Chapter 3: The World of Numbers (Number Systems)

12 bags make $12÷2=6$ groups of 2 bags. Each group gets 15 ingots.

$$6\times 15=90$$

He leaves with 90 copper ingots.

11, 13, 17 and 19 are all prime numbers (each has exactly two factors, 1 and itself). In fact they are exactly the primes between 10 and 20. The primes after 19 are 23, 29 and 31 (21 = 3 × 7, 25 = 5 × 5, 27 = 3 × 9 are not prime).

They are all prime numbers. The next three are 23, 29 and 31.

No. For closure, the difference of every pair of natural numbers would have to be a natural number, but:

• $3-5=-2$, which is not a natural number.
• $4-4=0$, which is not a natural number (natural numbers start at 1).

(Sometimes the answer is natural, e.g. $9-2=7$, but one failing example is enough to break closure.)

No, natural numbers are not closed under subtraction: for example, 3 − 5 = −2 and 4 − 4 = 0 are not natural numbers.

The thumb touches the joints (segments) of the other four fingers. Each finger has 3 segments, so one hand gives

$$4\times 3=12$$

So counting naturally goes in groups of 12, which is why base-12 (duodecimal) systems arose. We still see this in a dozen (12), 12 months, and 12 hours on a clock face. If the fingers of the other hand keep track of how many twelves have been counted (up to 5), we reach $12\times 5=60$, which matches the ancient base-60 system still used for minutes and seconds.

12 on one hand (4 fingers × 3 joints). This is the basis of counting in dozens, i.e. a base-12 system.

A drop is a subtraction:

$$4-15=-11$$

The midnight temperature is −11 °C.

Write debts and losses as negative numbers and fortunes as positive numbers:

$$(-850)+1200+(-450)=350-450=-100$$

(−850) + 1200 + (−450) = −100. He is left with a debt of ₹100.

• (i) A debt times a fortune is a debt: $(-12)\times 5=-60$.
• (ii) A debt times a debt is a fortune: $(-8)\times (-7)=56$.
• (iii) Subtracting a debt from zero gives a fortune: $0-(-14)=0+14=14$.
• (iv) A debt divided by a fortune is a debt: $(-20)÷4=-5$.

(i) −60 (ii) 56 (iii) 14 (iv) −5

Suppose your net worth is ₹10, but this includes a debt: you have ₹15 in hand and you owe a friend ₹5, so $15+(-5)=10$.

Now the friend says, "Forget the ₹5 you owe me." The debt of ₹5 is taken away (subtracted). You still have ₹15 in hand and owe nothing, so your net worth becomes ₹15.

$$10-(-5)=15=10+5$$

Removing a debt makes you richer by exactly the same amount as receiving that much money. So subtracting −5 has the same effect as adding +5.

Taking away a debt of ₹5 increases your worth by ₹5, just like receiving ₹5. Hence 10 − (−5) = 10 + 5 = 15.

Use the rule: $\frac{a}{b}=\frac{c}{d}$ if and only if $ad=bc$ (cross-multiplication).

• (i) $2\times 6=12$ and $3\times 4=12$. Equal, so $\frac{2}{3}=\frac{4}{6}$.
• (ii) $5\times 8=40$ and $4\times 10=40$. Equal, so $\frac{5}{4}=\frac{10}{8}$.
• (iii) $(-3)\times 10=-30$ and $5\times (-6)=-30$. Equal, so $-\frac{3}{5}=-\frac{6}{10}$.
• (iv) Write $3=\frac{3}{1}$: $9\times 1=9$ and $3\times 3=9$. Equal, so $\frac{9}{3}=3$.

In each pair the cross-products are equal (12 = 12, 40 = 40, −30 = −30, 9 = 9), so the numbers are equal.

