NCERT Solutions Class 9 Maths Ch 3: The World of Numbers

He will leave with 90 copper ingots.

1. Ingots for one bag2 bags give 15 ingots, so 1 bag gives 152 ingots.
2. Ingots for 12 bags12 × 152 = 6 × 15 = 90

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

Each of 11, 13, 17 and 19 has exactly two factors: 1 and itself, so each is a prime number. The next three primes after 19 are 23, 29, 31.

No. Natural numbers are not closed under subtraction.

Closure means the result must also be a natural number. Subtraction can fail this test:

Because 3 − 5 = −2 is not a natural number, the natural numbers are not closed under subtraction.

You can count up to 12 on one hand. This is why base-12 counting arose.

1. Count the jointsThe thumb does the counting, so the 4 fingers are used.4 × 3 = 12
2. Link with base 12One hand reaches 12, so a counting system built on groups of 12 (base 12, the duodecimal system) came naturally. The number 12 is still used today in a dozen and in 12 months.

The midnight temperature is −11 °C.

A fall of 15 °C from 4 °C takes the temperature 11 degrees below zero.

The equation is −850 + 1200 − 450 = −100. He is in debt by ₹100.

1. Write each event as an integerDebt = −850, profit = +1200, loss = −450.
2. Add them−850 + 1200 − 450 = 350 − 450 = −100

His final standing is −₹100: he still owes ₹100.

(i) −60 (ii) 56 (iii) 14 (iv) −5

Brahmagupta treated a debt as negative and a fortune as positive.

Taking away a debt makes you richer, so 10 − (−5) = 10 + 5 = 15.

1. Meaning of −5A negative number is a debt. −5 means you owe ₹5.
2. Meaning of 10 − (−5)You have ₹10 and someone removes a debt of ₹5 that you owed.
3. ResultRemoving a debt increases what you have by ₹5.10 − (−5) = 10 + 5 = 15

So subtracting a negative number is the same as adding the matching positive number.

23 = 46.

1. Simplify46 = 4 ÷ 26 ÷ 2 = 23
2. Cross-multiply check2 × 6 = 12   and   3 × 4 = 12The cross products are equal.

Hence 23 and 46 are equal.

54 = 108.

1. Simplify108 = 10 ÷ 28 ÷ 2 = 54
2. Cross-multiply check5 × 8 = 40   and   4 × 10 = 40

Hence 54 and 108 are equal.

−35 = −610.

1. Simplify−610 = −6 ÷ 210 ÷ 2 = −35
2. Cross-multiply check(−3) × 10 = −30   and   5 × (−6) = −30

Hence −35 and −610 are equal.

93 = 3.

1. Simplify93 = 9 ÷ 3 = 3
2. Write 3 as a fraction and cross-multiply3 = 31:   9 × 1 = 9   and   3 × 3 = 9

Hence 93 and 3 are equal.

The sum of 25 + 310 is 710.

1. Make the denominators equalLCM of 5 and 10 = 10.25 = 410
2. Add410 + 310 = 710

The sum of 712 + 58 is 2924.

1. Make the denominators equalLCM of 12 and 8 = 24.712 = 1424   58 = 1524
2. Add1424 + 1524 = 2924

The sum of −47 + 314 is −514.

1. Make the denominators equalLCM of 7 and 14 = 14.−47 = −814
2. Add−814 + 314 = −514

The difference of 56 − 14 is 712.

1. Make the denominators equalLCM of 6 and 4 = 12.56 = 1012   14 = 312
2. Subtract1012 − 312 = 712

The difference of 118 − 34 is 58.

1. Make the denominators equalLCM of 8 and 4 = 8.34 = 68
2. Subtract118 − 68 = 58

The difference of −79 − (−23) is −19.

1. Make the denominators equalLCM of 9 and 3 = 9.−23 = −69
2. Subtract−79 − (−69) = −19

Subtracting a negative number is the same as adding the positive number.

The product of 23 × 310 is 15.

1. Multiply the numerators and the denominators23 × 310 = 630 = 15

The product of 711 × 58 is 3588.

1. Multiply the numerators and the denominators711 × 58 = 3588

The product of −47 × 514 is −1049.

1. Multiply the numerators and the denominators−47 × 514 = −2098 = −1049

The quotient of 23 ÷ 310 is 209.

1. Multiply by the reciprocal of the second fraction23 ÷ 310 = 23 × 103 = 209

The quotient of 711 ÷ 58 is 5655.

