The World of Numbers — CBSE Class 9 Maths Textbook Solutions

The rate is $15$15 ingots per $2$2 bags, i.e. $\frac{15}{2}$15/2 ingots per bag.

For $12$12 bags, set up a proportion: $\frac{15}{2} = \frac{x}{12}$15/2 = x/12.

$x = \frac{15 \times 12}{2} = \frac{180}{2} = 90$x = (15 × 12)/2 = 180/2 = 90.

He will leave with 90 copper ingots.

All four numbers — $11, 13, 17, 19$11, 13, 17, 19 — are prime numbers; specifically, they are exactly the prime numbers strictly between 10 and 20.

Continuing the sequence of prime numbers beyond 19, the next three primes are $23, 29, 31$23, 29, 31.

No, the natural numbers $\mathbb{N} = \{1,2,3,\dots\}$N = {1,2,3,…} are not closed under subtraction.

Closure under subtraction would require $a - b \in \mathbb{N}$a - b ∈ N for every $a, b \in \mathbb{N}$a, b ∈ N. This fails whenever $b \geq a$b ≥ a.

Example 1: $3 - 5 = -2$3 - 5 = -2, which is not a natural number.

Example 2: $5 - 5 = 0$5 - 5 = 0, which is also not a natural number (natural numbers start at 1).

Since we can find natural numbers whose difference is not natural, the set is not closed under subtraction.

Excluding the thumb (which is used as the pointer), the remaining four fingers each have 3 joints, giving $4 \times 3 = 12$4 × 3 = 12 joints in total on one hand.

Using the thumb to touch each joint in turn, one can count all the way up to 12 on a single hand before running out of joints.

This is exactly why many ancient counting systems (and modern remnants of them, such as 12 inches in a foot, 12 hours on a clock face, and counting in dozens) are based on 12 (duodecimal/base-12) rather than on the fingers alone (which would naturally suggest base-10): the 12 finger-joints provided a ready-made counting frame of 12, and the other hand's fingers (up to 5, or the fingers again touching each set of 12) could then be used to track how many full dozens had been counted, similarly to how some cultures reached systems based on 60 (a multiple of 12).

Think of a negative number as a debt and multiplication by a positive number as adding copies of that debt (making you poorer), while multiplication by a negative number represents removing/taking away copies of that debt (making you richer).

So $(-3) \times 4$(-3) × 4 means "add four debts of ₹3 each", giving a total debt of $-12$-12.

But $(-3) \times (-4)$(-3) × (-4) means "remove four debts of ₹3 each". If each debt of ₹3 that is removed makes you ₹3 richer, then removing 4 such debts makes you $4 \times 3 = 12$4 × 3 = 12 richer.

So $(-3) \times (-4) = +12$(-3) × (-4) = +12.

In general, multiplying by a negative number reverses the direction of the action (addition becomes removal), so a debt (negative) that is removed (multiplied by a negative) results in a fortune (positive). This is why a negative times a negative is always positive.

Noon temperature $= 4\,°C$= 4 °C. It drops by $15\,°C$15 °C, so we subtract:

$4 - 15 = -11$4 - 15 = -11.

The midnight temperature is $-11\,°C$-11 °C.

Represent the loan as $-850$-850 (a debt), the profit as $+1200$+1200 (a fortune), and the loss as $-450$-450.

Equation: $-850 + 1200 - 450$-850 + 1200 - 450.

Step 1: $-850 + 1200 = 350$-850 + 1200 = 350.

Step 2: $350 - 450 = -100$350 - 450 = -100.

Final financial standing: $-₹100$-₹100, i.e. the trader ends up ₹100 in debt.

Using Brahmagupta's laws (product of a debt and a fortune is a debt; product of two debts is a fortune; subtracting a number leaves the value unchanged when that number is 0, and subtracting a negative is the same as adding):

(i) $(-12) \times 5 = -60$(-12) × 5 = -60 (debt × fortune = debt).

(ii) $(-8) \times (-7) = 56$(-8) × (-7) = 56 (debt × debt = fortune).

(iii) $0 - (-14) = 0 + 14 = 14$0 - (-14) = 0 + 14 = 14.

(iv) $(-20) \times 4 = -80$(-20) × 4 = -80 (debt × fortune = debt).

Suppose you have a fortune of ₹10, but you also owe someone a debt of ₹5 (this debt is recorded as $-5$-5 against you). If that debt of ₹5 is cancelled/forgiven (i.e. it is removed, or subtracted, from your situation), your net worth increases by ₹5, because you no longer have to pay it back.

So starting from ₹10 and removing (subtracting) a debt of ₹5 is mathematically $10 - (-5)$10 - (-5), and the real-world effect is that your wealth goes up by ₹5, i.e. $10 + 5 = 15$10 + 5 = 15.

