Rate $= \dfrac{15 \text{ ingots}}{2 \text{ bags}}$

For 12 bags: $12 \times \dfrac{15}{2} = 12 \times 7.5 = 90$ ingots

What they share: All four are prime numbers — they are divisible only by 1 and themselves. They are precisely the prime numbers between 10 and 20.

Next three primes: After 19, we check: 20 (not prime), 21 (not), 22 (not), 23 (prime ✓), 24 (no), …, 29 (prime ✓), 30 (no), 31 (prime ✓)

NO — Natural numbers are not closed under subtraction.

Closure means: if $a, b \in \mathbb{N}$, then $a – b$ must also be in $\mathbb{N}$.

Counter-examples:

4 fingers × 3 joints per finger $= 12$ countable joints on one hand.

The thumb acts as a pointer to track which joint is being counted.

Relation to base-12: This method naturally produces the number 12 as a counting unit. In ancient times, 12 was a very practical base (base-12 or duodecimal system) because 12 is divisible by 1, 2, 3, 4, and 6 — making it ideal for dividing quantities (e.g., 12 months, 12 inches in a foot, 24 hours).

Midnight temp $= 4 – 15 = -11\,°C$

Equation: $-850 + 1200 – 450$

$= 1200 – 850 – 450$
$= 1200 – 1300$
$= -100$

Real-world example:

Imagine you have ₹10 in your pocket (fortune = +10).
You owe your friend ₹5 (debt = −5).
Your friend cancels your debt and says “you don’t have to pay me back.”

Removing the debt of ₹5 is like gaining ₹5:
$10 – (-5) = 10 + 5 = 15$

In Brahmagupta’s language: if you remove (subtract) a debt (negative), you are effectively adding a fortune (positive).

LHS: $\left(\dfrac{1}{2}+\dfrac{3}{4}\right)\times\dfrac{8}{3} = \left(\dfrac{2}{4}+\dfrac{3}{4}\right)\times\dfrac{8}{3} = \dfrac{5}{4}\times\dfrac{8}{3} = \dfrac{40}{12} = \dfrac{10}{3}$

RHS: $\dfrac{1}{2}\times\dfrac{8}{3}+\dfrac{3}{4}\times\dfrac{8}{3} = \dfrac{8}{6}+\dfrac{24}{12} = \dfrac{4}{3}+2 = \dfrac{4}{3}+\dfrac{6}{3} = \dfrac{10}{3}$

LHS $=$ RHS $=\dfrac{10}{3}$ ✓

$\dfrac{7}{9}\!\left(\dfrac{6}{7}-\dfrac{3}{4}\right) = \dfrac{7}{9}\times\dfrac{6}{7} – \dfrac{7}{9}\times\dfrac{3}{4}$

$= \dfrac{42}{63} – \dfrac{21}{36} = \dfrac{6}{9} – \dfrac{7}{12} = \dfrac{2}{3} – \dfrac{7}{12}$

$= \dfrac{8}{12} – \dfrac{7}{12} = \dfrac{1}{12}$

Expand LHS using distributive law:
$\dfrac{5}{6}\cdot x + \dfrac{5}{6}\cdot\dfrac{3}{5} = \dfrac{5x}{6} + \dfrac{15}{30} = \dfrac{5x}{6} + \dfrac{1}{2}$

This equals the RHS $= \dfrac{5x}{6} + \dfrac{1}{2}$ for all values of $x$.

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

Convert to common denominator 8: $-\dfrac{1}{2}=-\dfrac{4}{8}$ and $\dfrac{1}{4}=\dfrac{2}{8}$

Integers strictly between $-4$ and $2$ in the 8ths: $-3, -2, -1, 0, 1$

Choose any three: $\dfrac{-3}{8},\ \dfrac{-1}{8},\ \dfrac{1}{8}$

LCD of 4 and 12 is 12:
$-\dfrac{1}{4}+\dfrac{5}{12} = -\dfrac{3}{12}+\dfrac{5}{12} = \dfrac{2}{12} = \dfrac{1}{6}$

Convert to improper fractions:
$15\dfrac{3}{4} = \dfrac{63}{4}$ m total silk
$2\dfrac{1}{4} = \dfrac{9}{4}$ m per kurta

Number of kurtas $= \dfrac{63}{4} \div \dfrac{9}{4} = \dfrac{63}{4}\times\dfrac{4}{9} = \dfrac{63}{9} = 7$

The gap = $3.1416 – 3.1415 = 0.0001$ — very tiny but infinitely many rationals exist.

Three examples: $3.14151,\; 3.14155,\; 3.14159$

Or as fractions: $\dfrac{314151}{100000},\; \dfrac{314155}{100000},\; \dfrac{314159}{100000}$

Method 1 — Average (used in chapter):
Between $a$ and $b$: $\dfrac{a+b}{2}$ is always rational and lies between them.