1. Multiply by the reciprocal of the second fraction711 ÷ 58 = 711 × 85 = 5655

The quotient of −47 ÷ 514 is −85.

1. Multiply by the reciprocal of the second fraction−47 ÷ 514 = −47 × 145 = −5635 = −85

Both sides equal 103, so the statement is true.

1. Left side(12 + 34) × 83 = (24 + 34) × 83 = 54 × 83 = 4012 = 103
2. Right side12 × 83 + 34 × 83 = 86 + 2412 = 43 + 63 = 103

LHS = RHS, hence proved.

The value is 112.

1. Distribute79 × 67 − 79 × 34 = 69 − 2136 = 23 − 712
2. Common denominator 12812 − 712 = 112

The equation is true for every rational number x, so x can be any rational number.

1. Expand the left side56x + 56 × 35 = 56x + 1530 = 56x + 12
2. Compare with the right sideBoth sides are exactly 56x + 12. The x terms cancel and we are left with 12 = 12, which is always true.

In order: −5/4 (= −1.25) < 2/3 (≈ 0.67) < 3/2 (= 1.5).

One valid set is −1/4, −1/8 and 1/8.

1. Write both numbers as decimals−12 = −0.5 and 14 = 0.25.
2. Pick numbers strictly between them−0.25, −0.125 and 0.125 lie between −0.5 and 0.25.−14 , −18 , 18

Answers may differ: there are infinitely many correct choices, such as 0 or 18.

The value is 16.

1. LCM of 4 and 12 = 12−14 = −312
2. Add−312 + 512 = 212 = 16

He can make exactly 7 kurtas.

1. Write the mixed numbers as improper fractions1534 = 634   and   214 = 94
2. Divide the silk by the silk needed for one kurta634 ÷ 94 = 634 × 49 = 639 = 7

For example: 3.14151, 3.14152 and 3.14153.

Write the two numbers with one more decimal place: 3.14150 and 3.14160. Any number from 3.14151 to 3.14159 lies between them.

Each of these is a terminating decimal, so it is a rational number: 3.14151 = 314151100000, 3.14152 = 314152100000, 3.14153 = 314153100000.

Answers may differ: there are infinitely many correct numbers.

Yes. Three simple methods are the average, equivalent fractions and extra decimal places.

1. 1. Take the average (midpoint)For two rational numbers a and b, the number a + b2 is rational and lies strictly between them.
2. 2. Use equivalent fractionsWrite both numbers with a common denominator, multiply top and bottom by a large number, and pick any numerator between the two new numerators.
3. 3. Use more decimal placesWrite both numbers as decimals and insert a number with more decimal places between them.

Repeating any of these methods gives as many rational numbers as you like.

720 and 13250 terminate (0.35 and 0.052). 415 repeats (0.26).

Rule: a fraction in lowest form terminates exactly when its denominator has only the prime factors 2 and 5.

Check for 4/15: 40 ÷ 15 = 2 remainder 10; 100 ÷ 15 = 6 remainder 10; the remainder 10 comes back, so the digit 6 repeats for ever.

1/13 = 0.076923 (block of 6 digits). 2/13 has a different block, 153846. The 12 fractions split into two cycles of 6 digits each.

1. Long division of 1/131 ÷ 13 gives 0.076923 076923 … The remainders repeat after 6 steps, so the repeating block is 076923.
2. 2/132 ÷ 13 = 0.153846. The digits 1, 5, 3, 8, 4, 6 are not a rotation of 0, 7, 6, 9, 2, 3.
3. All 12 fractions

   Fraction	Decimal	Group
   113	0.076923	A
   213	0.153846	B
   313	0.230769	A
   413	0.307692	A
   513	0.384615	B
   613	0.461538	B
   713	0.538461	B
   813	0.615384	B
   913	0.692307	A
   1013	0.769230	A
   1113	0.846153	B
   1213	0.923076	A

4. What we noticeGroup A (1/13, 3/13, 4/13, 9/13, 10/13, 12/13) uses the digits 0 7 6 9 2 3 in the same cyclic order, only starting at a different place. Group B (2/13, 5/13, 6/13, 7/13, 8/13, 11/13) does the same with 1 5 3 8 4 6. So 1/13 shows cyclic behaviour, but in two separate cycles.

(i) rational, 9. (ii) irrational. (iii) rational, 1/3. (iv) rational, 12345/99999. (v) irrational. (vi) rational, 23560185612239874790120/10²¹.

