The World of Numbers Worksheet

(c) −7/4

1. A rational number is p/q for integers p and q with q ≠ 0.
2. −7/4 has integer numerator and non-zero integer denominator.

(c) 0.272727…

1. The block 27 repeats indefinitely.
2. If x = 0.272727…, then 100x − x = 27, so x = 27/99 = 3/11.

(c) Every irrational number has a non-terminating, non-repeating decimal

1. Rational decimals terminate or eventually repeat.
2. Irrational real numbers have decimal expansions that neither terminate nor eventually repeat.

(b) Rational

1. Write a/b + c/d = (ad + bc)/(bd).
2. The numerator and denominator are integers and bd ≠ 0.

(c) 13/40

1. A reduced denominator with only factors 2 and 5 gives a terminating decimal.
2. 40 = 2³ × 5; 13/40 = 0.325.

(a) √2 and −√2

1. √2 and −√2 are both irrational.
2. Their sum is 0, which is rational.

(b) −5/8

1. Write the endpoints as −6/8 and −4/8.
2. The fraction −5/8 lies strictly between them.

−3/4 is three quarter-steps left of 0; 5/4 is five quarter-steps right. −3/4 is closer.

1. Divide each unit interval into four equal parts.
2. Count three parts to the left for −3/4 and five to the right for 5/4.
3. Distances from zero are 3/4 and 5/4; compare their magnitudes.

For example 6/15, 7/15, 8/15, 9/15; no smallest rational above 1/3 exists.

1. Write the endpoints as 5/15 and 10/15.
2. Choose the four fractions with numerators 6, 7, 8 and 9.
3. For any rational r > 1/3, the rational (r + 1/3)/2 lies strictly between them.

0.125 = 1/8; 0.1666… = 1/6.

1. 125/1000 reduces to 1/8.
2. For x = 0.1666…, 10x = 1.666… and 100x = 16.666…; subtract to obtain 90x = 15.
3. In long division record each remainder. A zero remainder terminates; a repeated remainder repeats the subsequent digits.

√2, √3 and √4 = 2 units.

1. Begin with a unit segment from O. At its end construct a perpendicular unit segment; join its end to O to get √2.
2. At the new endpoint construct a unit segment perpendicular to the latest radius; the next radius is √3.
3. Repeat once more. By Pythagoras the new squared radius is 3 + 1 = 4.

Construct a right triangle with legs 2 and 1; transfer its hypotenuse to the positive number line.

1. Mark O = 0 and A = 2. At A draw AB perpendicular to OA with AB = 1.
2. Pythagoras gives OB² = 2² + 1² = 5, so OB = √5.
3. With centre O and radius OB draw an arc meeting the positive number line. Its intersection represents √5.

2/7 = 0.285714285714… .

1. Divide 20 by 7: digit 2, remainder 6. Continue with ten times each remainder.
2. The subsequent digits/remainders are 8/4, 5/5, 7/1, 1/3, 4/2.
3. The original remainder 2 returns after six digits; the block 285714 repeats.

The sum of a rational number and an irrational number is irrational.

1. Let r be rational and s irrational.
2. If r + s were rational, then (r + s) − r would be rational because rationals are closed under subtraction.
3. But this difference equals s, a contradiction.

1/12 = 0.083333…; 08 is the non-repeating part and 3 repeats.

1. Divide 10 by 12: the first decimal digit is 0 and the remainder is 10.
2. Divide 100 by 12: digit 8, remainder 4.
3. Every subsequent step divides 40 by 12: digit 3, remainder 4 again. Therefore 3 repeats.

√2 is irrational.

1. Assume √2 = p/q, where p and q are coprime integers and q ≠ 0.
2. Then p² = 2q², so p² is even. An odd integer has odd square, hence p is even.
3. Put p = 2k. Then q² = 2k², so q is also even.
4. Both p and q are divisible by 2, contradicting lowest terms. The assumption is false.

(a + b)/2 is rational and lies strictly between a and b; repeat midpoint selection.

1. Sums of rationals and division by 2 preserve rationality.
2. a < b implies 2a < a + b < 2b, so a < (a + b)/2 < b.
3. Set r₁ = (a + b)/2 and rₙ₊₁ = (a + rₙ)/2.
4. Each rₙ₊₁ lies strictly between a and rₙ, so the sequence produces infinitely many distinct intermediate rationals.

√3 is irrational.

1. Squares of integers with remainders 0, 1, 2 on division by 3 have remainders 0, 1, 1 respectively.
2. Assume √3 = p/q in lowest terms. Then p² = 3q²; the remainder argument gives p = 3k.
3. Substitution gives q² = 3k², so 3 also divides q.
4. This contradicts coprimality. Therefore √3 is irrational.

2 + √3 is irrational, using the irrationality of √3.

1. Suppose 2 + √3 is rational.
2. Subtract the rational number 2. Closure of rationals under subtraction would make √3 rational.
3. This contradicts the proved irrationality of √3.
4. Therefore the supposition is false. The worksheet includes a separate proof of the required √3 result.