Make the denominators equal using the LCM, then add the numerators.

$$\text{(i) }\frac{4}{10}+\frac{3}{10}=\frac{7}{10}$$

$$\text{(ii) LCM}(12,8)=24:\frac{14}{24}+\frac{15}{24}=\frac{29}{24}$$

$$\text{(iii) }-\frac{8}{14}+\frac{3}{14}=-\frac{5}{14}$$

(i) $\frac{7}{10}$ (ii) $\frac{29}{24}$ (iii) $-\frac{5}{14}$

$$\text{(i) }\frac{10}{12}-\frac{3}{12}=\frac{7}{12}$$

$$\text{(ii) }\frac{11}{8}-\frac{6}{8}=\frac{5}{8}$$

$$\text{(iii) }-\frac{7}{9}+\frac{2}{3}=-\frac{7}{9}+\frac{6}{9}=-\frac{1}{9}$$

(i) $\frac{7}{12}$ (ii) $\frac{5}{8}$ (iii) $-\frac{1}{9}$

Multiply numerators and denominators: $\frac{a}{b}\times \frac{c}{d}=\frac{ac}{bd}$, then simplify.

$$\text{(i) }\frac{2\times 3}{3\times 10}=\frac{6}{30}=\frac{1}{5}\text{(ii) }\frac{7\times 5}{11\times 8}=\frac{35}{88}\text{(iii) }-\frac{4\times 5}{7\times 14}=-\frac{20}{98}=-\frac{10}{49}$$

(i) $\frac{1}{5}$ (ii) $\frac{35}{88}$ (iii) $-\frac{10}{49}$

Dividing by $\frac{c}{d}$ is multiplying by $\frac{d}{c}$.

$$\text{(i) }\frac{2}{3}\times \frac{10}{3}=\frac{20}{9}\text{(ii) }\frac{7}{11}\times \frac{8}{5}=\frac{56}{55}\text{(iii) }-\frac{4}{7}\times \frac{14}{5}=-\frac{56}{35}=-\frac{8}{5}$$

(i) $\frac{20}{9}$ (ii) $\frac{56}{55}$ (iii) $-\frac{8}{5}$

Left-hand side:

$$\left(\frac{2}{4}+\frac{3}{4}\right)\times \frac{8}{3}=\frac{5}{4}\times \frac{8}{3}=\frac{40}{12}=\frac{10}{3}$$

Right-hand side:

$$\frac{8}{6}+\frac{24}{12}=\frac{4}{3}+2=\frac{4}{3}+\frac{6}{3}=\frac{10}{3}$$

Both sides equal $\frac{10}{3}$. This is the distributive law $p(q+r)=pq+pr$.

LHS = RHS = $\frac{10}{3}$.

$$\frac{7}{9}\times \frac{6}{7}-\frac{7}{9}\times \frac{3}{4}=\frac{6}{9}-\frac{21}{36}=\frac{2}{3}-\frac{7}{12}=\frac{8}{12}-\frac{7}{12}=\frac{1}{12}$$

Check directly: $\frac{6}{7}-\frac{3}{4}=\frac{24-21}{28}=\frac{3}{28}$, and $\frac{7}{9}\times \frac{3}{28}=\frac{21}{252}=\frac{1}{12}$ ✓

$\frac{1}{12}$

Expand the left side using the distributive law:

$$\frac{5}{6}x+\frac{5}{6}\times \frac{3}{5}=\frac{5}{6}x+\frac{15}{30}=\frac{5}{6}x+\frac{1}{2}$$

So the left side is exactly the same as the right side, whatever x is. The equation is an identity: it is true for every rational number x. (Subtracting $\frac{5}{6}x$ from both sides leaves $\frac{1}{2}=\frac{1}{2}$.)

For example, $x=0$: LHS $=\frac{5}{6}\times \frac{3}{5}=\frac{1}{2}$ = RHS ✓; $x=6$: LHS $=\frac{5}{6}\times \frac{33}{5}=\frac{11}{2}$ and RHS $=5+\frac{1}{2}=\frac{11}{2}$ ✓

Every rational number x satisfies the equation (it is an identity, by the distributive law).