1. (i) √81√81 = 9 = 91, so it is rational.
2. (ii) √1212 is not a perfect square, so √12 is irrational.
3. (iii) 0.33333…Let x = 0.3. Then 10x = 3.3.10x − x = 3 ⇒ 9x = 3 ⇒ x = 39 = 13Rational.
4. (iv) 0.123451234512345…The block 12345 repeats, so it is rational. Let x = 0.12345.100000x − x = 12345 ⇒ 99999x = 12345 ⇒ x = 1234599999 = 411533333
5. (v) 1.01001000100001…The number of zeros keeps growing, so no block repeats and the decimal never ends. It is irrational.
6. (vi) 23.560185612239874790120It has 21 digits after the decimal point and ends, so it is a terminating decimal and rational.23.560185612239874790120 = 2356018561223987479012010²¹

The PDF’s final summary list gave 12345/90900 for part (iv) and a wrong numerator for part (vi). The working above is the correct one.

0.999… = 1 exactly.

1. Let x = 0.999…x = 0.99999…
2. Multiply by 1010x = 9.99999…
3. Subtract the first from the second10x − x = 9.99999… − 0.99999… ⇒ 9x = 9 ⇒ x = 1

So 0.9 = 1. There is no gap between 0.999… and 1, and decimal expansions are not always unique (1 = 1.000… = 0.999…).

1/17, 1/19, 1/23 and 1/29 also have cyclic blocks (like 1/7). They are primes whose block has n − 1 digits.

Idea: for a prime n, if 1/n repeats after exactly n − 1 digits, then every multiple k/n (k = 1 to n − 1) is a rotation of that same block.

Not every prime works. For example 1/13 repeats after only 6 digits (not 12), so its multiples split into two cycles, as seen in Q35. Other primes below 100 that give cyclic blocks are 47, 59, 61 and 97.

(i) 3/50 = 0.06 (terminating). (ii) 2/9 = 0.222… = 0.2 (non-terminating, repeating).

1. (i) 3 ÷ 5030 ÷ 50 = 0 (first decimal place 0), 300 ÷ 50 = 6 with remainder 0.350 = 0.06The remainder becomes 0, so the decimal terminates.
2. (ii) 2 ÷ 920 ÷ 9 = 2 remainder 2. The remainder 2 comes back again and again, so the digit 2 repeats for ever.29 = 0.2222… = 0.2The decimal is non-terminating and repeating.

√5 is irrational. The proof is by contradiction.

1. Assume the oppositeSuppose √5 is rational. Then √5 = pq, where p and q are integers with no common factor (co-prime) and q ≠ 0.
2. Square both sides5 = p²q² ⇒ 5q² = p²
3. p is a multiple of 55 divides p², so 5 divides p. Let p = 5k.
4. Substitute5q² = (5k)² = 25k² ⇒ q² = 5k²
5. q is a multiple of 5 too5 divides q², so 5 divides q.
6. ContradictionBoth p and q are divisible by 5, but they were co-prime. This is impossible, so our assumption was wrong.

Hence √5 is irrational.

12.6 = 635.

0.0120 = 3250.

Divide the top and bottom by 40.

3.052 = 1511495.

1. Write the numberx = 3.05252…   (1 non-repeating digit and 2 repeating digits)
2. Move the decimal point past the non-repeating digits10x = 30.5252…
3. Move it past one full repeating block1000x = 3052.5252…
4. Subtract1000x − 10x = 3052.5252… − 30.5252… ⇒ 990x = 3022
5. Solve for xx = 3022990 = 1511495

Check: 1511495 = 3.0525252525… when divided out.

1.235 = 1223990.

1. Write the numberx = 1.23535…   (1 non-repeating digit and 2 repeating digits)
2. Move the decimal point past the non-repeating digits10x = 12.3535…
3. Move it past one full repeating block1000x = 1235.3535…
4. Subtract1000x − 10x = 1235.3535… − 12.3535… ⇒ 990x = 1223
5. Solve for xx = 1223990

Check: 1223990 = 1.2353535353… when divided out.

0.23 = 2399.

1. Write the numberx = 0.2323…   (2 repeating digits)
2. Move the decimal point past one full repeating block100x = 23.2323…
3. Subtract100x − x = 23.2323… − 0.2323… ⇒ 99x = 23
4. Solve for xx = 2399

Check: 2399 = 0.2323232323… when divided out.

2.05 = 3718.