Hence $10 - (-5) = 15$10 - (-5) = 15: removing/subtracting a debt (a negative quantity) has exactly the same effect as adding a fortune (a positive quantity) of the same size. This is why subtracting a negative number is equivalent to adding the corresponding positive number.

A rational number is defined as $\frac{p}{q}$p/q, which represents $p$p divided by $q$q. Division by zero is not defined in mathematics — there is no number that, when multiplied by 0, gives back $p$p (unless $p=0$p=0 too, in which case $\frac{0}{0}$0/0 could be anything, so it is still undefined as a single value).

If $q$q were allowed to be $0$0, the expression $\frac{p}{q}$p/q would not represent any actual number at all, so it could not be used consistently in arithmetic (adding, comparing, placing on the number line, etc.). To make sure every symbol $\frac{p}{q}$p/q genuinely stands for one specific number, we must require $q \neq 0$q ≠ 0.

To add or subtract two rational numbers with different denominators, we convert each into an equivalent fraction with a common denominator — typically the LCM of the two denominators — by multiplying the numerator and denominator of each fraction by the same non-zero number (which doesn't change the fraction's value, since it is equivalent to multiplying by 1). Once both fractions share the same denominator $b$b, we use the rule $\frac{a}{b} + \frac{c}{b} = \frac{a+c}{b}$a/b + c/b = (a+c)/b (and similarly for subtraction) to combine them.

The distributive law states $p(q+r) = pq + pr$p(q+r) = pq + pr. Let us verify it with $p = \frac{1}{2}$p = 1/2, $q = \frac{1}{3}$q = 1/3, $r = \frac{1}{4}$r = 1/4.

LHS: $p(q+r) = \frac{1}{2}\left(\frac{1}{3}+\frac{1}{4}\right) = \frac{1}{2} \times \frac{4+3}{12} = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24}$p(q+r) = 1/2(1/3+1/4) = 1/2 × (4+3)/12 = 1/2 × 7/12 = 7/24.

RHS: $pq + pr = \frac{1}{2}\times\frac{1}{3} + \frac{1}{2}\times\frac{1}{4} = \frac{1}{6} + \frac{1}{8} = \frac{4}{24} + \frac{3}{24} = \frac{7}{24}$pq + pr = 1/2× 1/3 + 1/2× 1/4 = 1/6 + 1/8 = 4/24 + 3/24 = 7/24.

Since LHS $=$= RHS $= \frac{7}{24}$= 7/24, the distributive law $p(q+r) = pq+pr$p(q+r) = pq+pr is verified for these rational numbers (and this reasoning, using the general rules of fraction arithmetic, works for any rational $p,q,r$p,q,r).

Two rational numbers $\frac{a}{b}$a/b and $\frac{c}{d}$c/d are equal if and only if $ad = bc$ad = bc (cross-multiplication).

(i) $\frac{24}{36}$24/36 and $\frac{2}{3}$2/3: check $24 \times 3 = 72$24 × 3 = 72 and $36 \times 2 = 72$36 × 2 = 72. Equal, so $\frac{24}{36} = \frac{2}{3}$24/36 = 2/3 (also seen directly by dividing numerator and denominator of $\frac{24}{36}$24/36 by their common factor 12).

(ii) $\frac{5}{4}$5/4 and $\frac{10}{8}$10/8: check $5 \times 8 = 40$5 × 8 = 40 and $4 \times 10 = 40$4 × 10 = 40. Equal, so $\frac{5}{4} = \frac{10}{8}$5/4 = 10/8.

(iii) $-\frac{3}{5}$-3/5 and $-\frac{6}{10}$-6/10: check $(-3)\times 10 = -30$(-3)× 10 = -30 and $5 \times (-6) = -30$5 × (-6) = -30. Equal, so $-\frac{3}{5} = -\frac{6}{10}$-3/5 = -6/10.

(iv) $3 = \frac{3}{1}$3 = 3/1 and $\frac{9}{3}$9/3: check $3 \times 3 = 9$3 × 3 = 9 and $1 \times 9 = 9$1 × 9 = 9. Equal, so $3 = \frac{9}{3}$3 = 9/3.

(i) LCM of 5,10 is 10: $\frac{2}{5}+\frac{3}{10} = \frac{4}{10}+\frac{3}{10} = \frac{7}{10}$2/5+3/10 = 4/10+3/10 = 7/10.

(ii) LCM of 12,8 is 24: $\frac{7}{12}+\frac{5}{8} = \frac{14}{24}+\frac{15}{24} = \frac{29}{24}$7/12+5/8 = 14/24+15/24 = 29/24.

(iii) LCM of 7,14 is 14: $-\frac{4}{7}+\frac{3}{14} = -\frac{8}{14}+\frac{3}{14} = -\frac{5}{14}$-4/7+3/14 = -8/14+3/14 = -5/14.

(i) LCM of 6,4 is 12: $\frac{5}{6}-\frac{1}{4} = \frac{10}{12}-\frac{3}{12} = \frac{7}{12}$5/6-1/4 = 10/12-3/12 = 7/12.

