Worked examples — Real number system — ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ

This page is the drill hall for the parent topic. The parent showed you the ideas; here we hit every kind of question the topic can throw at you. First we lay out a matrix of every case-class, then we work examples that together fill in every cell.

The scenario matrix

Every cell below is covered by at least one worked example (its label is in the last column).

Case class	The trap / edge it tests	Covered by
Classify a "clean" number	Is $\sqrt{9}$ irrational? (perfect square in disguise)	Ex 1
Classify signs & zero	Is $0$ natural? Is a negative rational?	Ex 1
Terminating decimal → fraction	finite digits after the point	Ex 2
Pure repeating decimal → fraction	$0.\overline{d}$	Ex 3
Mixed (delayed) repeat → fraction	non-repeating head + repeating tail	Ex 3
Density: rational between two rationals	the averaging trick, common denominators	Ex 4
Density: rational between two irrationals	truncate the decimals	Ex 5
Prove irrationality	contradiction, parity	Ex 6
Prove sum/product stays rational or breaks	rational ± irrational, rational × irrational	Ex 7
Cardinality reasoning	countable vs uncountable, ℤ vs ℕ	Ex 8
Limiting behaviour (a "hole")	a rational sequence with irrational limit	Ex 9
Real-world word problem	rounding, exact vs approximate	Ex 10

Prerequisites you may want open: Decimal Representations, Number Theory Fundamentals, Cardinality and Infinity, Cantor's Diagonal Argument.

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Ex 1 — Classify signs, zero, and a disguised perfect square

Step 1 — Simplify anything hiding under a root. $\sqrt{16}=4$ because $4\times 4=16$. Why this step? $\sqrt{\cdot }$ is only irrational when the number under it is not a perfect square. Never classify a root before simplifying — that is exactly the trap.

Step 2 — Walk the ladder ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ for each.

Number	ℕ?	ℤ?	ℚ?	Irr?	ℝ?	reason
$5$	✓	✓	✓ ($5/1$)	✗	✓	a counting number
$-4$	✗	✓	✓ ($-4/1$)	✗	✓	negative → not natural
$0$	✗*	✓	✓ ($0/1$)	✗	✓	*not natural under $\mathbb{N}=\{1,2,\dots \}$
$\sqrt{16}=4$	✓	✓	✓	✗	✓	perfect square!
$\sqrt{5}$	✗	✗	✗	✓	✓	5 not a perfect square
$-7/2$	✗	✗	✓	✗	✓	not a whole number

Why this step? Each column is a "does it satisfy the definition" check. A ✓ in ℕ forces ✓ all the way down the ladder; an ✗ in ℚ forces "irrational."

Verify: $\sqrt{5}=2.2360\dots$ is not a whole number and 5 has no integer square root, so it is genuinely irrational; $-7/2=-3.5$ terminates, so it is rational. ✓

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Ex 2 — Terminating decimal to a fraction

Step 1 — Put the digits over the matching power of ten. There are 3 digits after the point, so $0.375=\frac{375}{1000}.$ Why this step? "0.375" literally means $375$ thousandths — that is the definition of decimal place value (Decimal Representations).

Step 2 — Reduce by the gcd. $\gcd (375,1000)=125$, so $\frac{375}{1000}=\frac{375÷125}{1000÷125}=\frac{3}{8}.$ Why this step? Lowest terms means dividing top and bottom by their greatest common divisor.

Verify: $3÷8=0.375$. ✓

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Ex 3 — Repeating decimals (pure and delayed) to fractions

Part (a) — pure repeat.

Step 1 — Name it and shift by one repeat-length. Let $x=0.\overline{7}=0.7777\dots$. The block "7" has length 1, so multiply by $10^1$: $10x=7.7777\dots$ Why this step? Shifting by exactly one block makes the infinite tails identical.