1. Write the numberx = 2.055…   (1 non-repeating digit and 1 repeating digit)
2. Move the decimal point past the non-repeating digits10x = 20.5555…
3. Move it past one full repeating block100x = 205.5555…
4. Subtract100x − 10x = 205.5555… − 20.5555… ⇒ 90x = 185
5. Solve for xx = 18590 = 3718

Check: 3718 = 2.0555555555… when divided out.

2.125 = 1913900.

1. Write the numberx = 2.1255…   (2 non-repeating digits and 1 repeating digit)
2. Move the decimal point past the non-repeating digits100x = 212.5555…
3. Move it past one full repeating block1000x = 2125.5555…
4. Subtract1000x − 100x = 2125.5555… − 212.5555… ⇒ 900x = 1913
5. Solve for xx = 1913900

Check: 1913900 = 2.1255555555… when divided out.

3.125 = 2813900.

1. Write the numberx = 3.1255…   (2 non-repeating digits and 1 repeating digit)
2. Move the decimal point past the non-repeating digits100x = 312.5555…
3. Move it past one full repeating block1000x = 3125.5555…
4. Subtract1000x − 100x = 3125.5555… − 312.5555… ⇒ 900x = 2813
5. Solve for xx = 2813900

Check: 2813900 = 3.1255555555… when divided out.

2.1625 = 216239999.

1. Write the numberx = 2.16251625…   (4 repeating digits)
2. Move the decimal point past one full repeating block10000x = 21625.1625…
3. Subtract10000x − x = 21625.1625… − 2.1625… ⇒ 9999x = 21623
4. Solve for xx = 216239999

Check: 216239999 = 2.1625162516… when divided out.

0.532 lies between 0.53 and 0.54, a little to the right of 0.53.

1. Step 1: whole number part0.532 lies between 0 and 1.
2. Step 2: first decimal placeThe first digit after the decimal point is 5, so it lies between 0.5 and 0.6. Divide that gap into 10 equal parts.
3. Step 3: second decimal placeThe next digit is 3, so it lies between 0.53 and 0.54. Divide that gap into 10 equal parts.
4. Step 4: third decimal placeThe last digit is 2, so count 2 small parts to the right of 0.53. That point is 0.532.

1.15 = 1.15555… = 5245 lies between 1.155 and 1.156.

1. Step 11.15555… lies between 1 and 2.
2. Step 2The first decimal digit is 1, so it lies between 1.1 and 1.2.
3. Step 3The second digit is 5, so it lies between 1.15 and 1.16.
4. Step 4The third digit is 5, so it lies between 1.155 and 1.156. The digit 5 repeats, so it keeps getting closer to 1.1556 but never reaches it. It sits a little more than halfway from 1.155 to 1.156.

Zooming in again and again never ends for a repeating decimal, so we place the point as accurately as the drawing allows.

For example: 3.1, 3.2, 3.3, 3.4, 3.5, 3.6.

Write 3 = 3010 and 4 = 4010. Six fractions between them are:

Answers may differ: there are infinitely many.

For example: 21/50, 22/50, 23/50, 24/50 and 25/50.

1. Make the denominators larger25 = 2050   and   35 = 3050
2. Pick numerators between 20 and 302150, 2250, 2350, 2450, 2550

For example: 6/30, 7/30, 8/30, 9/30 and 10/30.

1. Common denominator (LCM of 6 and 5 = 30)16 = 530   and   25 = 1230
2. Pick numerators between 5 and 12630, 730, 830, 930, 1030 = 15, 730, 415, 310, 13

x = 2.

1. LCM of 3 and 5 = 155x + 3x15 = 1615
2. Simplify8x15 = 1615
3. Multiply both sides by 158x = 16 ⇒ x = 2

Check: 23 + 25 = 1015 + 615 = 1615.

ab is negative (in fact ab = −1).

1. Rearrangea + 1b = 0 gives a = −1b.
2. Multiply both sides by b (b ≠ 0)ab = −1

Since −1 is negative, ab is negative. So a and b have opposite signs.

The number is p/10⁴ with p not divisible by 10. Yes, it is necessary: the denominator in lowest form is always divisible by 2⁴ or by 5⁴.

1. Writing the number as p/10⁴The decimal has 4 decimal places, so multiplying it by 10⁴ gives an integer p. Then the number = p10⁴. The last non-zero digit is in the 4th place, so p does not end in 0, that is, p is not divisible by 10.
2. The denominator in lowest formIn lowest form the denominator is 10⁴ ÷ gcd(p, 10⁴), and 10⁴ = 2⁴ × 5⁴.
3. Why 2⁴ or 5⁴ must remainSince p is not divisible by 10, p cannot be a multiple of both 2 and 5. If p is odd, no factor 2 cancels, so 2⁴ stays in the denominator. If p is not a multiple of 5, no factor 5 cancels, so 5⁴ stays. At least one of these two cases always holds.