(ii) LCM of 8,4 is 8: $\frac{11}{8}-\frac{3}{4} = \frac{11}{8}-\frac{6}{8} = \frac{5}{8}$11/8-3/4 = 11/8-6/8 = 5/8.

(iii) $-\frac{7}{9}-\left(-\frac{2}{3}\right) = -\frac{7}{9}+\frac{2}{3} = -\frac{7}{9}+\frac{6}{9} = -\frac{1}{9}$-7/9-(-2/3) = -7/9+2/3 = -7/9+6/9 = -1/9.

(i) $\frac{2}{3}\times\frac{3}{10} = \frac{6}{30} = \frac{1}{5}$2/3× 3/10 = 6/30 = 1/5.

(ii) $\frac{7}{11}\times\frac{5}{8} = \frac{35}{88}$7/11× 5/8 = 35/88 (already in lowest terms).

(iii) $-\frac{4}{7}\times\frac{5}{14} = -\frac{20}{98} = -\frac{10}{49}$-4/7× 5/14 = -20/98 = -10/49.

Dividing by a fraction means multiplying by its reciprocal.

(i) $\frac{2}{3}\div\frac{3}{10} = \frac{2}{3}\times\frac{10}{3} = \frac{20}{9}$2/3÷ 3/10 = 2/3× 10/3 = 20/9.

(ii) $\frac{7}{11}\div\frac{5}{8} = \frac{7}{11}\times\frac{8}{5} = \frac{56}{55}$7/11÷ 5/8 = 7/11× 8/5 = 56/55.

(iii) $-\frac{4}{7}\div\frac{5}{14} = -\frac{4}{7}\times\frac{14}{5} = -\frac{56}{35} = -\frac{8}{5}$-4/7÷ 5/14 = -4/7× 14/5 = -56/35 = -8/5.

LHS: $\left(\frac{1}{2}+\frac{3}{4}\right)\times\frac{8}{3}$(1/2+3/4)× 8/3.

First, $\frac{1}{2}+\frac{3}{4} = \frac{2}{4}+\frac{3}{4} = \frac{5}{4}$1/2+3/4 = 2/4+3/4 = 5/4.

So LHS $= \frac{5}{4}\times\frac{8}{3} = \frac{40}{12} = \frac{10}{3}$= 5/4× 8/3 = 40/12 = 10/3.

RHS: $\frac{1}{2}\times\frac{8}{3} + \frac{3}{4}\times\frac{8}{3} = \frac{8}{6} + \frac{24}{12} = \frac{4}{3} + 2 = \frac{4}{3}+\frac{6}{3} = \frac{10}{3}$1/2× 8/3 + 3/4× 8/3 = 8/6 + 24/12 = 4/3 + 2 = 4/3+6/3 = 10/3.

Since LHS $=$= RHS $= \frac{10}{3}$= 10/3, the statement is shown — this is simply the distributive law $p(q+r)=pq+pr$p(q+r)=pq+pr applied with $p=\frac{8}{3}$p=8/3, $q=\frac{1}{2}$q=1/2, $r=\frac{3}{4}$r=3/4.

Using the distributive property $p(q-r) = pq - pr$p(q-r) = pq - pr with $p=\frac{7}{9}$p=7/9, $q=\frac{6}{7}$q=6/7, $r=\frac{3}{4}$r=3/4:

$\frac{7}{9}\times\frac{6}{7} - \frac{7}{9}\times\frac{3}{4} = \frac{42}{63} - \frac{21}{36}$7/9× 6/7 - 7/9× 3/4 = 42/63 - 21/36.

Simplify each: $\frac{42}{63} = \frac{2}{3}$42/63 = 2/3 and $\frac{21}{36} = \frac{7}{12}$21/36 = 7/12.

$\frac{2}{3} - \frac{7}{12} = \frac{8}{12} - \frac{7}{12} = \frac{1}{12}$2/3 - 7/12 = 8/12 - 7/12 = 1/12.

(Check directly: $\frac{6}{7}-\frac{3}{4} = \frac{24}{28}-\frac{21}{28} = \frac{3}{28}$6/7-3/4 = 24/28-21/28 = 3/28, and $\frac{7}{9}\times\frac{3}{28} = \frac{21}{252} = \frac{1}{12}$7/9× 3/28 = 21/252 = 1/12 — matches.)

So the simplified value is $\frac{1}{12}$1/12.

Expand the LHS using the distributive property:

$\frac{5}{6}\left(x+\frac{3}{5}\right) = \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}$5/6(x+3/5) = 5/6x + 5/6× 3/5 = 5/6x + 15/30 = 5/6x + 1/2.

This is exactly the RHS, $\frac{5}{6}x + \frac{1}{2}$5/6x + 1/2, for every value of $x$x.