• $\frac{2}{3}$: divide the interval from 0 to 1 into 3 equal parts and take 2 parts to the right of 0.
• $-\frac{5}{4}=-1\frac{1}{4}$: it lies between −2 and −1. Divide that interval into 4 equal parts and take 1 part to the left of −1.
• $1\frac{1}{2}=\frac{3}{2}$: halfway between 1 and 2.

$-\frac{5}{4}$ lies between −2 and −1 (a quarter of the way from −1 to −2), $\frac{2}{3}$ between 0 and 1, and $1\frac{1}{2}$ halfway between 1 and 2.

Write both with denominator 8: $-\frac{1}{2}=-\frac{4}{8}$ and $\frac{1}{4}=\frac{2}{8}$. Any fraction with numerator −3, −2, −1, 0 or 1 over 8 lies between them, for example

$$-\frac{3}{8},-\frac{1}{4}\left(=-\frac{2}{8}\right),0$$

For example $-\frac{3}{8}$, $-\frac{1}{4}$ and $0$ (also $-\frac{1}{8}$, $\frac{1}{8}$).

$$-\frac{3}{12}+\frac{5}{12}=\frac{2}{12}=\frac{1}{6}$$

$\frac{1}{6}$

Convert to improper fractions: $15\frac{3}{4}=\frac{63}{4}$ and $2\frac{1}{4}=\frac{9}{4}$.

$$\frac{63}{4}÷\frac{9}{4}=\frac{63}{4}\times \frac{4}{9}=\frac{63}{9}=7$$

He can make exactly 7 kurtas (with no silk left over).

Add one more decimal place: 3.1415 = 3.14150 and 3.1416 = 3.14160. Any number from 3.14151 to 3.14159 lies between them.

For example 3.14151, 3.14155 and 3.14159.

Besides taking the average $\frac{a+b}{2}$, here are some other ways:

• Equal denominators, then enlarge: write both numbers with the same denominator and, if their numerators are consecutive, multiply the numerator and denominator by 10. Example: between $\frac{1}{3}=\frac{10}{30}$ and $\frac{1}{2}=\frac{15}{30}$ we get $\frac{11}{30},\frac{12}{30},\dots$
• Mediant (for positive fractions): between $\frac{a}{b}$ and $\frac{c}{d}$ lies $\frac{a+c}{b+d}$. Example: between $\frac{1}{3}$ and $\frac{1}{2}$ lies $\frac{2}{5}$.
• Decimals: convert both to decimals and pick a decimal in between. Example: between $\frac{1}{3}=0.333\dots$ and $\frac{1}{2}=0.5$ lies 0.4.

Yes: the mediant $\frac{a+c}{b+d}$, the common-denominator method, and the decimal method all give rational numbers between two given rationals.

Rule: a fraction in lowest terms has a terminating decimal exactly when its denominator has no prime factors other than 2 and 5.

• $\frac{7}{20}$: $20=2^2\times 5$. Terminating. Long division gives $0.35$.
• $\frac{4}{15}$: $15=3\times 5$ has the factor 3. Repeating. Long division gives $0.2666\dots =0.2\overline{6}$.
• $\frac{13}{250}$: $250=2\times 5^3$. Terminating. Long division gives $0.052$.

$\frac{7}{20}=0.35$ and $\frac{13}{250}=0.052$ terminate; $\frac{4}{15}=0.2\overline{6}$ repeats.

Long division of 1 by 13 gives remainders 10, 9, 12, 3, 4, 1, after which they repeat:

$$\frac{1}{13}=0.\overline{076923}$$

The repeating block is 076923 (6 digits). But $\frac{2}{13}=0.\overline{153846}$ uses completely different digits, so it is not a rotation of 076923. Continuing:

Observation: the twelve fractions split into two families of six. In family A (1, 3, 4, 9, 10, 12) every repeating block is a rotation of 076923; in family B (2, 5, 6, 7, 8, 11) every block is a rotation of 153846. So 076923 is not a full cyclic number like 142857 (which works for all of $\frac{1}{7}$ to $\frac{6}{7}$); its period, 6, is only half of 12.