Step 2 — Subtract to cancel the tail. $10x-x=7.777\dots -0.777\dots =7\Rightarrow 9x=7\Rightarrow x=\frac{7}{9}.$ Why this step? The endless tails match exactly, so their difference is $0$ — the infinity disappears.

Verify: $7÷9=0.7777\dots$ ✓

Part (b) — delayed repeat.

Step 1 — Shift the non-repeating head out. Let $x=0.41\overline{6}=0.41666\dots$. Two digits ("41") before the repeat, so $100x=41.666\dots$ Why this step? We move the fixed head "41" to the left of the point so only the repeating tail remains after the point.

Step 2 — Shift one more block and subtract. $1000x=416.666\dots ,1000x-100x=416.666\dots -41.666\dots =375.$ So $900x=375\Rightarrow x=\frac{375}{900}=\frac{5}{12}$ (divide by $\gcd =75$). Why this step? Subtracting two shifts that share the same tail cancels it again.

Verify: $5÷12=0.41666\dots$ ✓

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Ex 4 — Density: a rational between two rationals

Step 1 — Common denominator to compare. $\frac{2}{5}=\frac{14}{35}$ and $\frac{3}{7}=\frac{15}{35}$, so $\frac{2}{5}<\frac{3}{7}$. Why this step? Same denominator $\Rightarrow$ just compare numerators. In the figure, the two blue dots labelled $14/35$ and $15/35$ sit right next to each other — the whole picture is zoomed into that tiny interval.

Step 2 — Take the average. $c=\frac{\frac{14}{35}+\frac{15}{35}}{2}=\frac{29/35}{2}=\frac{29}{70}.$ Why this step? $c=\frac{a+b}{2}$ satisfies $a<c<b$ because $2a<a+b<2b$. The yellow dot in the figure is exactly this midpoint, dropped between the two blue dots.

Step 3 — Do it again (density never stops). Between $\frac{14}{35}=\frac{28}{70}$ and $\frac{29}{70}$: $c_2=\frac{\frac{28}{70}+\frac{29}{70}}{2}=\frac{57/70}{2}=\frac{57}{140}.$ Why this step? The same construction works on the new, smaller interval. The green dot in the figure is this second midpoint — showing you can keep splitting forever, so there are infinitely many rationals in any gap.

Verify: $\frac{2}{5}=0.4<\frac{29}{70}=0.4142\dots <\frac{3}{7}=0.4285\dots$ and $0.4<\frac{57}{140}=0.4071\dots <0.4142\dots$ ✓

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Ex 5 — Density: a rational between two irrationals

Step 1 — Get enough decimal digits. $\sqrt{2}=1.41421\dots$ and $\sqrt{3}=1.73205\dots$. Why this step? We need a place value where the two numbers already disagree so a "rounded" number can slip between them.

Step 2 — Pick a short terminating decimal in the gap. We need a rational $c$ with $\sqrt{2}<c<\sqrt{3}$. Since $\sqrt{2}=1.414\dots$ and $\sqrt{3}=1.732\dots$, the number $1.5$ sits above $\sqrt{2}$ and below $\sqrt{3}$. Take $c=1.5=\frac{3}{2}.$ Why this step? A terminating decimal is automatically rational; here $1.414\dots <1.5<1.732\dots$.

Verify: $\sqrt{2}=1.4142\dots <1.5<1.7320\dots =\sqrt{3}$, and $1.5=\frac{3}{2}\in \mathbb{Q}$. ✓

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Ex 6 — Prove an irrationality (parity by contradiction)

Step 1 — Assume it's rational in lowest terms. Suppose $\sqrt{3}=\frac{p}{q}$ with $\gcd (p,q)=1$. Square: $3q^2=p^2$. Why this step? Contradiction proofs start by assuming the opposite; "lowest terms" is the fact we will violate.

