The Real Number Hierarchy: $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$
- Natural Numbers ($\mathbb{N}$): The counting numbers $\mathbb{N} = \{0, 1, 2, 3, 4, \dots\}$ (by IB MYP/DP convention, $0 \in \mathbb{N}$; the strictly positive set is written $\mathbb{Z}^+ = \{1, 2, 3, \dots\}$).
- Integers ($\mathbb{Z}$): Whole numbers including positive, negative, and zero: $\mathbb{Z} = \{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}$.
- Rational Numbers ($\mathbb{Q}$): Any number that can be expressed as a ratio of two integers in the form: $$\mathbb{Q} = \left\{ \frac{p}{q} \;\middle|\; p, q \in \mathbb{Z},\, q \ne 0 \right\}$$ Rational numbers produce either terminating decimals (e.g., $\frac{3}{8} = 0.375$) or recurring decimals (e.g., $\frac{5}{11} = 0.\overline{45}$).
- Irrational Numbers ($\mathbb{Q}'$ or $\mathbb{R} \setminus \mathbb{Q}$): Numbers that cannot be expressed as $\frac{p}{q}$. Their decimal expansions are non-terminating and non-repeating (e.g., $\sqrt{2}, \sqrt{3}, \pi, e, \sqrt{7}$).
- Real Numbers ($\mathbb{R}$): The union of all rational and irrational numbers ($\mathbb{R} = \mathbb{Q} \cup \mathbb{Q}'$). Every real number corresponds to a unique point on the continuous real number line.
- Nested Subset Inclusion: $$\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$$
Figure 1.1: Strict Venn hierarchy of numerical sets: $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$ with disjoint irrational set $\mathbb{Q}'$.
Example 1 · Criterion A
Classifying Numbers into the Smallest Appropriate Set
Determine the most specific (smallest) number set ($\mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{Q}'$) for each of the following:
(a) $\sqrt{49}$ (b) $-\frac{36}{4}$ (c) $3.14159$ (d) $\sqrt{18}$ (e) $0.\overline{81}$
- (a) $\sqrt{49} = 7$: Since $7$ is a non-negative whole counting number, its most specific set is the Natural numbers ($\mathbb{N}$).
- (b) $-\frac{36}{4} = -9$: Since $-9$ is a negative whole number, it is not in $\mathbb{N}$, but belongs to the Integers ($\mathbb{Z}$).
- (c) $3.14159$: This is a terminating decimal ($=\frac{314159}{100000}$), hence a ratio of integers. Its smallest set is the Rational numbers ($\mathbb{Q}$) (Note: it is an approximation of $\pi$, but as written it is terminating).
- (d) $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$: Since $2$ is not a perfect square, $\sqrt{2}$ cannot be expressed as $\frac{p}{q}$. Therefore, $\sqrt{18}$ is an Irrational number ($\mathbb{Q}'$).
- (e) $0.\overline{81}$: A repeating decimal with period 2. All periodic decimals are rational: $0.\overline{81} = \frac{81}{99} = \frac{9}{11}$. Smallest set: Rational numbers ($\mathbb{Q}$).
Summary: (a) $\mathbb{N}$; (b) $\mathbb{Z}$; (c) $\mathbb{Q}$; (d) $\mathbb{Q}'$; (e) $\mathbb{Q}$.
Converting Recurring Decimals to Exact Fractions $\frac{p}{q}$
- Why Fractions? Floating-point decimals lose precision in computations. In MYP Criterion A and DP AA examinations, exact values in terms of $\frac{p}{q}$ are required.
- Pure Recurring Decimals: Where the repeating period begins immediately after the decimal point.
• Multiply by $10^k$, where $k$ is the period length (number of digits in repeating cycle).
• Subtract the original equation $x$ to eliminate the infinite recurring tail. $$\text{If } x = 0.\overline{a_1 a_2 \dots a_k}, \quad 10^k x - x = a_1 a_2 \dots a_k \implies x = \frac{a_1 a_2 \dots a_k}{10^k - 1}$$ - Mixed Recurring Decimals: Where a non-repeating prefix precedes the periodic cycle (e.g., $0.5\overline{24}$).
• Step 1: Multiply by $10^m$ (where $m$ is the number of non-repeating digits) to align the repeat at the decimal boundary.
• Step 2: Multiply by $10^{m+k}$ to shift one full repeating period.