$\frac{1}{13}=0.\overline{076923}$. $\frac{2}{13}=0.\overline{153846}$ is not a rotation of it. The thirteenths form two cyclic families: rotations of 076923 and rotations of 153846.

• (i) $\sqrt{81}=9=\frac{9}{1}$. Rational.
• (ii) $\sqrt{12}=\sqrt{4\times 3}=2\sqrt{3}$. Since $\sqrt{3}$ is irrational, $2\sqrt{3}$ is irrational (if $2\sqrt{3}=\frac{p}{q}$ then $\sqrt{3}=\frac{p}{2q}$ would be rational). Irrational.
• (iii) $0.3333\dots =0.\overline{3}$. Let $x=0.\overline{3}$; $10x=3.\overline{3}$; $9x=3$; $x=\frac{1}{3}$. Rational.
• (iv) $0.\overline{12345}$ (block of 5 digits). Let $x=0.\overline{12345}$; $100000x=12345.\overline{12345}$; $99999x=12345$; $x=\frac{12345}{99999}=\frac{4115}{33333}$. Rational.
• (v) $1.0100100010000\dots$: the number of zeros between the 1s keeps increasing (1, 2, 3, 4, ...), so no single block repeats and it never terminates. Irrational.
• (vi) It terminates (after 21 decimal places), so it is rational: $\frac{23560185612239874790120}{10^{21}}=\frac{589004640305996869753}{25000000000000000000}$ in lowest terms.

Rational: (i) 9, (iii) $\frac{1}{3}$, (iv) $\frac{4115}{33333}$, (vi) $\frac{589004640305996869753}{25\times 10^{18}}$. Irrational: (ii) and (v).

$$x=0.999\dots$$

$$10x=9.999\dots$$

Subtracting the first line from the second, the endless tails of 9s cancel exactly:

$$9x=9\Rightarrow x=1$$

Another way to see it: $\frac{1}{3}=0.333\dots$, so $3\times \frac{1}{3}=0.999\dots$, but $3\times \frac{1}{3}=1$. There is no number between 0.999... and 1, so they are the same point on the number line.

$9x=9$ gives $x=1$, so $0.\overline{9}=1$ exactly.

The block is cyclic when the decimal of $\frac{1}{n}$ has the longest possible period, $n-1$ digits (then every remainder 1, 2, ..., n − 1 appears in the long division). This happens for the primes 7, 17, 19, 23, 29, 47, 59, 61, 97, ...

Example, $n=17$:

$$\frac{1}{17}=0.\overline{0588235294117647}(16\text{ digits})$$

$$\frac{2}{17}=0.\overline{1176470588235294},\frac{3}{17}=0.\overline{1764705882352941},\dots$$

Each is the same block started at a different place, so 0588235294117647 is cyclic, just like 142857. ($\frac{1}{19}$ gives the 18-digit cyclic block 052631578947368421.) For 13 this fails (see Q2), since its period is only 6.

For example n = 17, 19, 23, 29, 47, 59, 61, 97 (primes for which $\frac{1}{n}$ has period n − 1). E.g. $\frac{1}{17}=0.\overline{0588235294117647}$ is cyclic.

(i) $50=2\times 5^2$, so it terminates: $3÷50=0.06$ (since $\frac{3}{50}=\frac{6}{100}$).

(ii) $2÷9$: each step leaves remainder 2 again, so the digit 2 repeats forever: $0.222\dots$

(i) $\frac{3}{50}=0.06$ (terminating) (ii) $\frac{2}{9}=0.\overline{2}$ (non-terminating, repeating)

We use proof by contradiction, as for $\sqrt{2}$.