Step 2 — Deduce $3\mid p$. $p^2=3q^2$ is a multiple of 3, so $3\mid p^2$, hence $3\mid p$ (because 3 is prime — if 3 divides a product $p\cdot p$, it divides a factor). Write $p=3k$. Why this step? The whole proof hinges on 3 being prime, which powers this divisibility step (Number Theory Fundamentals).

Step 3 — Deduce $3\mid q$. $3q^2=(3k{)}^2=9k^2\Rightarrow q^2=3k^2\Rightarrow 3\mid q$. Why this step? Same divisibility logic applied to $q$.

Step 4 — Collide with lowest terms. Both $p$ and $q$ are divisible by 3, contradicting $\gcd (p,q)=1$. So $\sqrt{3}\notin \mathbb{Q}$. $\square$

Verify: $\sqrt{3}=1.7320508\dots$ never repeats or terminates, consistent with irrationality. ✓

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Ex 7 — Combining rationals with irrationals: sum AND product

Step 1 — (a) Sum: suppose $3+\sqrt{2}$ were rational, call it $r$. Then $\sqrt{2}=r-3$. But $r-3$ is rational (rationals are closed under subtraction), which would make $\sqrt{2}$ rational — contradiction (Ex 6-style fact). So $3+\sqrt{2}$ is irrational. Why this step? Isolate the irrational part; if the rest is rational, the irrational is forced to be rational — impossible.

Step 2 — (b) Sum that cancels: simplify first. $\sqrt{2}+(5-\sqrt{2})=5$. The two irrational pieces cancel! Why this step? Never assume; a sum of two irrationals can be rational. Here $5\in \mathbb{Q}$.

Step 3 — (c) Product by a non-zero rational: suppose $7\sqrt{2}=r$ were rational. Then $\sqrt{2}=\frac{r}{7}$, and dividing a rational by the non-zero rational $7$ gives a rational — again forcing $\sqrt{2}\in \mathbb{Q}$, impossible. So $7\sqrt{2}$ is irrational. Why this step? Multiplying an irrational by a non-zero rational always keeps it irrational, by the same isolate-and-contradict move.

Step 4 — (d) Product by zero: the exception. $0\cdot \sqrt{2}=0$, which is rational. The "non-zero" clause in Step 3 mattered — multiplying by $0$ collapses everything to $0$. Why this step? Zero is the degenerate case: it wipes out the irrational, so the product lands back in ℚ.

Verify: $3+\sqrt{2}=4.4142\dots$ (non-repeating); $\sqrt{2}+5-\sqrt{2}=5$; $7\sqrt{2}=9.8994\dots$ (non-repeating); $0\cdot \sqrt{2}=0$. ✓

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Ex 8 — Cardinality: is ℤ the same "size" as ℕ?

Step 1 — Zig-zag through the integers. List them $0,1,-1,2,-2,3,-3,\dots$ In the figure, the green arcs show this bounce: start at $0$, hop to $1$, back to $-1$, out to $2$, and so on. Each blue dot is an integer; the yellow label above it is its position $n$ in the list. Why this step? "Countable" means there exists a one-to-one list indexed by ℕ (size ${\aleph }_0$). Bouncing out from 0 guarantees every integer appears at a finite step — nothing is skipped and nothing repeats.

Step 2 — Formula for the pairing. $f(n)=\{\begin{array}{ll}n/2& n\text{ even}\\ -(n-1)/2& n\text{ odd}\end{array}(n=1,2,3,\dots )$ gives $f(1)=0,f(2)=1,f(3)=-1,f(4)=2,\dots$ — exactly the yellow-labelled order in the figure. Why this step? An explicit bijection (a rule that pairs each natural with exactly one integer and vice versa) is airtight proof of "same size."