• Step 3: Subtract Equation 1 from Equation 2 and simplify to lowest terms. - Terminating Decimals Condition: A fraction $\frac{p}{q}$ in irreducible form terminates if and only if the prime factorization of the denominator $q$ contains only $2$ and/or $5$: $$q = 2^a \cdot 5^b \quad (a, b \in \mathbb{N})$$
Example 2 · Criterion A & C
Converting Pure and Mixed Recurring Decimals
Express each decimal as an irreducible fraction $\frac{p}{q}$ where $p, q \in \mathbb{Z}$:
(a) $0.\overline{54}$ (b) $2.1\overline{36}$
- (a) Pure repeating ($k = 2$ digits): $$\text{Let } x = 0.545454\dots \quad \text{--- (1)}$$ Multiply by $10^2 = 100$: $$100x = 54.545454\dots \quad \text{--- (2)}$$ Subtract (1) from (2): $$100x - x = 54.5454\dots - 0.5454\dots \implies 99x = 54$$ $$x = \frac{54}{99} = \frac{54 \div 9}{99 \div 9} = \frac{6}{11}$$
- (b) Mixed repeating ($1$ non-repeating digit, period $k = 2$): $$\text{Let } y = 2.1363636\dots$$ Multiply by $10^1 = 10$ to shift the non-repeating digit: $$10y = 21.363636\dots \quad \text{--- (1)}$$ Multiply by $10^3 = 1000$ to shift one full repeat cycle: $$1000y = 2136.363636\dots \quad \text{--- (2)}$$ Subtract (1) from (2): $$1000y - 10y = 2136.36\dots - 21.36\dots \implies 990y = 2115$$ $$y = \frac{2115}{990}$$ Divide numerator and denominator by $\gcd(2115, 990) = 45$: $$y = \frac{2115 \div 45}{990 \div 45} = \frac{47}{22} = 2\frac{3}{22}$$
Answers: (a) $\frac{6}{11}$; (b) $\frac{47}{22}$.
Proof of Irrationality by Contradiction (Extended eAssessment)
- Logic of Proof by Contradiction (Criterion C):
1. Assume the negation of the proposition to be proved (e.g., assume $\sqrt{2} \in \mathbb{Q}$).
2. Deduce logical consequences through rigorous algebraic steps.
3. Reach an impossible statement or violation of initial definitions (a contradiction).
4. Conclude that the initial assumption must be false; thus the proposition is true. - Classic Euclid Proof for $\sqrt{2}$ (Key Theorem):
Assume $\sqrt{2} = \frac{p}{q}$, where $p, q \in \mathbb{Z}^+$ and $\gcd(p, q) = 1$ (coprime, i.e., in simplest form). $$\sqrt{2} = \frac{p}{q} \implies 2 = \frac{p^2}{q^2} \implies p^2 = 2q^2$$ Since $p^2$ is a multiple of $2$, $p^2$ is even, which implies $p$ is even.
Let $p = 2k$ for some integer $k$. Substitute into $p^2 = 2q^2$: $$(2k)^2 = 2q^2 \implies 4k^2 = 2q^2 \implies q^2 = 2k^2$$ This implies $q^2$ is even, so $q$ must also be even.
The Contradiction: Both $p$ and $q$ are even, so they share a common factor of $2$. This contradicts the assumption that $\gcd(p, q) = 1$. Hence, $\sqrt{2}$ cannot be rational; $\sqrt{2} \in \mathbb{Q}'$. - Operations with Rationals and Irrationals:
• $\text{Rational} \pm \text{Irrational} = \text{Irrational}$ (e.g., $3 + \sqrt{5} \in \mathbb{Q}'$).
• $\text{Non-zero Rational} \times \text{Irrational} = \text{Irrational}$ (e.g., $\frac{2}{3}\pi \in \mathbb{Q}'$).
• $\text{Irrational} \times \text{Irrational}$ can be rational or irrational (e.g., $\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4 \in \mathbb{Q}$, but $\sqrt{2} \times \sqrt{3} = \sqrt{6} \in \mathbb{Q}'$).
Interval Notation, Inequalities & Number Line Graphs
- Interval Notation vs Set-Builder:
• Open Interval $(a, b)$: $\{x \in \mathbb{R} \mid a < x < b\}$ — endpoints excluded (open circles on number line).
• Closed Interval $[a, b]$: $\{x \in \mathbb{R} \mid a \le x \le b\}$ — endpoints included (solid dots on number line).