Step 1: Assume $\sqrt{5}$ is rational. Then $\sqrt{5}=\frac{p}{q}$, where p and q are integers, $q\ne 0$, and p, q have no common factor other than 1.

Step 2: Squaring, $5=\frac{p^2}{q^2}$, so $p^2=5q^2$.

Step 3: So $p^2$ is a multiple of 5. Since 5 is prime, if 5 divides $p\times p$ it must divide p. So $p=5k$ for some integer k.

Step 4: Substitute: $25k^2=5q^2$, so $q^2=5k^2$. Hence $q^2$ is a multiple of 5, and as before q is a multiple of 5.

Step 5: Now p and q are both multiples of 5, which contradicts the assumption that they have no common factor.

So the assumption was wrong, and $\sqrt{5}$ is irrational.

Assuming $\sqrt{5}=\frac{p}{q}$ in lowest terms forces both p and q to be multiples of 5, a contradiction. Hence $\sqrt{5}$ is irrational.

Terminating decimals:

$$\text{(i) }12.6=\frac{126}{10}=\frac{63}{5}\text{(ii) }0.0120=\frac{120}{10000}=\frac{3}{250}$$

(iii) $x=3.0\overline{52}$: one non-repeating digit, two repeating digits.

$$10x=30.\overline{52},1000x=3052.\overline{52}\Rightarrow 990x=3022\Rightarrow x=\frac{3022}{990}=\frac{1511}{495}$$

(iv) $x=1.2\overline{35}$:

$$10x=12.\overline{35},1000x=1235.\overline{35}\Rightarrow 990x=1223\Rightarrow x=\frac{1223}{990}$$

(v) $x=0.\overline{23}$:

$$100x=23.\overline{23}\Rightarrow 99x=23\Rightarrow x=\frac{23}{99}$$

(vi) $x=2.0\overline{5}$:

$$10x=20.\overline{5},100x=205.\overline{5}\Rightarrow 90x=185\Rightarrow x=\frac{185}{90}=\frac{37}{18}$$

(vii) $x=2.12\overline{5}$:

$$100x=212.\overline{5},1000x=2125.\overline{5}\Rightarrow 900x=1913\Rightarrow x=\frac{1913}{900}$$

(viii) $x=3.12\overline{5}$:

$$100x=312.\overline{5},1000x=3125.\overline{5}\Rightarrow 900x=2813\Rightarrow x=\frac{2813}{900}$$

(ix) $x=2.\overline{1625}$:

$$10000x=21625.\overline{1625}\Rightarrow 9999x=21623\Rightarrow x=\frac{21623}{9999}$$

(Each answer is in lowest terms; e.g. check (iii): $1511÷495=3.05252\dots$ ✓)

(i) $\frac{63}{5}$ (ii) $\frac{3}{250}$ (iii) $\frac{1511}{495}$ (iv) $\frac{1223}{990}$ (v) $\frac{23}{99}$ (vi) $\frac{37}{18}$ (vii) $\frac{1913}{900}$ (viii) $\frac{2813}{900}$ (ix) $\frac{21623}{9999}$

We use successive magnification: zoom into the right interval one decimal place at a time.

(i) 0.532 lies between 0.5 and 0.6; within that, between 0.53 and 0.54; and it is the 2nd of the ten thousandth marks after 0.53.

(ii) $1.1\overline{5}=1.1555\dots$ lies between 1.1 and 1.2; then between 1.15 and 1.16; then between 1.155 and 1.156, and so on, always closer to the right end. As a fraction, $1.1\overline{5}=\frac{104}{90}=\frac{52}{45}$: divide the interval from 1 to 2 into 45 equal parts and take 7 parts after 1.

0.532 is 2 thousandths after 0.53; $1.1\overline{5}$ $(=\frac{52}{45})$ lies between 1.155 and 1.156.