The PDF answered “not necessary”, using 0.1250 = 1/8 as an example. That example is wrong: 0.1250 has its last non-zero digit in the 3rd decimal place, so it does not fit the question. The correct answer is “yes”.

It terminates, with 3 decimal places.

1. Factorise the denominator125 = 5³gcd(18, 125) = 1, so the fraction is already in lowest form. The denominator has only the prime factor 5, so the decimal terminates.
2. Count the decimal placesMultiply the top and bottom by 2³ = 8 to get a power of 10:18125 = 18 × 8125 × 8 = 1441000 = 0.144The number of decimal places is 3.

It will have 3 decimal places.

1. Make the denominator a power of 102³ × 5 needs 5² more to become 2³ × 5³ = 10³.
2. Multiply top and bottom by 5²p2³ × 5 = p × 252³ × 5 × 25 = 25p10³
3. CountThe denominator is 10³, so there are 3 decimal places. In general the number of decimal places is the larger of the two powers, here max(3, 1) = 3.

Example: 140 = 0.025 has 3 decimal places.

a = 21/36 and b = 30/36 (so k₂ − k₁ = 9 > 6). Five numbers between them: 22/36, 23/36, 24/36, 25/36, 26/36.

1. Same denominatorLCM of 12 and 6 = 12. Then a = 712 and b = 1012, so k₂ − k₁ = 10 − 7 = 3, which is not greater than 6.
2. Make the gap biggerMultiply the top and bottom of both by 3:a = 2136   and   b = 3036Now k₁ = 21, k₂ = 30, m = 36 and k₂ − k₁ = 9 > 6.
3. Five numbers with the same denominatorThe integers 22, 23, 24, 25, 26 lie between 21 and 30:2236, 2336, 2436, 2536, 2636
4. Why k₂ − k₁ > n + 1?The integers strictly between k₁ and k₂ number k₂ − k₁ − 1. To get n numerators we need k₂ − k₁ − 1 ≥ n, that is k₂ − k₁ ≥ n + 1. Making the gap more than n + 1 guarantees that there are enough integers. Here n = 5, so we need a gap of at least 6; the gap 9 is enough.

x = y = z = 0.

1. Use an identity(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
2. Substitute the given values0 = x² + y² + z² + 2(0) ⇒ x² + y² + z² = 0
3. A sum of squares is zeroEach square is at least 0, so the sum is 0 only if every square is 0. Hence x² = 0, y² = 0, z² = 0.

So x = 0, y = 0 and z = 0 at the same time.

If a < b, then a < (a + b) ÷ 2 < b.

1. Assume a < bAdd a to both sides: 2a < a + b. Add b to both sides of a < b: a + b < 2b.
2. Combine2a < a + b < 2b
3. Divide by 2a < a + b2 < b

Hence (a + b)2 lies between a and b. It is rational because sums and quotients of rational numbers are rational.

The ten hypotenuses are √2, √3, 2, √5, √6, √7, 2√2, 3, √10 and √11.

1. How the spiral growsThe first triangle has both short sides equal to 1. In every next triangle one short side is 1 and the other is the previous hypotenuse.
2. The ruleIf the short sides are √k and 1, the hypotenuse is √(k + 1) (since (√k)² + 1² = k + 1)
3. All ten triangles

   Triangle	Short sides	Hypotenuse
   1	1 and 1	√2
   2	√2 and 1	√3
   3	√3 and 1	√4 = 2
   4	2 and 1	√5
   5	√5 and 1	√6
   6	√6 and 1	√7
   7	√7 and 1	√8 = 2√2
   8	√8 and 1	√9 = 3
   9	3 and 1	√10
   10	√10 and 1	√11

The figure has 10 triangles. The first radial side of length 1 is a short side of triangle 1, not a hypotenuse, so “1” is not in the list (the PDF listed it by mistake and missed √11).

Because division by zero has no meaning.

1. The idea of pqThe fraction pq asks: “what number times q gives p?”
2. If q = 0 and p ≠ 0No number multiplied by 0 gives p (every product with 0 is 0), so 60 has no value.
3. If q = 0 and p = 0Every number times 0 gives 0, so 00 could be any number. It is not a single, definite number.

So we write q ≠ 0 to make sure every rational number is one definite point on the number line.

