End-of-Chapter Exercises

(i) $\dfrac{3}{50}$:
$50 = 2 \times 5^2$ — only factors 2 and 5, so the decimal terminates.
Long division: $3 \div 50 = 0.06$ $$\frac{3}{50} = 0.06$$

(ii) $\dfrac{2}{9}$:
$9 = 3^2$ — contains the prime 3, so the decimal is non-terminating and repeating.
Long division: $2 \div 9 = 0.222\ldots$ $$\frac{2}{9} = 0.\overline{2}$$

Proof by contradiction:
Assume, for contradiction, that $\sqrt{5}$ is rational. Then we can write: $$\sqrt{5} = \frac{p}{q}$$ where $p$ and $q$ are integers, $q \neq 0$, and $\gcd(p, q) = 1$ (i.e., the fraction is in its lowest terms).

Squaring both sides: $$5 = \frac{p^2}{q^2} \implies p^2 = 5q^2$$ This means $5 \mid p^2$ (5 divides $p^2$). Since 5 is prime, by Euclid's lemma, $5 \mid p$.
So we can write $p = 5k$ for some integer $k$.

Substituting $p = 5k$: $$(5k)^2 = 5q^2 \implies 25k^2 = 5q^2 \implies q^2 = 5k^2$$ This means $5 \mid q^2$, so by Euclid's lemma again, $5 \mid q$.

But now we have $5 \mid p$ and $5 \mid q$, which means $\gcd(p, q) \geq 5$. This contradicts our assumption that $\gcd(p, q) = 1$.

Therefore our initial assumption is false, and $\sqrt{5}$ is irrational. $\blacksquare$

(i) $12.6$ — terminating decimal: $$12.6 = \frac{126}{10} = \frac{63}{5}$$

(ii) $0.0120$ — terminating decimal: $$0.0120 = \frac{120}{10000} = \frac{12}{1000} = \frac{3}{250}$$

(iii) $3.0\overline{52}$ — 2 non-repeating decimal places, block of 2:
Let $x = 3.05252\ldots$
$100x = 305.252\ldots$ and $10x = 30.5252\ldots$ $$100x - 10x = 305.252\ldots - 30.5252\ldots = 274.727... $$ Let me redo carefully. The non-repeating part is "30", repeating is "52":
Multiply by $10$ (shift past non-repeating): $10x = 30.5252...$
Multiply by $1000$ (shift past non-repeating + one period): $1000x = 3052.5252...$ $$1000x - 10x = 3052.5252\ldots - 30.5252\ldots = 3022$$ $$990x = 3022 \implies x = \frac{3022}{990} = \frac{1511}{495}$$

(iv) $1.\overline{235}$ — repeating block of 3 from the start:
Let $x = 1.235235\ldots$ $$1000x = 1235.235\ldots$$ $$1000x - x = 1234 \implies 999x = 1234 \implies x = \frac{1234}{999}$$

(v) $0.\overline{23}$ — repeating block of 2:
Let $x = 0.232323\ldots$ $$100x = 23.232323\ldots$$ $$99x = 23 \implies x = \frac{23}{99}$$

(vi) $2.0\overline{5}$ — 1 non-repeating place, block of 1:
Let $x = 2.0555\ldots$
$10x = 20.555\ldots$ and $100x = 205.555\ldots$ $$100x - 10x = 205.555\ldots - 20.555\ldots = 185$$ $$90x = 185 \implies x = \frac{185}{90} = \frac{37}{18}$$

(vii) $2.1\overline{25}$ — 1 non-repeating place, block of 2:
Let $x = 2.12525\ldots$
$10x = 21.2525\ldots$ and $1000x = 2125.2525\ldots$ $$1000x - 10x = 2125.2525\ldots - 21.2525\ldots = 2104$$ $$990x = 2104 \implies x = \frac{2104}{990} = \frac{1052}{495}$$

(viii) $3.1\overline{25}$ — 1 non-repeating place, block of 2:
Let $x = 3.12525\ldots$
$10x = 31.2525\ldots$ and $1000x = 3125.2525\ldots$ $$990x = 3125.2525\ldots - 31.2525\ldots = 3094$$ $$x = \frac{3094}{990} = \frac{1547}{495}$$

(ix) $2.\overline{1625}$ — repeating block of 4 from start:
Let $x = 2.16251625\ldots$ $$10000x = 21625.16251625\ldots$$ $$10000x - x = 21623 \implies 9999x = 21623 \implies x = \frac{21623}{9999}$$ Check if reducible: $\gcd(21623, 9999)$. $21623 = 2(9999) + 1625$; $9999 = 6(1625) + 249$; … $\gcd = 1$. So $\dfrac{21623}{9999}$ is already in lowest terms.