So the equation is not a normal equation with one solution — it is an identity: it holds true for every rational number $x$x, simply because it is just the distributive law written out. There is no unique $x$x to find; rather, this exercise shows that distributing $\frac{5}{6}$5/6 over $\left(x+\frac{3}{5}\right)$(x+3/5) automatically produces the right-hand side for any $x$x.

$\frac{8}{5}$8/5: Write it as a mixed number: $\frac{8}{5} = 1\frac{3}{5} = 1.6$8/5 = 13/5 = 1.6. So it lies between the integers 1 and 2. To locate it precisely, divide the unit interval from 1 to 2 into 5 equal parts, and move 3 parts to the right of 1 (since $\frac{8}{5} = 1 + \frac{3}{5}$8/5 = 1 + 3/5).

$-\frac{7}{4}$-7/4: Write it as a mixed number: $-\frac{7}{4} = -1\frac{3}{4} = -1.75$-7/4 = -13/4 = -1.75. So it lies between the integers $-2$-2 and $-1$-1. To locate it, divide the unit interval from $-2$-2 to $-1$-1 into 4 equal parts, and move 3 parts to the right of $-2$-2 (equivalently, 1 part to the left of $-1$-1), since $-\frac{7}{4} = -2 + \frac{1}{4}$-7/4 = -2 + 1/4.

In general, to plot $\frac{p}{q}$p/q, divide the unit interval containing it into $q$q equal parts and count $p$p parts from 0 (or from the appropriate integer).

Convert each to a mixed number/decimal to locate it:

$\frac{3}{2} = 1\frac{1}{2} = 1.5$3/2 = 11/2 = 1.5: lies exactly halfway between 1 and 2.

$-\frac{5}{4} = -1\frac{1}{4} = -1.25$-5/4 = -11/4 = -1.25: lies between $-2$-2 and $-1$-1, one quarter of the way from $-1$-1 towards $-2$-2 (i.e. divide $[-2,-1]$[-2,-1] into 4 parts and mark the point 1 part to the left of $-1$-1).

$\frac{11}{2} = 5\frac{1}{2} = 5.5$11/2 = 51/2 = 5.5: lies exactly halfway between 5 and 6.

On one number line marked with integers from $-2$-2 to $6$6, place a point at $-1.25$-1.25 (between $-2$-2 and $-1$-1), a point at $1.5$1.5 (between 1 and 2), and a point at $5.5$5.5 (between 5 and 6).

Convert to a common denominator. $-\frac{1}{2} = -\frac{4}{8}$-1/2 = -4/8 and $\frac{1}{4} = \frac{2}{8}$1/4 = 2/8.

We need rational numbers strictly between $-\frac{4}{8}$-4/8 and $\frac{2}{8}$2/8. Three such numbers (using eighths, choosing integers strictly between $-4$-4 and $2$2) are:

$-\frac{3}{8}$-3/8, $-\frac{1}{8}$-1/8, and $0$0 (i.e. $\frac{0}{8}$0/8).

Check: $-\frac{1}{2} = -0.5 < -0.375 = -\frac{3}{8} < -0.125 = -\frac{1}{8} < 0 < 0.25 = \frac{1}{4}$-1/2 = -0.5 < -0.375 = -3/8 < -0.125 = -1/8 < 0 < 0.25 = 1/4. All three lie strictly between $-\frac{1}{2}$-1/2 and $\frac{1}{4}$1/4.

LCM of 4 and 12 is 12.

$-\frac{1}{4} = -\frac{3}{12}$-1/4 = -3/12.

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

Total silk $= 15\frac{3}{4} = \frac{63}{4}$= 153/4 = 63/4 metres.

Silk per kurta $= 1\frac{3}{4} = \frac{7}{4}$= 13/4 = 7/4 metres.

Number of kurtas $= \dfrac{63/4}{7/4} = \frac{63}{4}\times\frac{4}{7} = \frac{63}{7} = 9$= 63/4/7/4 = 63/4× 4/7 = 63/7 = 9.

He can make 9 kurtas.

We need numbers strictly between $3.1415$3.1415 and $3.1416$3.1416. Since these differ only in the fourth decimal place, we can insert extra decimal digits to find numbers in between, for example:

$3.14151$3.14151, $3.14153$3.14153, and $3.14158$3.14158.

Check: $3.1415 < 3.14151 < 3.14153 < 3.14158 < 3.1416$3.1415 < 3.14151 < 3.14153 < 3.14158 < 3.1416. All three are valid rational numbers strictly between the given values (in fact, infinitely many such numbers exist, by the density property of rational numbers).

Besides taking the average $\frac{a+b}{2}$(a+b)/2, other methods include:

1. Common denominator method: Rewrite both numbers as fractions with the same (large enough) denominator, then pick any integer numerator strictly between the two given numerators.

2. Decimal insertion: Write both numbers as decimals and insert an extra digit at a place where they first differ (as done in Q5 above).