Step 3 — Why ℝ escapes: Cantor's diagonal argument. Suppose someone claims a complete list of all reals in $[0,1]$, written as decimals: $r_1=0.d_{11}{d}_{12}{d}_{13}\dots ,r_2=0.d_{21}{d}_{22}{d}_{23}\dots ,r_3=0.d_{31}{d}_{32}{d}_{33}\dots$ Now build a new number $x=0.x_1{x}_2{x}_3\dots$ by walking down the diagonal and changing each digit: let $x_n=5$ if $d_{nn}\ne 5$, and $x_n=4$ if $d_{nn}=5$. Then $x$ differs from $r_1$ in place 1, from $r_2$ in place 2, from $r_n$ in place $n$ — so $x$ is not on the list, yet it is a real number in $[0,1]$. The list was supposed to be complete: contradiction. So no list can hold all reals — ℝ is uncountable, $|\mathbb{R}|=2^{{\aleph }_0}>{\aleph }_0$. See Cantor's Diagonal Argument. Why this step? Countability is exactly the claim "a complete list exists." The diagonal number defeats any such list, which is precisely why ℝ is a strictly bigger infinity than ℤ or ℕ.

Verify: $f(1)=0,f(2)=1,f(3)=-1,f(4)=2,f(5)=-2$ — a valid start of an every-integer-once list; and the diagonal $x$ differs from each $r_n$ in position $n$ by construction (choosing digits 4/5 avoids the $0.4999\dots =0.5$ ambiguity), so it genuinely lies off any list. ✓

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Ex 9 — A limiting "hole": rational sequence, irrational limit

Step 1 — Identify the target. Each term is $\sqrt{2}$ truncated to more places: $1.4142\dots$. The limit is $\sqrt{2}$. Why this step? The truncations of a number's decimal expansion converge to that number — each extra digit pins it down 10× tighter.

Step 2 — Note the punchline. $\sqrt{2}$ is irrational (Ex 6-style), yet every sequence term is rational. This is a hole in ℚ: the sequence "should" have a limit sitting right there, but in ℚ that point is missing. ℝ fills it — that's the Completeness Axiom. Why this step? It shows density (rationals everywhere) is weaker than completeness (no gaps), exactly the parent's warning.

Verify: $1.4142^2=1.99996\dots \to 2$, confirming the terms approach $\sqrt{2}$. ✓

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Ex 10 — Real-world: exact vs rounded

Step 1 — Exact side. Side $=\sqrt{20}=\sqrt{4\cdot 5}=2\sqrt{5}$ metres. Why this step? Area of a square is $s^2$, so $s=\sqrt{\text{area}}$; factor out the perfect square 4.

Step 2 — Is it rational? $5$ is not a perfect square, so $\sqrt{5}$ (hence $2\sqrt{5}$) is irrational. The decimal $4.47$ cannot be exact — it's a rounding. Why this step? Same disguised-root check as Ex 1: only perfect-square radicands give rationals.

Step 3 — Error of the rounding. $2\sqrt{5}=4.472135\dots$, so the worker's $4.47$ is short by $4.472135\dots -4.47=0.002135\dots \text{ m}\approx 2.1\text{ mm}.$ Why this step? In real measurement, rationals approximate reals to any needed precision — the practical steel-man from the parent's mistake callout.

Verify: $(2\sqrt{5}{)}^2=4\cdot 5=20$ exactly; and $4.47^2=19.9809\ne 20$, confirming it's only approximate. ✓

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Recall Self-test (reveal after answering)

Turn $0.\overline{18}$ into a fraction. ::: $x=0.\overline{18},100x-x=18\Rightarrow x=\frac{18}{99}=\frac{2}{11}$. Is $\pi -\pi$ rational? ::: Yes — it equals $0\in \mathbb{Q}$; two irrationals can cancel. Is $5\sqrt{3}$ rational? ::: No — a non-zero rational times an irrational is irrational. Why is $\sqrt{16}$ rational but $\sqrt{5}$ not? ::: 16 is a perfect square ($=4$); 5 is not, so its root can't be written $p/q$. Is ℤ bigger than ℕ? ::: No — both are countable (${\aleph }_0$) via the zig-zag bijection.