Method 2 — Common denominator:
Express both fractions with the same denominator $m$. If $\dfrac{k_1}{m} < \dfrac{k_2}{m}$ and $k_2-k_1 > 1$, pick any integer numerator between $k_1$ and $k_2$.

Method 3 — Decimal method:
Convert to decimals and pick any finite or repeating decimal between them.

$\dfrac{1}{13} = 0.\overline{076923}$ (repeating block: 076923, length 6)

Cyclic pattern:

The number $\frac{1}{7}$ has a repeating block of length 6 (= $7-1$). Such primes are called full-reptend primes.

For a prime $p$, the repeating block of $\frac{1}{p}$ has length $p-1$ when 10 is a primitive root modulo $p$.

Method: Use common denominator. $3=\dfrac{21}{7}$, $4=\dfrac{28}{7}$. Integers from 22 to 27 give 6 rationals:
$\dfrac{22}{7},\dfrac{23}{7},\dfrac{24}{7},\dfrac{25}{7},\dfrac{26}{7},\dfrac{27}{7}$

Multiply: $\dfrac{2}{5}=\dfrac{14}{35}$, $\dfrac{3}{5}=\dfrac{21}{35}$. Integers 15–20 give:
$\dfrac{15}{35}=\dfrac{3}{7},\;\dfrac{16}{35},\;\dfrac{17}{35},\;\dfrac{18}{35},\;\dfrac{19}{35}$

$\dfrac{1}{6}=\dfrac{5}{30}$, $\dfrac{2}{5}=\dfrac{12}{30}$. Integers 6–11 between 5 and 12:
$\dfrac{6}{30}=\dfrac{1}{5},\;\dfrac{7}{30},\;\dfrac{8}{30}=\dfrac{4}{15},\;\dfrac{9}{30}=\dfrac{3}{10},\;\dfrac{10}{30}=\dfrac{1}{3}$

LCD of 3 and 5 is 15:
$\dfrac{5x}{15}+\dfrac{3x}{15}=\dfrac{16}{15}$
$\dfrac{8x}{15}=\dfrac{16}{15}$
$8x=16 \Rightarrow x=2$

$a + \dfrac{1}{b} = 0 \Rightarrow a = -\dfrac{1}{b}$

Therefore: $ab = \left(-\dfrac{1}{b}\right)\times b = -1$

$ab = -1 < 0$

A terminating decimal with last digit at the 4th place $= 0.d_1d_2d_3d_4$ (where $d_4\neq0$).
This equals $\dfrac{d_1d_2d_3d_4}{10^4} = \dfrac{p}{10^4}$ where $p$ is an integer.
Since $d_4\neq0$, $p$ is not divisible by 10.

In lowest form: cancel common factors between $p$ and $10^4 = 2^4\times5^4$.
The denominator divides $10^4$, so it is of the form $2^a\times5^b$ where $a\leq4$ and $b\leq4$.

Is divisibility by $2^4$ or $5^4$ necessary? No — the denominator must only divide $2^4\times5^4$, not necessarily equal it. For example, $0.0001=\dfrac{1}{10000}=\dfrac{1}{2^4\times5^4}$ (yes), but $0.1250=\dfrac{125}{1000}=\dfrac{1}{8}=\dfrac{1}{2^3}$ (no $5^4$).

$125 = 5^3$ — only prime factor is 5 → terminating.

To find decimal places, make denominator a power of 10:
$\dfrac{18}{5^3} = \dfrac{18\times2^3}{5^3\times2^3} = \dfrac{18\times8}{10^3} = \dfrac{144}{1000} = 0.144$

Denominator $= 2^3\times5^1$. To make it a power of 10, multiply by $5^2$:
$2^3\times5^1\times5^2 = 2^3\times5^3 = 10^3$

So the decimal terminates after 3 places.
In general: for denominator $2^m\times5^n$, decimal places $= \max(m,n)$.

We need common denominator $m$ such that $k_2-k_1 > 6$ (to fit 5 integers strictly between $k_1$ and $k_2$, need $k_2-k_1\geq7$).

$a=\dfrac{7}{12}$, $b=\dfrac{5}{6}=\dfrac{10}{12}$. With $m=12$: $k_2-k_1=10-7=3 < 6$. Not enough.

Use $m=84$: $a=\dfrac{7}{12}=\dfrac{49}{84}$, $b=\dfrac{5}{6}=\dfrac{70}{84}$
$k_2-k_1=70-49=21>6$ ✓

Five distinct rationals with integer numerators between 49 and 70:
$\dfrac{51}{84}=\dfrac{17}{28},\;\dfrac{55}{84},\;\dfrac{60}{84}=\dfrac{5}{7},\;\dfrac{65}{84},\;\dfrac{68}{84}=\dfrac{17}{21}$

Triangle $n$ has legs of length $\sqrt{n}$ and $1$.
Its hypotenuse $= \sqrt{(\sqrt{n})^2+1^2} = \sqrt{n+1}$