3. Weighted averages: For any positive integer $n$n, the number $a + \frac{b-a}{n}$a + (b-a)/n (for $a<b$a<b) lies strictly between $a$a and $b$b; varying $n$n gives infinitely many different rational numbers in between.

No. $\sqrt{2}$√2 cannot be written as $\frac{p}{q}$p/q for any integers $p,q$p,q (with $q \neq 0$q ≠ 0) — that is, $\sqrt{2}$√2 is irrational.

No matter which fraction $\frac{p}{q}$p/q we try, squaring it never gives exactly 2 (we can get closer and closer, e.g. $1.41421356\ldots^2 \approx 2$1.41421356…² ≈ 2, but never exactly 2, for any fraction). This is proved rigorously in the next section using proof by contradiction (Section 3.5.1).

Proof that $\sqrt{3}$√3 is irrational:

Step 1: Assume, for contradiction, that $\sqrt{3}$√3 is rational, so $\sqrt{3} = \frac{p}{q}$√3 = p/q in lowest terms ($p,q$p,q coprime integers, $q \neq 0$q ≠ 0).

Step 2: Square both sides: $3 = \frac{p^2}{q^2}$3 = p²/q², so $p^2 = 3q^2$p² = 3q².

Step 3: Since $p^2$p² is 3 times an integer, $p^2$p² is divisible by 3. Since 3 is prime, $p$p itself must be divisible by 3. Write $p = 3k$p = 3k.

Step 4: Substitute: $(3k)^2 = 3q^2 \Rightarrow 9k^2 = 3q^2 \Rightarrow q^2 = 3k^2$(3k)² = 3q² ⇒ 9k² = 3q² ⇒ q² = 3k².

Step 5: So $q^2$q² is divisible by 3, and hence $q$q is divisible by 3.

Step 6: But now both $p$p and $q$q are divisible by 3, contradicting the assumption that $p,q$p,q share no common factor other than 1.

Therefore $\sqrt{3}$√3 cannot be written as $\frac{p}{q}$p/q; it is irrational.

Does the same approach work for $\sqrt{5}$√5, $\sqrt{7}$√7, $\sqrt{10}$√10?

Yes for $\sqrt{5}$√5 and $\sqrt{7}$√7: since 5 and 7 are also prime, the identical argument (if a prime divides $p^2$p², it divides $p$p) applies directly, giving contradictions the same way.

For $\sqrt{10}$√10: 10 is not prime ($10 = 2\times5$10 = 2×5), but the method still works with a small adjustment. If $10 = \frac{p^2}{q^2}$10 = p²/q², then $p^2 = 10q^2$p² = 10q² is even, so $p$p is even; write $p=2k$p=2k. Then $4k^2 = 10q^2 \Rightarrow 2k^2 = 5q^2$4k² = 10q² ⇒ 2k² = 5q², so $q^2$q² is even and hence $q$q is even — again both $p,q$p,q share the common factor 2, a contradiction. So $\sqrt{10}$√10 is also irrational, and in general this contradiction method works for $\sqrt{n}$√n whenever $n$n is not a perfect square.

We use the Baudhayana–Pythagoras theorem on a right triangle with rational (often unit) legs — the hypotenuse of such a triangle is frequently irrational.

For example, a right triangle with both legs of length 1 unit has a hypotenuse of length $\sqrt{1^2+1^2} = \sqrt{2}$√(1²+1²) = √2, which is irrational. Once we have physically constructed a segment of this length (using a set square/perpendicular and a ruler), we can transfer that exact length onto the number line using a compass: place the compass point at the origin $O$O, open it to the length of the constructed segment ($\sqrt{2}$√2), and draw an arc that cuts the number line at a point $P$P. That point $P$P then represents the irrational number $\sqrt{2}$√2 on the number line (this is exactly the construction shown in Fig. 3.11).

Constructing $\sqrt{3}$√3: Starting from the segment $OB$OB of length $\sqrt{2}$√2 (constructed as in Fig. 3.11), draw a new perpendicular at $B$B of length 1 unit, meeting at a point $C$C. By Pythagoras, $OC = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2+1} = \sqrt{3}$OC = (√2)² + 1² = √(2+1) = √3. Then, with centre $O$O and radius $OC$OC, draw an arc to cut the number line at the point representing $\sqrt{3}$√3.

Constructing $\sqrt{5}$√5: One way is to continue the same spiral one more step from $OC=\sqrt{3}$OC=√3: erect a unit perpendicular at $C$C, reaching a point $D$D with $OD = \sqrt{3+1} = \sqrt{4} = 2$OD = √(3+1) = √4 = 2; then a further unit perpendicular gives a hypotenuse of $\sqrt{4+1}=\sqrt{5}$√(4+1)=√5. (Alternatively, directly construct a right triangle with legs 2 and 1: the hypotenuse is $\sqrt{2^2+1^2} = \sqrt{5}$√(2²+1²) = √5.) Transfer this length to the number line with a compass centred at $O$O.

