The World of Numbers

12 bags contain 12/2 = 6 groups of two.

Each group earns 15 ingots, so 6×15 = 90.

Each listed number has exactly two positive divisors.

The next primes above 19 are 23, 29 and 31.

3 − 5 = −2 is not natural, and 5 − 5 = 0 is not in the book’s natural-number set {1, 2, …}.

The thumb is the counter, leaving four fingers with three joints each.

4×3 = 12 distinct positions can represent a dozen before starting another group.

Subtract the drop from the noon reading: 4 − 15 = −11.

Represent the loan as −850, profit as +1200 and loss as −450.

−850 + 1200 − 450 = −100.

Opposite signs give a negative product; two negatives give a positive product.

Subtracting −14 adds 14; dividing −20 by +4 gives a negative quotient.

Suppose net wealth is ₹10, including a ₹5 debt. Cancelling that debt removes −₹5 from the account.

Cross-products agree: (i) 2×6 = 3×4 = 12; (ii) 5×8 = 4×10 = 40.

(iii) (−3)×10 = 5×(−6) = −30; (iv) 9×1 = 3×3 = 9.

Use denominators 10, 24 and 14 respectively.

(i) 4/10 + 3/10; (ii) 14/24 + 15/24; (iii) −8/14 + 3/14.

Use denominators 12, 8 and 9 respectively.

(i) 10/12 − 3/12; (ii) 11/8 − 6/8; (iii) −7/9 + 6/9.

Multiply numerators and denominators, then reduce.

(i) 6/30; (ii) 35/88; (iii) −20/98.

Multiply the first fraction by the reciprocal of the second.

(i) (2/3)(10/3); (ii) (7/11)(8/5); (iii) (−4/7)(14/5).

Left side: (1/2 + 3/4)(8/3) = (5/4)(8/3) = 10/3.

Right side: (1/2)(8/3) + (3/4)(8/3) = 4/3 + 2 = 10/3.

(7/9)(6/7 − 3/4) = 2/3 − 7/12.

8/12 − 7/12 = 1/12.

Expand the left side: (5/6)x + (5/6)(3/5) = (5/6)x + 1/2.

This is exactly the right side, independent of x.

2/3 is two thirds to the right of 0; −5/4 is one quarter left of −1; 11/2 is midway between 5 and 6.

−1/4, 0 and 1/8 are each greater than −1/2 and less than 1/4.

−1/4 + 5/12 = −3/12 + 5/12.

15¾ = 63/4 and 2¼ = 9/4.

(63/4)/(9/4) = 7.

Write both bounds with five decimal places: 3.14150 and 3.14160.

Pick values strictly between them.

For a < b, use a + t(b − a) for any rational t with 0 < t < 1.

Choosing t = 1/3 gives (2a + b)/3; t = 2/3 gives (a + 2b)/3.

20 = 2²×5 and 250 = 2×5³: their reduced fractions terminate.

15 contains factor 3, not cancelled by 4: the decimal repeats.

1/13 = 0.(076923), where parentheses denote the repeating block.

For numerators 2 through 12 the blocks are 153846, 230769, 307692, 384615, 461538, 538461, 615384, 692307, 769230, 846153, 923076.

The rotations form two groups: rotations of 076923 and rotations of 153846. Unlike sevenths, the first block does not rotate for every multiplier.

(i) √81 = 9 = 9/1. (ii) √12 = 2√3 is irrational.

(iii) 0.333… = 1/3. (iv) repeating 12345 gives 12345/99999 = 4115/33333.

(v) The gaps between 1s grow, so no fixed block eventually repeats: irrational.

(vi) The written decimal ends, so it is rational: 589004640305996869753/25000000000000000000.

Let x = 0.999…; then 10x = 9.999….

Subtract: 9x = 9, hence x = 1.

For 1/17, long division produces 0.(0588235294117647). For 1/19 it gives 0.(052631578947368421).

Keeping leading zeros, multiplying the block by 1 through p − 1 permutes it by cyclic rotation for these denominators.

3/50 = 6/100 = 0.06.

For 2/9, long division returns remainder 2 after every digit 2.

Suppose √5 = p/q in lowest terms, q ≠ 0. Squaring gives p² = 5q².

Since 5 is prime and divides p², it divides p. Write p = 5k.

Then q² = 5k², so 5 also divides q, contradicting the fraction being reduced.

Parentheses mark the PDF’s repeating bars: (iii) 3.0(52), (iv) 1.2(35), (v) 0.(23), (vi) 2.0(5), (vii) 2.12(5), (viii) 3.1(25), (ix) 2.(1625).