Write $3=\frac{30}{10}$ and $4=\frac{40}{10}$. The numbers $\frac{31}{10},\frac{32}{10},\dots ,\frac{39}{10}$ lie between them.

For example 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 (i.e. $\frac{31}{10},\frac{16}{5},\frac{33}{10},\frac{17}{5},\frac{7}{2},\frac{18}{5}$).

Multiply numerator and denominator by 10: $\frac{2}{5}=\frac{20}{50}$ and $\frac{3}{5}=\frac{30}{50}$. So $\frac{21}{50},\frac{22}{50},\dots ,\frac{29}{50}$ lie between them.

For example $\frac{21}{50},\frac{11}{25},\frac{23}{50},\frac{12}{25},\frac{1}{2}$.

LCM of 6 and 5 is 30: $\frac{1}{6}=\frac{5}{30}$ and $\frac{2}{5}=\frac{12}{30}$. The numerators 6, 7, 8, 9, 10, 11 lie between 5 and 12.

For example $\frac{6}{30}=\frac{1}{5}$, $\frac{7}{30}$, $\frac{8}{30}=\frac{4}{15}$, $\frac{9}{30}=\frac{3}{10}$, $\frac{10}{30}=\frac{1}{3}$.

$$\frac{5x+3x}{15}=\frac{16}{15}\Rightarrow 8x=16\Rightarrow x=2$$

Check: $\frac{2}{3}+\frac{2}{5}=\frac{10+6}{15}=\frac{16}{15}$ ✓

x = 2

From $a+\frac{1}{b}=0$ we get $a=-\frac{1}{b}$. Multiply both sides by b (allowed, since $b\ne 0$):

$$ab=-\frac{1}{b}\times b=-1$$

ab = −1, so ab is negative.

Let the number be $x=N.d_1{d}_2{d}_3{d}_4$, where N is the integer part and $d_4\ne 0$ is the last non-zero digit. Multiplying by $10^4$ moves the decimal point 4 places, giving an integer:

$$p=10^{4}x=N{d}_1{d}_2{d}_3{d}_4\Rightarrow x=\frac{p}{10^4}$$

The last digit of p is $d_4\ne 0$, so p is not divisible by 10.

Now $\frac{p}{10^4}=\frac{p}{2^4\times 5^4}$. To reach lowest terms we can cancel only factors of 2 or 5 that divide p. Since p is not divisible by 10, p cannot be divisible by both 2 and 5:

• If p is odd, no factor 2 cancels, so the denominator keeps $2^4$.
• If p is not a multiple of 5, no factor 5 cancels, so the denominator keeps $5^4$.

At least one of these is always true. Examples: $0.0002=\frac{2}{10^4}=\frac{1}{5000}=\frac{1}{2^3\times 5^4}$ (divisible by $5^4$ but not $2^4$); $0.0005=\frac{1}{2000}=\frac{1}{2^4\times 5^3}$ (divisible by $2^4$ but not $5^4$).

Yes. The lowest-terms denominator is always divisible by $2^4$ or by $5^4$ (at least one of them, not necessarily both), because p cannot cancel factors of both 2 and 5.

$\frac{18}{125}$ is in lowest terms and $125=5^3$ has only the prime factor 5, so the decimal terminates. To make the denominator a power of 10, multiply by $2^3$:

$$\frac{18}{125}=\frac{18\times 8}{5^3\times 2^3}=\frac{144}{1000}=0.144$$

It terminates, with 3 decimal places ($\frac{18}{125}=0.144$).

Let the number be $\frac{p}{2^3\times 5}=\frac{p}{40}$ in lowest terms, so p has no factor 2 or 5. Multiply numerator and denominator by $5^2$ to get a power of 10:

$$\frac{p}{2^3\times 5}=\frac{25p}{2^3\times 5^3}=\frac{25p}{1000}$$

So there are at most 3 decimal places. The last digit of $25p$ is not 0 (p is odd, so $25p$ ends in 25 or 75), so the third decimal place is non-zero. Example: $\frac{1}{40}=0.025$, $\frac{3}{40}=0.075$.

