numerator), multiply both numerator and denominator by 2 to create room: $$\frac{a}{d} = \frac{2a}{2d}, \quad \frac{b}{d} = \frac{2b}{2d}$$ Now $\dfrac{2a+1}{2d}$ lies strictly between them.

Method 3 – Extending the Decimal Expansion:
Write both rationals as decimals and add extra decimal digits. Any finite or repeating decimal in between is a rational number.
Example: Between $0.3$ and $0.4$, pick $0.35 = \dfrac{7}{20}$.