For m nonrepeating and k repeating digits, subtract 10ᵐx from 10ᵐ⁺ᵏx to cancel the repeating tail.

Example (iii): (1000 − 10)x = 3052 − 30, so x = 3022/990 = 1511/495.

For terminating decimals, write the value over a power of 10 and reduce.

0.532 = 532/1000: subdivide 0–1 into thousandths and count 532 parts.

1.15 = 1 + 15/100: subdivide 1–2 into hundredths and count 15 parts.

Use tenths strictly between 3.0 and 4.0.

Rewrite the bounds as 20/50 and 30/50.

The bounds are 5/30 and 12/30. Choose five interior integer numerators.

x/3 + x/5 = 16/15 becomes 5x + 3x = 16 after multiplying by 15.

a/b + 1 = 0 gives a/b = −1 and a = −b.

Thus ab = −b² < 0 because b is nonzero.

Multiplying by 10⁴ gives an integer p. Since the fourth digit is nonzero, p is not divisible by 10.

The denominator 10⁴ = 2⁴×5⁴. Reduction cannot cancel both a factor 2 and a factor 5 from p, since that would make p divisible by 10.

Thus the reduced denominator retains all four factors of at least one prime: 2⁴ or 5⁴, but not necessarily both.

For example 0.0625 = 1/16 and 0.0016 = 1/625.

125 = 5³; multiply numerator and denominator by 2³.

18/125 = 144/1000.

2³×5 = 40. To make a power of 10, multiply denominator by 5² to obtain 1000.

Because the fraction is reduced, the factor 2³ cannot cancel, so fewer than three places cannot suffice.

Use m = 36: a = 21/36 and b = 30/36. The numerator gap is 9 > 6.

Five choices are 22/36 through 26/36.

There are k₂ − k₁ − 1 interior integer numerators. To select n of them, the exact condition is k₂ − k₁ ≥ n + 1.

The book’s strict condition k₂ − k₁ > n + 1 is sufficient but stronger than necessary; equality still leaves n interior numerators.

Square x + y + z = 0: x² + y² + z² + 2(xy + yz + zx) = 0.

The second condition leaves x² + y² + z² = 0. Each square is nonnegative.

Assume a < b. Then (a + b)/2 − a = (b − a)/2 > 0 and b − (a + b)/2 = (b − a)/2 > 0.

The average is rational because rational numbers are closed under addition and division by nonzero 2.

If a = b, the average equals them rather than lying strictly between distinct values.

Start with perpendicular unit legs: first hypotenuse √2.

Each new triangle uses the previous hypotenuse and a perpendicular unit leg. If the old squared length is n, the new one is n + 1.

The nine triangles in the figure therefore produce square roots of 2 through 10.

By distributivity, (−3)(4 + (−4)) = 0.

Since (−3)4 = −12, the other product must be +12 to make the sum zero.

If p/0 meant a number x, then 0x = p. For p ≠ 0 this is impossible; for p = 0 every x would satisfy it, so it defines no unique quotient.

Choose a common multiple, usually the least common multiple, and scale each numerator by the same factor as its denominator.

For p = a/b, q = c/d and r = e/f, p(q + r) = a(cf + ed)/(bdf) = ac/(bd) + ae/(bf).

8/5 = 1.6, so mark three fifths past 1.

−7/4 = −1.75, so mark three quarters left of −1.

For a < b, subtract a from the average and the average from b: both differences equal (b − a)/2 > 0.

Repeating the operation gives another interior rational each time.

If √2 = p/q in lowest terms, p² = 2q² implies p is even.

Writing p = 2k gives q² = 2k², so q is even too, contradicting lowest terms.

For √3, p² = 3q² forces both p and q divisible by 3; for √5 or √7 use the corresponding prime.

For √10, p² = 10q² forces p even. Substituting p = 2k gives 2k² = 5q², forcing q even too.

Construct a right triangle with unit legs to obtain √2. Transfer its hypotenuse to the number line using a compass.

To obtain √3, use perpendicular legs √2 and 1; for √5 use legs 2 and 1.

In general, a right triangle with legs √(n − 1) and 1 has hypotenuse √n.

10/3 = 3.333… has repeating digit 3 immediately after the decimal.

11/12 = 0.91666… has nonrepeating digits 91 followed by repeating 6.

A terminating decimal can be written over 10ᵏ = 2ᵏ5ᵏ, so its reduced denominator has no other primes.

Conversely, for q = 2ᵃ5ᵇ, multiply by missing powers to make q a power of 10.

The square of a real number is nonnegative.

In the complex number system, i is defined by i² = −1; the two solutions of z² = −1 are i and −i.