The Lumenary · Grades 9–12

Real numbers and irrationality

Name: ______________________

Date: ______________________

You will learn to

Prove that $\sqrt 2$ is irrational, and that $\sqrt n$ is irrational for every positive integer $n$ that is not a perfect square
Describe real numbers by decimal expansions, order them, and explain why a decimal represents a fraction exactly when it terminates or repeats
Work with intervals and absolute value, and prove the triangle inequality $|x + y| \le |x| + |y|$
Prove that a rational plus an irrational is irrational, and find irrational numbers whose product is rational
State the completeness property of $\R$ informally, and prove that $\{q \in \Q : q^2 < 2\}$ has no least upper bound in $\Q$

1.
Which of these numbers are irrational? (Choose all that apply.)
- A. $\sqrt{12}$
- B. $\sqrt{49}$
- C. $0.121212\ldots$ (repeating forever)
- D. $1 + \sqrt 2$
- E. $\sqrt 2 \cdot \sqrt 8$
- F. $\tfrac{22}{7}$
2.
Write $0.363636\ldots$ (the block $36$ repeating forever) as a fraction $\tfrac pq$ in lowest terms. What is the denominator $q$ ?
3.
How many integers $x$ satisfy $|x - 3| \le 2$ ?
4.
In the proof that $p^2 = 2q^2$ has no solution with $q \ne 0$ , we write: "Since $p^2$ is even, $p$ is even." Which fact justifies this step?
- A.The square of an odd number is odd (Lemma 1, by contrapositive)
- B.Every even number is a perfect square
- C.If $p$ is even then $p^2$ is even
- D. $\sqrt 2$ is irrational
5.
A student copies the proof for $\sqrt 2$ to "prove" that $\sqrt 4$ is irrational: "Suppose $p^2 = 4q^2$ with $q \ge 1$ as small as possible. Then $p^2$ is divisible by $4$ , so $p$ is divisible by $4$ ; write $p = 4k$ . Then $16k^2 = 4q^2$ , so $q^2 = 4k^2$ — a smaller solution." Since $\sqrt 4 = 2$ , something is wrong. Which step fails?
- A." $p^2$ divisible by $4$ implies $p$ divisible by $4$ "
- B.Choosing $q$ as small as possible
- C.Dividing $16k^2 = 4q^2$ by $4$
- D.Nothing fails: $\sqrt 4$ is irrational
6.
Let $S = \{1 - \tfrac1n : n = 1, 2, 3, \dots\}$ . Which statement is true?
- A. $\sup S = 1$ , and $1 \notin S$
- B. $S$ has no supremum because it has no largest element
- C. $\sup S = 0.999$
- D. $\sup S = 1$ , and $1$ is the maximum of $S$
7.
Which pair $(x, y)$ is a counterexample to the claim "if $x$ and $y$ are irrational, then $x + y$ is irrational"?
- A. $x = \sqrt 2$ , $y = 3 - \sqrt 2$
- B. $x = \sqrt 2$ , $y = \sqrt 3$
- C. $x = \sqrt 2$ , $y = 1$
- D. $x = \pi$ , $y = \pi$
8.
Which Lean statement says that $\Q$ is not complete, using the set $S = \{q \in \Q : q^2 < 2\}$ ? (IsLUB S b means " $b$ is a least upper bound of $S$ ".)
- A.¬ ∃ b : ℚ, IsLUB {q : ℚ | q ^ 2 < 2} b
- B.¬ ∃ b : ℚ, b ∈ upperBounds {q : ℚ | q ^ 2 < 2}
- C.¬ ∃ b : ℝ, IsLUB {x : ℝ | x ^ 2 < 2} b
- D.∀ q : ℚ, q ^ 2 ≠ 2
9.
Write repetend_length(p, q) for integers $0 \le p$ and $q \ge 1$ : the length of the repeating block of the decimal expansion of $\tfrac pq$ , or 0 if the decimal terminates. Use long division and remember the position at which each remainder first appeared. For example $\tfrac17 = 0.\overline{142857}$ gives 6, $\tfrac{5}{12} = 0.41\overline{6}$ gives 1, and $\tfrac38 = 0.375$ gives 0.
Computational — Practice

Answer key — Real numbers and irrationality

1.
Irrational: $\sqrt{12} = 2\sqrt 3$ (as $12$ is not a square) and $1 + \sqrt 2$ (rational plus irrational). Rational: $\sqrt{49} = 7$ , $0.\overline{12} = \tfrac{4}{33}$ , $\sqrt 2 \cdot \sqrt 8 = 4$ and $\tfrac{22}{7}$ .
2.
Let $x = 0.\overline{36}$ . Then $100x = 36.\overline{36}$ , so $99x = 36$ and $x = \tfrac{36}{99} = \tfrac{4}{11}$ . The denominator is $11$ . (Check: $4 \div 11 = 0.3636\ldots$ )
3.
$|x - 3| \le 2$ means $-2 \le x - 3 \le 2$ , i.e. $1 \le x \le 5$ . The integers are $1, 2, 3, 4, 5$ : there are $5$ .
4.
The step uses Lemma 1: if $p^2$ is even then $p$ is even, proved by its contrapositive — the square of an odd number $2k + 1$ is $4k^2 + 4k + 1$ , which is odd. Option (c) is the converse, and option (d) would make the argument circular.
5.
The false step is " $4 \mid p^2 \Rightarrow 4 \mid p$ ": for $p = 2$ , $4 \mid 4$ but $4 \nmid 2$ . For $\sqrt 2$ the analogous step " $2 \mid p^2 \Rightarrow 2 \mid p$ " is true because $2$ is prime (Lemma 1). Theorem 2's proof shows where non-squareness is really needed: in $m < \sqrt n$ , which fails for $n = 4$ .
6.
Every element is $< 1$ , so $1$ is an upper bound. If $c < 1$ , pick $n$ with $\tfrac1n < 1 - c$ ; then $1 - \tfrac1n > c$ , so $c$ is not an upper bound. Hence $\sup S = 1$ . Since $1 - \tfrac1n < 1$ for every $n$ , the supremum is not attained: $S$ has no maximum.
7.
Take $x = \sqrt 2$ and $y = 3 - \sqrt 2$ . Both are irrational (by Theorem 3, $3 + (-\sqrt 2)$ is irrational), but $x + y = 3$ . So the claim is false. Theorem 3 only covers rational $+$ irrational.
8.
Option (a) is the formal version of Theorem 5. Option (b) is false (every number $\ge \sqrt 2$ , e.g. $2$ , is a rational upper bound). Option (c) is false because $\R$ is complete. Option (d) is a true lemma used in the proof, but it is not the statement that $\Q$ lacks least upper bounds.
9.
```
def repetend_length(p, q):
    r = p % q
    seen = {}
    step = 0
    while r != 0 and r not in seen:
        seen[r] = step
        r = (10 * r) % q
        step += 1
    if r == 0:
        return 0
    return step - seen[r]
```
There are only $q$ possible remainders $0, 1, \dots, q - 1$ , so the loop stops after at most $q$ steps. If remainder $0$ appears the decimal terminates; otherwise some non-zero remainder repeats and the digits repeat from there, with a block of at most $q - 1$ digits. This is exactly the argument that every fraction has a terminating or repeating decimal.