The Lumenary · Grades 6–8

Rational numbers

Name: ______________________

Date: ______________________

You will learn to

Add, subtract, multiply and divide rational numbers, including negative ones
Explain why every non-zero rational number has a reciprocal, while most integers do not
Prove that between any two different rational numbers there is another
Convert between fractions and decimals, and prove that the decimal expansion of a fraction terminates or eventually repeats
Explain why $0.999\ldots = 1$ and why no fraction squares to $2$

1.
Compute $-\tfrac34 + \tfrac56$ .
- A. $\tfrac1{12}$
- B. $\tfrac{2}{10}$
- C. $-\tfrac1{12}$
- D. $-\tfrac{19}{12}$
2.
A water seller in Kano has $7$ litres of water and fills bottles that hold $\tfrac12$ litre each. How many bottles can she fill? (Compute $7 \div \tfrac12$ .)
3.
Compute $\left(-\tfrac56\right) \div \left(-\tfrac{10}{9}\right)$ .
- A. $\tfrac34$
- B. $-\tfrac34$
- C. $\tfrac{25}{27}$
- D. $\tfrac43$
4.
Which of these numbers have a reciprocal in $\Q$ ? (Choose all that apply.)
- A. $-\tfrac25$
- B. $0$
- C. $7$
- D. $0.25$
5.
Which of these fractions have terminating decimal expansions? (Choose all that apply.)
- A. $\tfrac38$
- B. $\tfrac7{12}$
- C. $\tfrac{9}{40}$
- D. $\tfrac{6}{15}$
- E. $\tfrac13$
6.
Which fraction equals $0.\overline{36} = 0.363636\ldots$ ?
- A. $\tfrac4{11}$
- B. $\tfrac{9}{25}$
- C. $\tfrac25$
- D. $\tfrac13$
7.
How many rational numbers lie strictly between $\tfrac13$ and $\tfrac12$ ?
- A.None — they are next to each other
- B.Exactly one
- C.Finitely many, but a lot
- D.Infinitely many
8.
In the proof that $\sqrt2$ is irrational we have $a^2 = 2b^2$ and conclude that $a$ is even. Which reason is correct?
- A. $a^2$ is even, and the square of an odd number is odd, so $a$ cannot be odd
- B.Because $2$ is even
- C.Because $b$ is even
- D.Because $\tfrac ab$ is in simplest form
9.
Which Lean statement says correctly that between any two different rational numbers there is another?
- A.∀ a b : Rat, a < b → ∃ c, a < c ∧ c < b
- B.∃ c : Rat, ∀ a b : Rat, a < c ∧ c < b
- C.∀ a b : Rat, ∃ c, a < c ∧ c < b
- D.∀ a b : Rat, a < b → a < a + b ∧ a + b < b
10.
Write period_length(a, b) for whole numbers $a \ge 0$ and $b > 0$ . Do long division: start with the remainder a % b, and at each step replace the remainder r by (10 * r) % b. Remember the step at which each remainder first appears. Return the length of the repeating block of $\tfrac ab$ , or 0 if the decimal terminates (that is, if the remainder 0 appears).
Computational — Practice

Answer key — Rational numbers

1.
Over the common denominator $12$ : $-\tfrac34 + \tfrac56 = -\tfrac{9}{12} + \tfrac{10}{12} = \tfrac{1}{12}$ .
2.
$7 \div \tfrac12 = 7 \cdot 2 = 14$ . Each litre fills two half-litre bottles, so $7$ litres fill $14$ . Dividing by a number between $0$ and $1$ makes a positive number bigger.
3.
$\left(-\tfrac56\right) \div \left(-\tfrac{10}{9}\right) = \left(-\tfrac56\right) \cdot \left(-\tfrac{9}{10}\right) = \tfrac{45}{60} = \tfrac34$ .
4.
Every non-zero rational number $\tfrac ab$ has the reciprocal $\tfrac ba$ . So $-\tfrac25$ , $7$ and $0.25$ have reciprocals ( $-\tfrac52$ , $\tfrac17$ and $4$ ); $0$ does not, because $0 \cdot y = 0 \ne 1$ .
5.
In simplest form the denominators are $8 = 2^3$ , $12 = 2^2 \cdot 3$ , $40 = 2^3 \cdot 5$ , $5$ (since $\tfrac6{15} = \tfrac25$ ) and $3$ . By Theorem 5, $\tfrac38$ , $\tfrac9{40}$ and $\tfrac6{15}$ terminate; $\tfrac7{12}$ and $\tfrac13$ repeat.
6.
Let $x = 0.\overline{36}$ . Then $100x = 36 + x$ , so $99x = 36$ and $x = \tfrac{36}{99} = \tfrac{4}{11}$ . Check: $4 \div 11 = 0.3636\ldots$
7.
The average $\tfrac{5}{12}$ lies between them. Averaging $\tfrac13$ with $\tfrac5{12}$ gives $\tfrac38$ , then $\tfrac{17}{48}$ , and so on: each step produces a new rational number closer to $\tfrac13$ . So there are infinitely many.
8.
From $a^2 = 2b^2$ , $a^2$ is even. If $a = 2k+1$ were odd, then $a^2 = 4k^2 + 4k + 1$ would be odd. So $a$ is even. (In Lean this step is the lemma even_of_sq_even.)
9.
Only ∀ a b : Rat, a < b → ∃ c, a < c ∧ c < b is right. Swapping ∀ and ∃ changes the meaning completely, dropping a < b makes the statement false, and the sum $a + b$ is not between $a$ and $b$ in general.

10.

def period_length(a, b):
    r = a % b
    seen = {}
    step = 0
    while r not in seen:
        seen[r] = step
        r = (10 * r) % b
        step += 1
    if r == 0:
        return 0
    return step - seen[r]

The loop must stop within $b + 1$ steps, because there are only $b$ possible remainders — the pigeonhole argument of Theorem 4. If the repeated remainder is $0$ , every later digit is $0$ and the decimal terminates.