• Half-open $[a, b)$: $\{x \in \mathbb{R} \mid a \le x < b\}$ — $a$ included, $b$ excluded.
• Unbounded Intervals: $(-\infty, b] \iff x \le b$; $(a, \infty) \iff x > a$; $(-\infty, \infty) \iff x \in \mathbb{R}$. Note that $\pm \infty$ always take round parentheses $($ and $)$. - Absolute Value as Distance on $\mathbb{R}$:
$$|x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}$$
Geometrically, $|x - c|$ represents the distance between $x$ and $c$ on the real number line.
• $|x - c| \le d \iff c - d \le x \le c + d \iff x \in [c - d, c + d]$ (closed neighborhood).
• $|x - c| > d \iff x < c - d \text{ or } x > c + d \iff x \in (-\infty, c - d) \cup (c + d, \infty)$.
Example 3 · Criterion A & D
Interval Notation and Tolerance Neighborhoods
An precision engineering ball bearing is manufactured with a nominal radius of $15\text{ mm}$ and an allowable error tolerance of $\pm 0.04\text{ mm}$.
(a) Write an absolute value inequality for the acceptable radius $r$.
(b) Express the acceptable range of $r$ in interval notation.
(c) State whether a measured ball bearing with $r = 15.038\text{ mm}$ is accepted.
- (a) Absolute value inequality: The distance between the actual radius $r$ and $15\text{ mm}$ must not exceed $0.04\text{ mm}$: $$|r - 15| \le 0.04$$
- (b) Interval notation: $$-0.04 \le r - 15 \le 0.04 \implies 15 - 0.04 \le r \le 15 + 0.04$$ $$14.96 \le r \le 15.04 \implies r \in [14.96, 15.04]\text{ mm}$$
- (c) Verification: Since $14.96 \le 15.038 \le 15.04$, the ball bearing satisfies the tolerance constraint and is accepted.
Answers: (a) $|r - 15| \le 0.04$; (b) $[14.96, 15.04]$; (c) Accepted.
Prime Factorization, HCF & LCM Applications
- Fundamental Theorem of Arithmetic: Every integer $n > 1$ can be uniquely factored as a product of prime powers: $$n = p_1^{a_1} \cdot p_2^{a_2} \cdots p_k^{a_k}$$
- Finding HCF and LCM via Index Form:
Let $A = 2^3 \cdot 3^2 \cdot 5^1$ and $B = 2^2 \cdot 3^3 \cdot 7^1$:
• $\text{HCF}(A, B)$ = product of common prime factors with lowest exponents: $$\text{HCF} = 2^{\min(3,2)} \cdot 3^{\min(2,3)} = 2^2 \cdot 3^2 = 4 \cdot 9 = 36$$
• $\text{LCM}(A, B)$ = product of all prime factors with highest exponents: $$\text{LCM} = 2^{\max(3,2)} \cdot 3^{\max(2,3)} \cdot 5^1 \cdot 7^1 = 2^3 \cdot 3^3 \cdot 5 \cdot 7 = 8 \cdot 27 \cdot 35 = 7560$$ - Fundamental Product Identity: $$\text{HCF}(A, B) \times \text{LCM}(A, B) = A \times B$$
State whether each of the following statements is True or False. If false, provide a counterexample:
(a) All integers are rational numbers ($\mathbb{Z} \subset \mathbb{Q}$).
(b) The sum of any two irrational numbers is always irrational.
(c) $0$ is a positive integer.
Show Full Markscheme
(a) True: [1 mark] Any integer $z \in \mathbb{Z}$ can be written as the ratio $\frac{z}{1}$ where $z, 1 \in \mathbb{Z}$ and $1 \ne 0$. Hence $\mathbb{Z} \subset \mathbb{Q}$. [A1]
(b) False: [1 mark] Counterexample: Let $a = \sqrt{3}$ and $b = -\sqrt{3}$. Both $a, b \in \mathbb{Q}'$, but $a + b = \sqrt{3} + (-\sqrt{3}) = 0 \in \mathbb{Q}$ (or $(2 + \sqrt{5}) + (2 - \sqrt{5}) = 4 \in \mathbb{Q}$). [R1]
(c) False: [1 mark] Zero is neither positive nor negative ($\mathbb{Z}^+ = \{1, 2, 3, \dots\}$). [A1]
Convert the recurring decimal $0.2\overline{7}$ into an irreducible fraction $\frac{p}{q}$ where $p, q \in \mathbb{Z}$. Show complete algebraic working.