(i) $0.532$:
$0.532$ lies between 0 and 1. More precisely, it lies between $0.5$ and $0.6$ (between $\frac{1}{2}$ and $\frac{3}{5}$).
To locate precisely: divide the unit segment $[0, 1]$ into 1000 equal parts; $0.532$ is at the 532nd mark from 0.
Alternatively: divide $[0.5, 0.6]$ into 10 equal parts; $0.532$ is 3.2 parts along that sub-segment.

(ii) $1.15$:
$1.15$ lies between 1 and 2. More precisely between $1.1$ and $1.2$.
To locate precisely: divide the segment $[1, 2]$ into 100 equal parts; $1.15$ is at the 15th mark from 1.
Or note that $1.15 = 1\frac{3}{20}$ — three-twentieths of the way past 1.

Number line sketch:

Multiply numerator and denominator of both bounds to get room for 6 integers between them: $$3 = \frac{3 \times 8}{8} = \frac{24}{8} \qquad 4 = \frac{4 \times 8}{8} = \frac{32}{8}$$ Integers strictly between 24 and 32: $25, 26, 27, 28, 29, 30, 31$ — we need only 6.

Six rational numbers between 3 and 4: $$\frac{25}{8},\quad \frac{26}{8},\quad \frac{27}{8},\quad \frac{28}{8},\quad \frac{29}{8},\quad \frac{30}{8}$$ Simplified: $\dfrac{25}{8},\ \dfrac{13}{4},\ \dfrac{27}{8},\ \dfrac{7}{2},\ \dfrac{29}{8},\ \dfrac{15}{4}$

Scale up the denominator to 50 (multiply by 10): $$\frac{2}{5} = \frac{20}{50} \qquad \frac{3}{5} = \frac{30}{50}$$ Integers strictly between 20 and 30: $21, 22, 23, 24, 25, 26, 27, 28, 29$.
Pick any 5: $21, 22, 23, 24, 25$.

Five rational numbers: $$\frac{21}{50},\quad \frac{22}{50},\quad \frac{23}{50},\quad \frac{24}{50},\quad \frac{25}{50}$$ Simplified: $\dfrac{21}{50},\ \dfrac{11}{25},\ \dfrac{23}{50},\ \dfrac{12}{25},\ \dfrac{1}{2}$

Find the LCD of 6 and 5: LCD $= 30$. $$\frac{1}{6} = \frac{5}{30} \qquad \frac{2}{5} = \frac{12}{30}$$ Integers strictly between 5 and 12: $6, 7, 8, 9, 10, 11$. We need only 5.

Five rational numbers: $$\frac{6}{30},\quad \frac{7}{30},\quad \frac{8}{30},\quad \frac{9}{30},\quad \frac{10}{30}$$ Simplified: $\dfrac{1}{5},\ \dfrac{7}{30},\ \dfrac{4}{15},\ \dfrac{3}{10},\ \dfrac{1}{3}$

Find the LCD of 3 and 5, which is 15. Rewrite the left side: $$\frac{x}{3} + \frac{x}{5} = \frac{5x}{15} + \frac{3x}{15} = \frac{8x}{15}$$

Set equal to the right side: $$\frac{8x}{15} = \frac{16}{15}$$ $$8x = 16 \implies x = 2$$

From the given condition $a + \dfrac{1}{b} = 0$: $$a = -\frac{1}{b}$$

Now compute $ab$: $$ab = \left(-\frac{1}{b}\right) \times b = -\frac{b}{b} = -1$$

Since $ab = -1$ regardless of the actual values of $a$ and $b$ (as long as they are non-zero), $ab$ is always negative (equal to $-1$).