Find the LCM of the denominators and rewrite each fraction with that denominator.

1. Find the LCMThe least common multiple of the two denominators.
2. Multiply top and bottomMultiply the numerator and denominator of each fraction by the number that turns its denominator into the LCM.

Example: 23 + 34. The LCM of 3 and 4 is 12, so 23 = 812 and 34 = 912, and the sum is 1712.

a(b + c) = ab + ac holds for all rational numbers a, b, c.

1. A numerical checkTake a = 12, b = 13, c = 14.a(b + c) = 12(412 + 312) = 12 × 712 = 724ab + ac = 16 + 18 = 424 + 324 = 724Both sides are 724.
2. In generalWrite a = pq, b = mn, c = kn (common denominator n).a(b + c) = pq × m + kn = p(m + k)qn = pm + pkqn = pmqn + pkqn = ab + acThis uses the distributive law for integers: p(m + k) = pm + pk.

8/5 = 1.6 lies between 1 and 2. −7/4 = −1.75 lies between −1 and −2.

1. 8/5Divide each unit into 5 equal parts and count 8 parts to the right of 0. This lands 3 parts after 1, at 1.6.
2. −7/4Divide each unit into 4 equal parts and count 7 parts to the left of 0. This lands 3 parts after −1 on the left side, at −1.75.

No. √2 cannot be written as p/q, so it is irrational.

1. Assume the oppositeSuppose √2 = pq with p, q co-prime integers, q ≠ 0.
2. Square2 = p²q² ⇒ p² = 2q²
3. p is evenp² is even, so p is even. Let p = 2k.
4. Substitute(2k)² = 2q² ⇒ 4k² = 2q² ⇒ q² = 2k²
5. q is even tooq² is even, so q is even.
6. ContradictionBoth p and q are even, but they were co-prime. So the assumption is false.

Hence √2 is irrational.

√3 is irrational. The same approach works for √5, √7 and √10.

1. Assume the oppositeSuppose √3 = pq with p, q co-prime. Squaring: 3q² = p².
2. p is a multiple of 33 divides p², so 3 divides p. Let p = 3k.
3. Substitute3q² = 9k² ⇒ q² = 3k²So 3 divides q² and hence q.
4. ContradictionBoth p and q are multiples of 3, but they were co-prime. So √3 is irrational.

For √5 and √7: 5 and 7 are prime, so the argument is identical with 3 replaced by 5 or 7.

For √10: from p² = 10q², both 2 and 5 divide p², so both divide p, so 10 divides p. Put p = 10k: 100k² = 10q² ⇒ q² = 10k², so 10 divides q as well. This contradicts co-primeness.

The method works whenever n is not a perfect square.

Build each new length as the hypotenuse of a right triangle whose short sides are the previous length and 1.

1. √2 firstA right triangle with both short sides 1 has hypotenuse √2.
2. √3At the end of the segment of length √2 draw a perpendicular of length 1. The hypotenuse is √(2 + 1) = √3.
3. √5Start with a segment of length 2 (which is √4). At its end draw a perpendicular of length 1. The hypotenuse is √(4 + 1) = √5.
4. Any √nOnce you have √(n − 1), draw a perpendicular of length 1 at its end. The hypotenuse is √((n − 1) + 1) = √n. Repeating this builds the whole square root spiral.

10/3 = 3.3 and 11/12 = 0.916. In both, a block of digits repeats for ever.

The remainders in the long division repeat, so the digits repeat. In 10 ÷ 3 the remainder 1 returns at once; in 11 ÷ 12 the remainder 8 returns after two digits (12 = 2² × 3 has the factor 3, so it cannot terminate).

A decimal that ends is a fraction with denominator 10ᵏ, and 10ᵏ = 2ᵏ × 5ᵏ has only the prime factors 2 and 5.

1. If the decimal terminatesSuppose it has k decimal places. Then the number = an integer10ᵏ. Since 10ᵏ = 2ᵏ × 5ᵏ, after reducing to lowest form the denominator can contain only the primes 2 and 5.
2. If q = 2ⁿ × 5ᵐMultiply the top and bottom by a power of 2 or 5 to make the powers equal (to 2ᵏ × 5ᵏ with k = the larger of n and m). The denominator becomes 10ᵏ, so the number is an integer divided by 10ᵏ, which is a terminating decimal.
3. If q has another prime factorNo multiplication by 2s and 5s can turn the denominator into a power of 10, so the long division never ends. The decimal is non-terminating and repeating.