General method (the "square root spiral"): Start at $O$O with $OA = 1$OA = 1 unit along the number line. At $A$A, erect a perpendicular of length 1 to reach $B_2$B₂, so $OB_2 = \sqrt{2}$OB₂ = √2. At $B_2$B₂, erect a new perpendicular of length 1 (perpendicular to $OB_2$OB₂) to reach $B_3$B₃, so $OB_3 = \sqrt{3}$OB₃ = √3. Continuing this process, after $(n-1)$(n-1) such steps, the segment from $O$O to the final point $B_n$B_n has length $\sqrt{n}$√n (since each step satisfies $OB_k^2 = OB_{k-1}^2 + 1^2$OB_k² = OB_k-1² + 1², so $OB_k^2 = k$OB_k² = k). Finally, draw an arc of radius $OB_n$OB_n centred at $O$O to mark the point representing $\sqrt{n}$√n on the number line.

A fraction in lowest terms has a terminating decimal exactly when its denominator's only prime factors are 2 and/or 5.

$\frac{7}{20}$7/20: $20 = 2^2 \times 5$20 = 2² × 5 — only 2's and 5's, so terminating.

$\frac{4}{15}$4/15: $15 = 3 \times 5$15 = 3 × 5 — has a factor of 3, so repeating (non-terminating).

$\frac{13}{250}$13/250: $250 = 2 \times 5^3$250 = 2 × 5³ — only 2's and 5's, so terminating.

Long division check:

$\frac{7}{20} = 0.35$7/20 = 0.35 (terminates).

$\frac{4}{15} = 0.2\overline{6} = 0.2666\ldots$4/15 = 0.26 = 0.2666… (repeating).

$\frac{13}{250} = 0.052$13/250 = 0.052 (terminates).

All three predictions match the actual long division results.

$\frac{1}{3} = 0.333\ldots = 0.\overline{3}$1/3 = 0.333… = 0.3. The repeating block is just the single digit "3" (period 1).

$\frac{1}{13} = 0.076923076923\ldots = 0.\overline{076923}$1/13 = 0.076923076923… = 0.076923. The repeating block is "076923", which has 6 digits (period 6).

Observation: Even though both are unit fractions with a repeating decimal, the length of the repeating block is very different — 1 digit for $\frac{1}{3}$1/3 versus 6 digits for $\frac{1}{13}$1/13. In general, the length of the repeating block of $\frac{1}{n}$1/n (for $n$n coprime to 10) depends on how many digits it takes for powers of 10 to first repeat a remainder when divided by $n$n — different denominators can give very different (and sometimes quite long) repeating-block lengths.

A decimal terminates exactly when the fraction can be rewritten with a denominator that is a power of 10, since $10^k = 2^k \times 5^k$10^k = 2^k × 5^k, and a fraction $\frac{m}{10^k}$m/10^k is simply the decimal formed by placing the digits of $m$m with the decimal point $k$k places from the right.

If the only prime factors of $q$q (in lowest terms) are 2 and/or 5, we can always multiply the numerator and denominator by whatever extra factors of 2 or 5 are needed so that the denominator becomes exactly $2^k \times 5^k = 10^k$2^k × 5^k = 10^k for some $k$k — this gives an exact terminating decimal (as shown with $\frac{3}{20} = \frac{3\times5}{20\times5} = \frac{15}{100} = 0.15$3/20 = 3×5/20×5 = 15/100 = 0.15).

However, if $q$q has any other prime factor (such as 3, 7, 11, ...), no amount of multiplying by extra factors of 2 or 5 can ever turn that prime factor into a power of 10 — since 10's only prime factors are 2 and 5. So the denominator can never be made into a pure power of 10, meaning the division never terminates exactly; instead, the remainders in long division must eventually repeat (since there are only finitely many possible remainders), producing a repeating decimal.

A fraction in lowest terms has a terminating decimal exactly when its denominator's only prime factors are 2 and/or 5.

$\frac{7}{20}$7/20: $20 = 2^2 \times 5$20 = 2² × 5 — terminating.

$\frac{4}{15}$4/15: $15 = 3 \times 5$15 = 3 × 5 — repeating (has a factor of 3).

$\frac{13}{250}$13/250: $250 = 2 \times 5^3$250 = 2 × 5³ — terminating.

Long division check: $\frac{7}{20} = 0.35$7/20 = 0.35; $\frac{4}{15} = 0.2\overline{6}$4/15 = 0.26; $\frac{13}{250} = 0.052$13/250 = 0.052. All predictions confirmed.

$\frac{1}{3} = 0.\overline{3}$1/3 = 0.3 (repeating block "3", period 1).

$\frac{1}{13} = 0.\overline{076923}$1/13 = 0.076923 (repeating block "076923", period 6).