Alternatively: $a = -\dfrac{1}{b}$ means $a$ and $b$ must have opposite signs (one positive, one negative), so their product $ab$ must be negative.

Part 1 — Writing in the form $\dfrac{p}{10^n}$:
A decimal terminating at the 4th place has the form $0.d_1 d_2 d_3 d_4$ where $d_4 \neq 0$.
Multiplying numerator and denominator by $10^4$: $$x = \frac{d_1 d_2 d_3 d_4}{10^4} = \frac{p}{10^4}$$ where $p = d_1 d_2 d_3 d_4$ is an integer. Since $d_4 \neq 0$ (the last digit is non-zero), $p$ is not divisible by 10.

Part 2 — Must the denominator (in lowest terms) be divisible by $2^4$ or $5^4$?
Not necessarily. Consider two cases:

Example 1: $x = 0.0625 = \dfrac{625}{10000} = \dfrac{1}{16} = \dfrac{1}{2^4}$. Here the denominator in lowest terms is $2^4$ — divisible by $2^4$ but not $5^4$.

Example 2: $x = 0.1250 = \dfrac{1250}{10000} = \dfrac{1}{8} = \dfrac{1}{2^3}$. Here the last non-zero digit is in the 3rd place, not the 4th.

Example 3: $x = 0.3125 = \dfrac{3125}{10000} = \dfrac{5}{16} = \dfrac{5}{2^4}$. The denominator is $2^4$, not $5^4$.

In general, $\dfrac{p}{10^4} = \dfrac{p}{2^4 \times 5^4}$. After simplifying, the denominator will be of the form $2^a \times 5^b$ where $0 \leq a \leq 4$ and $0 \leq b \leq 4$. It is not necessary that both $2^4$ and $5^4$ divide the denominator — one or both powers could be less than 4 after simplification.

First check if the fraction is in lowest terms: $$\gcd(18, 125): \quad 18 = 2 \times 3^2, \quad 125 = 5^3$$ No common factors, so $\dfrac{18}{125}$ is already in lowest terms.

Now factorise the denominator: $$125 = 5^3$$ The denominator contains only the prime 5 (no factor of 3 or any other prime), so the decimal terminates.

To find the number of decimal places, multiply to get a power of 10 in the denominator: $$\frac{18}{125} = \frac{18}{5^3} = \frac{18 \times 2^3}{5^3 \times 2^3} = \frac{18 \times 8}{10^3} = \frac{144}{1000} = 0.144$$ The decimal terminates in 3 decimal places.

The denominator is $2^3 \times 5^1$. Since it contains only the primes 2 and 5, the decimal terminates.

To express as a terminating decimal, we convert the denominator to a power of 10. We need equal powers of 2 and 5: $$2^3 \times 5^1 \longrightarrow \text{multiply top and bottom by } 5^2 = 25$$ $$\frac{p}{2^3 \times 5} \times \frac{5^2}{5^2} = \frac{25p}{2^3 \times 5^3} = \frac{25p}{10^3}$$ The denominator becomes $10^3$, giving a decimal with 3 decimal places.

The key is: the number of decimal places equals $\max(a, b)$ where the denominator is $2^a \times 5^b$.
Here $a = 3$, $b = 1$, so $\max(3, 1) = 3$.

Step 1 — Express with a common denominator where $k_2 - k_1 > 6$:
LCD of 12 and 6 is 12, but $\dfrac{7}{12}$ and $\dfrac{10}{12}$ give $k_2 - k_1 = 3$, which is not $> 6$.
Scale up to $m = 84$ (multiply by 7): $$a = \frac{7}{12} = \frac{49}{84}, \qquad b = \frac{5}{6} = \frac{70}{84}$$ $k_1 = 49$, $k_2 = 70$, $k_2 - k_1 = 21 > 6$. ✓

Step 2 — Five distinct rationals with integer numerators between $\dfrac{49}{84}$ and $\dfrac{70}{84}$:
Choose any 5 integers strictly between 49 and 70: $50, 55, 60, 65, 69$ (for variety). $$\frac{50}{84} = \frac{25}{42},\quad \frac{55}{84},\quad \frac{60}{84} = \frac{5}{7},\quad \frac{65}{84},\quad \frac{69}{84} = \frac{23}{28}$$