Show Full Markscheme
Step 1: Set up the equations [1 mark]
Let $x = 0.2777\dots$
Multiply by $10$ to shift non-repeating digit: $10x = 2.777\dots$ [M1]
Step 2: Multiply to shift repeating block [1 mark]
Multiply by $100$: $100x = 27.777\dots$ [M1]
Step 3: Eliminate recurring decimal tail [1 mark]
$100x - 10x = 27.777\dots - 2.777\dots \implies 90x = 25$ [A1]
Step 4: Reduce to irreducible fraction [1 mark]
$x = \frac{25}{90} = \frac{25 \div 5}{90 \div 5} = \frac{5}{18}$ [A1]
Without long division, determine whether each fraction produces a terminating or recurring decimal. Justify your answer using prime factorization:
(a) $\frac{21}{60}$ (b) $\frac{13}{14}$ (c) $\frac{7}{40}$
Show Full Markscheme
(a) $\frac{21}{60}$: [1 mark]
First simplify: $\frac{21 \div 3}{60 \div 3} = \frac{7}{20}$. Prime factorization of denominator: $20 = 2^2 \times 5^1$. Since the prime factors consist solely of $2$ and $5$, the decimal terminates ($0.35$). [A1]
(b) $\frac{13}{14}$: [1 mark]
Irreducible fraction; denominator $14 = 2 \times 7$. The presence of prime factor $7 \ne 2, 5$ means the decimal recurs. [A1]
(c) $\frac{7}{40}$: [1 mark]
Denominator $40 = 2^3 \times 5^1$. Only contains primes $2$ and $5$, hence it terminates ($0.175$). [A1]
Given two numbers $X = 360$ and $Y = 840$:
(a) Express $X$ and $Y$ in prime factor index form. [2 marks]
(b) Find $\text{HCF}(X, Y)$ and $\text{LCM}(X, Y)$. [2 marks]
Show Full Markscheme
(a) Prime factorization: [2 marks]
$360 = 36 \times 10 = 2^3 \times 3^2 \times 5^1$ [A1]
$840 = 84 \times 10 = 2^3 \times 3^1 \times 5^1 \times 7^1$ [A1]
(b) HCF and LCM: [2 marks]
$\text{HCF} = 2^{\min(3,3)} \times 3^{\min(2,1)} \times 5^{\min(1,1)} = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120$ [A1]
$\text{LCM} = 2^{\max(3,3)} \times 3^{\max(2,1)} \times 5^{\max(1,1)} \times 7^1 = 2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 35 = 2520$ [A1]
Consider the decimal expansions of fractions with denominator $7$:
$\frac{1}{7} = 0.\overline{142857}$, $\frac{2}{7} = 0.\overline{285714}$, $\frac{3}{7} = 0.\overline{428571}$.
(a) Describe the pattern observed in the repeating digits across these fractions. [2 marks]
(b) Predict the exact decimal expansion of $\frac{5}{7}$ without using long division. Justify your answer. [2 marks]
(c) State the general rule for the period of the decimal expansion of $\frac{1}{p}$ for any prime $p$. [1 mark]
Show Full Markscheme
(a) Pattern description: [2 marks]
The repeating sequence consists of the exact same 6 cyclic digits $\{1, 4, 2, 8, 5, 7\}$ in identical cyclic order, but starting from different positions depending on the numerator value. [R2]
(b) Prediction for $\frac{5}{7}$: [2 marks]
Since $5 \div 7 \approx 0.71\dots$, the sequence must start with $7$.
Following the cyclic sequence $1 \to 4 \to 2 \to 8 \to 5 \to 7 \to 1$, starting at $7$ gives:
$\frac{5}{7} = 0.\overline{714285}$ [A2]
(c) Period rule: [1 mark]
The maximum repeating period length for $\frac{1}{p}$ cannot exceed $p - 1$ digits (here $7 - 1 = 6$). [A1]
Given the subsets $A = \{x \in \mathbb{R} \mid -3 \le x < 5\}$ and $B = \{x \in \mathbb{R} \mid 1 \le x \le 9\}$:
(a) Express $A$ and $B$ in interval notation. [2 marks]
(b) Find $A \cap B$ and $A \cup B$ in interval notation. [2 marks]
Show Full Markscheme
(a) Interval notation: [2 marks]
$A = [-3, 5)$ [A1]
$B = [1, 9]$ [A1]
(b) Intersection and Union: [2 marks]
$A \cap B$ is the overlap from $1$ to $5$: $[1, 5)$ ($5$ is excluded because $5 \notin A$) [A1]
$A \cup B$ is the entire combined span from $-3$ to $9$: $[-3, 9]$ [A1]
Prove by contradiction that $\sqrt{3}$ is an irrational number. Present a complete, logically structured mathematical argument.