Observation: the repeating-block length varies a lot between denominators — $\frac{1}{13}$1/13's block is six times longer than $\frac{1}{3}$1/3's, showing that the period of a repeating decimal depends heavily on the specific denominator.

(i) $\sqrt{81} = 9$√81 = 9, which is a whole number, so it is rational: $9 = \frac{9}{1}$9 = 9/1.

(ii) $\sqrt{12} = 2\sqrt{3}$√12 = 2√3. Since 12 is not a perfect square and $\sqrt{3}$√3 is irrational, $\sqrt{12}$√12 is irrational.

(iii) $0.33333\ldots = 0.\overline{3}$0.33333… = 0.3 is a repeating decimal, so it is rational: it equals $\frac{1}{3}$1/3.

(iv) $0.123451234512345\ldots$0.123451234512345… has the repeating block "12345", so it is rational: $0.\overline{12345} = \frac{12345}{99999} = \frac{4115}{33333}$0.12345 = 12345/99999 = 4115/33333 (dividing numerator and denominator by their common factor 3).

(v) $1.01001000100001\ldots$1.01001000100001…: the number of zeros between successive 1's keeps increasing (1 zero, then 2 zeros, then 3 zeros, ...), so there is no fixed repeating block — it neither terminates nor settles into a repeating cycle. This is irrational.

(vi) $23.560185612239874790120$23.560185612239874790120 has a finite number of digits shown with no indicated repetition, i.e. it terminates, so it is rational: it can be written as $\frac{23560185612239874790120}{10^{21}}$2356018561223987479012010²¹ (a fraction with a power of 10 as denominator, which can be reduced by cancelling trailing zeros/common factors of 2 and 5).

Let $x = 0.\overline{9} = 0.999\ldots$x = 0.9 = 0.999…

Multiply both sides by 10: $10x = 9.999\ldots = 9.\overline{9}$10x = 9.999… = 9.9.

Subtract the first equation from the second: $10x - x = 9.\overline{9} - 0.\overline{9}$10x - x = 9.9 - 0.9.

The infinite "999..." tails after the decimal point are identical and cancel exactly, leaving: $9x = 9$9x = 9.

So $x = 1$x = 1.

Therefore $0.\overline{9} = 1$0.9 = 1 exactly — not "almost 1" or "approximately 1", but precisely equal to 1.

Cyclic-number reciprocals occur for certain prime denominators known as full reptend primes — primes $p$p for which 10 is a "primitive root" modulo $p$p, so that $\frac{1}{p}$1/p has a repeating block of length $p-1$p-1 whose digit-cyclings under multiplication by $1,2,\ldots,p-1$1,2,…,p-1 just permute the same digits.

Besides $n=7$n=7 (giving $142857$142857), other examples include:

$n = 17$n = 17: $\frac{1}{17} = 0.\overline{0588235294117647}$1/17 = 0.0588235294117647 (period 16, cyclic).

$n = 19$n = 19: $\frac{1}{19} = 0.\overline{052631578947368421}$1/19 = 0.052631578947368421 (period 18, cyclic).

$n = 23$n = 23, $29$29, $47$47, $59$59, $61$61, $97$97 are further examples of denominators whose reciprocals are cyclic numbers.

There is genuinely no real number whose square is $-1$-1: for any real number $x$x, if $x \geq 0$x ≥ 0 then $x^2 \geq 0$x² ≥ 0, and if $x < 0$x < 0 (so $x$x is negative) then $x^2 > 0$x² > 0 as well (a negative times a negative is positive, by Brahmagupta's rule). So in every case $x^2 \geq 0$x² ≥ 0, meaning $x^2$x² can never equal $-1$-1.

This shows $\sqrt{-1}$√-1 has no place anywhere on the real number line — it is not rational, and it is not irrational either; it simply falls outside the real numbers altogether.

To make sense of such square roots, mathematicians introduced an entirely new kind of number, called an imaginary number, denoted $i$i, defined so that $i^2 = -1$i² = -1. This steps completely off the real number line into a new dimension of numbers (complex numbers), which turn out to be essential in fields like electrical engineering and quantum mechanics, though their full study is left for later grades.

(i) $\frac{3}{50}$3/50: Long division of 3 by 50 gives $3.000 \div 50 = 0.06$3.000 ÷ 50 = 0.06. So $\frac{3}{50} = 0.06$3/50 = 0.06 (terminating), consistent with $50 = 2\times5^2$50 = 2×5² having only 2 and 5 as prime factors.

(ii) $\frac{2}{9}$2/9: Long division of 2 by 9 gives repeated remainder 2, producing $0.2222\ldots = 0.\overline{2}$0.2222… = 0.2 (repeating), consistent with $9=3^2$9=3² having a prime factor other than 2 or 5.