Step 3 — Why $k_2 - k_1 = n + 1$ is necessary:
When using a common denominator $m$, rational numbers with integer numerators between $a = \dfrac{k_1}{m}$ and $b = \dfrac{k_2}{m}$ are exactly: $$\frac{k_1 + 1}{m},\ \frac{k_1 + 2}{m},\ \ldots,\ \frac{k_2 - 1}{m}$$ The number of such fractions is $(k_2 - 1) - (k_1 + 1) + 1 = k_2 - k_1 - 1$.
For this to equal $n$, we need $k_2 - k_1 - 1 = n$, i.e., $k_2 - k_1 = n + 1$.
If $k_2 - k_1 \leq n$, there are not enough integer numerators available between $k_1$ and $k_2$ to find $n$ distinct rationals using this method alone.

Key identity: For any three numbers, $$(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$$

Substitute the given conditions $x + y + z = 0$ and $xy + yz + zx = 0$: $$0^2 = x^2 + y^2 + z^2 + 2(0)$$ $$0 = x^2 + y^2 + z^2$$

Since $x$, $y$, $z$ are rational numbers (hence real numbers), their squares are each $\geq 0$: $$x^2 \geq 0, \quad y^2 \geq 0, \quad z^2 \geq 0$$ The only way three non-negative quantities can sum to zero is if each is individually zero: $$x^2 = 0,\quad y^2 = 0,\quad z^2 = 0$$ $$\implies x = 0,\quad y = 0,\quad z = 0 \qquad \blacksquare$$

Without loss of generality, assume $a \lt b$ (the case $a > b$ is symmetric).
We need to show: $a \lt \dfrac{a+b}{2} \lt b$.

Left inequality: $a \lt \dfrac{a+b}{2}$ $$a \lt \frac{a+b}{2}$$ $$2a \lt a + b \quad \text{(multiply both sides by 2, positive)}$$ $$a \lt b \quad \checkmark \text{ (this is our assumption)}$$

Right inequality: $\dfrac{a+b}{2} \lt b$ $$\frac{a+b}{2} \lt b$$ $$a + b \lt 2b$$ $$a \lt b \quad \checkmark \text{ (again our assumption)}$$

Both inequalities hold whenever $a \lt b$, so $a \lt \dfrac{a+b}{2} \lt b$ is proven.

Note: since $a$ and $b$ are rational, $a + b$ is rational, and $\dfrac{a+b}{2}$ is also rational (division by 2 of a rational number is rational). So the midpoint is both between $a$ and $b$ and rational. $\blacksquare$

Pattern: Each right triangle has one leg of length 1 and the other leg equal to the hypotenuse of the previous triangle.
Using the Pythagorean theorem $h_n = \sqrt{h_{n-1}^2 + 1}$ with $h_1 = \sqrt{1^2 + 1^2} = \sqrt{2}$:

Triangle 1: legs = 1, 1   $\Rightarrow$   $h_1 = \sqrt{1^2 + 1^2} = \sqrt{2}$
Triangle 2: legs = $\sqrt{2}$, 1   $\Rightarrow$   $h_2 = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{3}$
Triangle 3: legs = $\sqrt{3}$, 1   $\Rightarrow$   $h_3 = \sqrt{3 + 1} = \sqrt{4} = 2$
Triangle 4: legs = 2, 1   $\Rightarrow$   $h_4 = \sqrt{4 + 1} = \sqrt{5}$
Triangle 5: legs = $\sqrt{5}$, 1   $\Rightarrow$   $h_5 = \sqrt{5 + 1} = \sqrt{6}$
Triangle 6: $h_6 = \sqrt{7}$
Triangle 7: $h_7 = \sqrt{8} = 2\sqrt{2}$
Triangle 8: $h_8 = \sqrt{9} = 3$
Triangle 9: $h_9 = \sqrt{10}$
Triangle 10: $h_{10} = \sqrt{11}$
Triangle 11: $h_{11} = \sqrt{12} = 2\sqrt{3}$
Triangle 12: $h_{12} = \sqrt{13}$

General pattern: $$h_n = \sqrt{n+1}$$ The $n$-th hypotenuse (counting from the first triangle with two legs of 1) equals $\sqrt{n+1}$.