Show Full Markscheme
Step 1: Assumption for contradiction [1 mark]
Assume $\sqrt{3}$ is rational. Then $\sqrt{3} = \frac{a}{b}$ where $a, b \in \mathbb{Z}^+$, $b \ne 0$, and $\gcd(a, b) = 1$ (irreducible form). [M1]
Step 2: Square both sides [1 mark]
$3 = \frac{a^2}{b^2} \implies a^2 = 3b^2$. [A1]
Step 3: Deduce divisibility of $a$ [1 mark]
Since $a^2$ is a multiple of $3$, $3$ divides $a^2$. Because $3$ is prime, $3$ must divide $a$.
Let $a = 3k$ for some integer $k$. [R1]
Step 4: Deduce divisibility of $b$ [1 mark]
Substitute $a = 3k$: $(3k)^2 = 3b^2 \implies 9k^2 = 3b^2 \implies b^2 = 3k^2$.
Thus $3$ divides $b^2$, which implies $3$ divides $b$. [R1]
Step 5: Identify contradiction & conclusion [1 mark]
Both $a$ and $b$ are divisible by $3$, contradicting the condition that $\gcd(a, b) = 1$.
Therefore, the initial assumption is false, and $\sqrt{3} \in \mathbb{Q}'$. $\blacksquare$ [R1]
Let $r$ be a non-zero rational number and $x$ be an irrational number.
Prove by contradiction that their product $P = r \cdot x$ is an irrational number.
Show Full Markscheme
Step 1: Assumption [1 mark]
Assume for contradiction that $P = r \cdot x$ is rational, so $P \in \mathbb{Q}$. [M1]
Step 2: Express in rational form [1 mark]
Since $r \in \mathbb{Q}$ and $r \ne 0$, we can write $r = \frac{a}{b}$ and $P = \frac{c}{d}$ where $a, b, c, d \in \mathbb{Z}$ and $a, b, d \ne 0$. [A1]
Step 3: Solve for $x$ [1 mark]
$$x = \frac{P}{r} = \frac{\frac{c}{d}}{\frac{a}{b}} = \frac{bc}{ad}$$
Since the integers are closed under multiplication, $bc \in \mathbb{Z}$ and $ad \in \mathbb{Z}$ with $ad \ne 0$.
Thus $x \in \mathbb{Q}$. [M1]
Step 4: Contradiction [1 mark]
This contradicts the premise that $x$ is irrational ($x \in \mathbb{Q}'$).
Hence, $r \cdot x$ must be irrational. $\blacksquare$ [R1]
A solar-powered automated greenhouse has two independent irrigation systems:
• System A activates every $45\text{ minutes}$.
• System B activates every $60\text{ minutes}$.
Both systems activate simultaneously at 08:00 AM on Monday morning.
(a) Find the Lowest Common Multiple (LCM) of $45$ and $60$ using prime factor trees or index form. [2 marks]
(b) Determine the exact time when both systems will next activate together. [1 mark]
(c) Between 08:00 AM and 08:00 PM (a 12-hour window), how many times will both systems trigger simultaneously? [2 marks]
Show Full Markscheme
(a) LCM calculation: [2 marks]
$45 = 3^2 \times 5^1$
$60 = 2^2 \times 3^1 \times 5^1$
$\text{LCM}(45, 60) = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = 180\text{ minutes}$. [A2]
(b) Next simultaneous time: [1 mark]
$180\text{ minutes} = 3\text{ hours}$.
$08:00\text{ AM} + 3\text{ hours} = \mathbf{11:00\text{ AM}}$. [A1]
(c) Total simultaneous triggers in 12 hours: [2 marks]
Total span: $12\text{ hours} = 720\text{ minutes}$.
Intervals: $\frac{720}{180} = 4$ intervals.