Exactly 3 decimal places, because the larger of the powers of 2 and 5 in the denominator is 3.

With $m=12$: $a=\frac{7}{12}$, $b=\frac{10}{12}$, so $k_2-k_1=3$, too small. Multiply by 3, i.e. take $m=36$:

$$a=\frac{21}{36},b=\frac{30}{36},k_2-k_1=30-21=9>6$$

Five rational numbers between them:

$$\frac{22}{36},\frac{23}{36},\frac{24}{36},\frac{25}{36},\frac{26}{36}$$

Why the condition: the fractions $\frac{k}{m}$ strictly between $\frac{k_1}{m}$ and $\frac{k_2}{m}$ are those with $k=k_1+1,k_1+2,\dots ,k_2-1$. There are exactly $k_2-k_1-1$ such integers. To get n numbers we need $k_2-k_1-1\ge n$, that is $k_2-k_1\ge n+1$. The condition $k_2-k_1>n+1$ (here $9>6$) guarantees this, with at least one numerator to spare. If the gap is too small (as with m = 12), we simply enlarge m.

$a=\frac{21}{36}$, $b=\frac{30}{36}$; five numbers: $\frac{22}{36},\frac{23}{36},\frac{24}{36},\frac{25}{36},\frac{26}{36}$. There are only $k_2-k_1-1$ integers between $k_1$ and $k_2$, so the gap must exceed n for n numerators to fit.

Square the first equation:

$$(x+y+z{)}^2=x^2+y^2+z^2+2(xy+yz+zx)$$

Substituting $x+y+z=0$ and $xy+yz+zx=0$:

$$0=x^2+y^2+z^2+2\times 0\Rightarrow x^2+y^2+z^2=0$$

A square of a rational number is never negative, so a sum of three squares can be 0 only if each square is 0. Hence $x^2=y^2=z^2=0$, i.e. $x=y=z=0$.

$x^2+y^2+z^2=(x+y+z{)}^2-2(xy+yz+zx)=0$, which forces x = y = z = 0.

Suppose $a<b$, so $b-a>0$. Then

$$\frac{a+b}{2}-a=\frac{b-a}{2}>0\Rightarrow \frac{a+b}{2}>a$$

$$b-\frac{a+b}{2}=\frac{b-a}{2}>0\Rightarrow \frac{a+b}{2}<b$$

So $a<\frac{a+b}{2}<b$. It is rational because the sum of two rationals is rational and dividing a rational by 2 gives a rational. In fact it is exactly the midpoint, at distance $\frac{b-a}{2}$ from each end.

$\frac{a+b}{2}-a=b-\frac{a+b}{2}=\frac{b-a}{2}>0$, so $a<\frac{a+b}{2}<b$.

In the spiral, the first triangle has both legs 1. Every next triangle uses the previous hypotenuse as one leg and a new leg of length 1, with a right angle between them. By the Baudhayana-Pythagoras theorem:

$$h_1=\sqrt{1^2+1^2}=\sqrt{2},h_2=\sqrt{(\sqrt{2}{)}^2+1^2}=\sqrt{3},h_3=\sqrt{3+1}=\sqrt{4}=2,\dots$$

Each new hypotenuse is $\sqrt{n+1}$ when the previous one is $\sqrt{n}$. Fig. 3.14 has 10 triangles, so the hypotenuses are

$$\sqrt{2},\sqrt{3},\sqrt{4}=2,\sqrt{5},\sqrt{6},\sqrt{7},\sqrt{8}=2\sqrt{2},\sqrt{9}=3,\sqrt{10},\sqrt{11}$$

√2, √3, 2, √5, √6, √7, 2√2, 3, √10 and √11.