Proof by contradiction:

Step 1: Suppose $\sqrt{5}$√5 is rational. Then $\sqrt{5} = \frac{p}{q}$√5 = p/q in lowest terms, where $p,q$p,q are coprime integers, $q \neq 0$q ≠ 0.

Step 2: Squaring, $5 = \frac{p^2}{q^2}$5 = p²/q², so $p^2 = 5q^2$p² = 5q².

Step 3: This means $p^2$p² is divisible by 5. Since 5 is prime, $p$p must be divisible by 5. Write $p = 5k$p = 5k.

Step 4: Substitute: $(5k)^2 = 5q^2 \Rightarrow 25k^2 = 5q^2 \Rightarrow q^2 = 5k^2$(5k)² = 5q² ⇒ 25k² = 5q² ⇒ q² = 5k².

Step 5: So $q^2$q² is divisible by 5, and hence $q$q is divisible by 5.

Step 6: Now both $p$p and $q$q are divisible by 5, contradicting our assumption that they are coprime (share no common factor other than 1).

This contradiction shows our original assumption was false. Therefore $\sqrt{5}$√5 cannot be expressed as $\frac{p}{q}$p/q — it is irrational.

Each terminating decimal is converted by writing it over the matching power of 10, then reducing to lowest terms.

(i) $12.6 = \frac{126}{10} = \frac{63}{5}$12.6 = 126/10 = 63/5.

(ii) $0.0120 = \frac{120}{10000} = \frac{3}{250}$0.0120 = 120/10000 = 3/250.

(iii) $3.052 = \frac{3052}{1000} = \frac{763}{250}$3.052 = 3052/1000 = 763/250.

(iv) $1.235 = \frac{1235}{1000} = \frac{247}{200}$1.235 = 1235/1000 = 247/200.

(v) $0.23 = \frac{23}{100}$0.23 = 23/100 (already lowest terms, since 23 is prime).

(vi) $2.05 = \frac{205}{100} = \frac{41}{20}$2.05 = 205/100 = 41/20.

(vii) $2.125 = \frac{2125}{1000} = \frac{17}{8}$2.125 = 2125/1000 = 17/8.

(viii) $3.125 = \frac{3125}{1000} = \frac{25}{8}$3.125 = 3125/1000 = 25/8.

(ix) $2.1625 = \frac{21625}{10000} = \frac{173}{80}$2.1625 = 21625/10000 = 173/80.

(i) $0.532 = \frac{532}{1000} = \frac{133}{250}$0.532 = 532/1000 = 133/250, which lies between 0 and 1 (closer to 0.5). To plot it, divide the unit interval $[0,1]$[0,1] into 1000 equal parts and move 532 of those parts to the right of 0 (in practice, mark it just past the halfway point between 0.5 and 0.6).

(ii) $1.15 = \frac{115}{100} = \frac{23}{20}$1.15 = 115/100 = 23/20, which lies between 1 and 2, specifically just after 1 (between 1.1 and 1.2). To plot it, divide the interval $[1,2]$[1,2] into 100 equal parts and move 15 parts to the right of 1 (or, dividing $[1,2]$[1,2] into 20 equal parts, move 3 parts to the right of 1, since $1.15 = 1 + \frac{3}{20}$1.15 = 1 + 3/20).

Six rational numbers strictly between 3 and 4 are:

$3.1, 3.2, 3.3, 3.4, 3.5, 3.6$3.1, 3.2, 3.3, 3.4, 3.5, 3.6

(equivalently, $\frac{31}{10}, \frac{32}{10}, \frac{33}{10}, \frac{34}{10}, \frac{35}{10}, \frac{36}{10}$31/10, 32/10, 33/10, 34/10, 35/10, 36/10).

$\frac{2}{5} = \frac{12}{30}$2/5 = 12/30 and $\frac{3}{5} = \frac{18}{30}$3/5 = 18/30 (using denominator 30).

Five rational numbers strictly between $\frac{12}{30}$12/30 and $\frac{18}{30}$18/30 are:

$\frac{13}{30}, \frac{14}{30} = \frac{7}{15}, \frac{15}{30} = \frac{1}{2}, \frac{16}{30} = \frac{8}{15}, \frac{17}{30}$13/30, 14/30 = 7/15, 15/30 = 1/2, 16/30 = 8/15, 17/30.

LCM of 6 and 5 is 30. $\frac{1}{6} = \frac{5}{30}$1/6 = 5/30 and $\frac{2}{5} = \frac{12}{30}$2/5 = 12/30.

Five rational numbers strictly between $\frac{5}{30}$5/30 and $\frac{12}{30}$12/30 are:

$\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}$6/30 = 1/5, 7/30, 8/30 = 4/15, 9/30 = 3/10, 10/30 = 1/3.

Common denominator of 3 and 5 is 15:

$\frac{5x}{15} + \frac{3x}{15} = \frac{16}{15}$5x/15 + 3x/15 = 16/15