Triggers occur at $t = 0\text{ h}$ (08:00), $3\text{ h}$ (11:00), $6\text{ h}$ (14:00), $9\text{ h}$ (17:00), and $12\text{ h}$ (20:00).
Total times = $\mathbf{5\text{ times}}$ (including start and end). [A2]
A civil engineer designs an expansion joint for a steel bridge span. Due to seasonal temperature fluctuations from $-15^\circ\text{C}$ to $45^\circ\text{C}$, the span length $L$ (in metres) satisfies the condition:
$$|L - 120.000| \le 0.035$$
(a) Interpret the physical meaning of $120.000$ and $0.035$ in this engineering context. [2 marks]
(b) Solve the inequality and state the range of possible span lengths in interval notation. [2 marks]
(c) During an extreme heat wave, a laser sensor measures $L = 120.038\text{ m}$. Determine whether the bridge span exceeds safe design tolerances. [1 mark]
Show Full Markscheme
(a) Contextual interpretation: [2 marks]
• $120.000\text{ m}$ is the nominal (designed equilibrium) length of the bridge span at baseline temperature. [A1]
• $0.035\text{ m}$ ($35\text{ mm}$) is the maximum allowable thermal expansion or contraction tolerance. [A1]
(b) Interval solution: [2 marks]
$-0.035 \le L - 120.000 \le 0.035 \implies 120.000 - 0.035 \le L \le 120.000 + 0.035$
$119.965 \le L \le 120.035 \implies L \in [119.965, 120.035]\text{ m}$. [A2]
(c) Evaluation: [1 mark]
Since $120.038\text{ m} > 120.035\text{ m}$, the bridge span exceeds the safe design tolerance by $0.003\text{ m}$ ($3\text{ mm}$). [R1]
Number Systems Quick Reference
Number Set Classifications
Decimal Conversions & Conditions
Intervals & Absolute Values
HCF & LCM Identities
Top 5 MYP Examination Pitfalls
1. Forgetting to Simplify Before Checking Denominators
When testing if $\frac{21}{60}$ terminates, looking at $60 = 2^2 \times 3 \times 5$ leads students to falsely claim it recurs because of factor $3$. Always simplify first: $\frac{21}{60} = \frac{7}{20}$; $20 = 2^2 \times 5$, so it terminates!
2. Confusing $\pi$ and $\frac{22}{7}$
$\frac{22}{7}$ is a rational approximation ($0.\overline{142857}$), while $\pi$ is strictly irrational ($3.14159265\dots$ never recurs or terminates). Never classify $\pi$ as rational.
3. Infinity with Square Brackets
Writing $[0, \infty]$ or $(-\infty, 4]$ is a strict mathematical error. Since infinity is a concept of unbound growth and not a reachable real number, $\pm \infty$ must always take round brackets: $[0, \infty)$ and $(-\infty, 4]$.
4. Mixed Recurring Algebraic Multipliers
For $x = 0.1\overline{6}$, students often compute $10x - x = 9x = 1.5$, leaving a decimal in the fraction $\frac{1.5}{9}$. Better practice: use $100x = 16.6\dots$ and $10x = 1.6\dots$, giving $90x = 15 \implies x = \frac{15}{90} = \frac{1}{6}$.
Number Systems Knowledge Check
Test your understanding with 5 criteria-aligned questions. Choose an answer to see feedback and a worked solution.
Which of the following numbers belongs to the set of irrational numbers ($\mathbb{Q}'$)?
• $\sqrt{0.49} = 0.7 = \frac{7}{10} \in \mathbb{Q}$.
• $0.3\overline{18}$ is a repeating decimal $=\frac{318-3}{990} = \frac{35}{110} = \frac{7}{22} \in \mathbb{Q}$.
• $-\frac{22}{7}$ is a quotient of two integers $\in \mathbb{Q}$.
• $\sqrt{8} = 2\sqrt{2}$. Since $2$ is not a square, $\sqrt{2}$ and $2\sqrt{2}$ are irrational ($\mathbb{Q}'$).
What is the exact fraction representation of the recurring decimal $0.\overline{135}$ in lowest terms?
Which inequality corresponds exactly to the interval notation $(-4, 7]$?
If the HCF of two numbers is $18$ and their product is $19440$, what is their LCM?
A sensor measures a temperature reading $T$ that satisfies $|T - 22| \le 3$. Which interval contains all acceptable temperature readings?