Exercises — Real number system — ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ

"Smallest set" means: go as far left in the chain as the number still fits. Locate each on the number-line figure above — orange dot, teal tick, or plum-style gap?

• $12$: a counting number (an orange-dot type) → smallest set is $==\mathbb{N}==$.
• $-7$: negative, so not natural, but whole → $==\mathbb{Z}==$.
• $\frac{9}{4}=2.25$: a ratio of integers, not whole (a teal-tick type) → $==\mathbb{Q}==$.
• $\sqrt{16}=4$: simplify first! $4$ is a counting number → $==\mathbb{N}==$.
• $\sqrt{5}=2.2360\dots$: $5$ is not a perfect square, so this cannot be $p/q$ (a plum-cross type, in a gap) → smallest set is $==\mathbb{R}==$ (it is irrational, and irrationals live only in $\mathbb{R}$).

• (a) False under the convention $\mathbb{N}=\{1,2,3,\dots \}$ used in the parent note (0 starts $\mathbb{Z}$).
• (b) True: $-3=\frac{-3}{1}$, a ratio of integers.
• (c) True: any integer $n=\frac{n}{1}\in \mathbb{Q}$.
• (d) False: $\frac{1}{2}\in \mathbb{Q}$ but $\frac{1}{2}\notin \mathbb{Z}$.

Why the midpoint? For any two numbers with $a<b$, adding $a$ to both sides of $a<b$ gives $2a<a+b$, and adding $b$ gives $a+b<2b$; dividing by $2$ (which preserves order since $2>0$): $a<\frac{a+b}{2}<b.$ So the average always lands strictly between — that is the on-page proof. Now compute it: $c=\frac{\frac{2}{5}+\frac{1}{2}}{2}=\frac{\frac{4}{10}+\frac{5}{10}}{2}=\frac{\frac{9}{10}}{2}=\frac{9}{20}.$ A sum/quotient of rationals is rational, so $c\in \mathbb{Q}$. Check: $\frac{2}{5}=\frac{8}{20}<\frac{9}{20}<\frac{10}{20}=\frac{1}{2}$. ✓ So $c=\frac{9}{20}$.

The block "27" has length $2$, so multiply by $10^2=100$ to shift one whole block left: $100x=27.2727\dots ,x=0.2727\dots$ Subtract to kill the infinite tail: $100x-x=27\Rightarrow 99x=27\Rightarrow x=\frac{27}{99}=\frac{3}{11}.$ Lowest terms since $\gcd (3,11)=1$. So $0.\overline{27}=\frac{3}{11}$. (See Decimal Representations.)

Closure = every result of the operation stays inside $S$.

• Addition: $1+1=2\notin S$. Not closed.
• Multiplication: all products $(\pm 1)(\pm 1)=\pm 1$, and anything $\times 0=0$; every result is in $S$. Closed.

(This mirrors the parent's example — see Properties of Operations.)

Tool: "rational + irrational = irrational" — why? If $r$ is rational and $r+i$ were rational, then $i=(r+i)-r$ would be a difference of rationals, hence rational — contradiction. So the sum is irrational.

• (a) $\sqrt{2}+(3-\sqrt{2})=3$. The irrational parts cancel → $==\text{rational}==$ (equals $3$).
• (b) Suppose $\sqrt{2}+\sqrt{3}=q\in \mathbb{Q}$. Square: $5+2\sqrt{6}=q^2$, so $\sqrt{6}=\frac{q^2-5}{2}\in \mathbb{Q}$. But $6$ is not a perfect square, so $\sqrt{6}$ is irrational — contradiction. → $==\text{irrational}==$.
• (c) $\frac{\sqrt{8}}{\sqrt{2}}=\sqrt{\frac{8}{2}}=\sqrt{4}=2$ → $==\text{rational}==$.

Moral: never trust the surface form — simplify and reason.

Recall from the definition box: $a\mid b$ means "$a$ divides $b$", and lowest terms means $\gcd (p,q)=1$. Assume $\sqrt{3}=\frac{p}{q}$ in lowest terms. Then $3q^2=p^2$, so $3\mid p^2$. Key fact (the tool): $3$ is prime, so $3\mid p^2\Rightarrow 3\mid p$. (Why: a prime dividing a product must divide one of the factors — here the two factors are both $p$; this is Euclid's lemma, a consequence of the FTA cited above.) Write $p=3m$: $3q^2=(3m{)}^2=9m^2\Rightarrow q^2=3m^2\Rightarrow 3\mid q.$ So $3\mid p$ and $3\mid q$, contradicting $\gcd (p,q)=1$. Hence $\sqrt{3}$ is irrational. □ What replaced "even/odd"? The step "$2\mid p^2\Rightarrow 2\mid p$" becomes "$3\mid p^2\Rightarrow 3\mid p$", which works because $3$ is prime. This same argument works for $\sqrt{n}$ whenever $n$ is not a perfect square (see Number Theory Fundamentals).

(i) Rational: take the midpoint $c=\frac{1.41+1.42}{2}=1.415$. Why is $1.415$ rational? A terminating decimal is a fraction over a power of ten: $1.415=\frac{1415}{1000}=\frac{283}{200}(\text{divide top and bottom by }\gcd (1415,1000)=5),$ so $c=\frac{283}{200}\in \mathbb{Q}$, and by the midpoint inequality $1.41<1.415<1.42$. ✓ (ii) Irrational: we claim $\sqrt{2}$ lies in $(1.41,1.42)$. Justify by squaring (squaring preserves order for positive numbers): $1.41^2=1.9881<2$ and $1.42^2=2.0164>2$. Since $1.41^2<2<1.42^2$ and everything is positive, $1.41<\sqrt{2}<1.42$. And $\sqrt{2}$ is irrational (parent proof). ✓ See the figure: the interval $(1.41,1.42)$ is blown up, the teal tick at $1.415$ and the plum cross at $\sqrt{2}$ both sit inside. Big idea: both $\mathbb{Q}$ and the irrationals are dense — every interval, however short, contains infinitely many of each.

Sum: $x=\pi$ and $y=1-\pi$. Each is irrational (if $1-\pi$ were rational, $\pi =1-(1-\pi )$ would be rational — false). Their sum $x+y=1\in \mathbb{Q}$. Product: $x=\sqrt{2}$ and $y=\sqrt{2}$ (or $y=3\sqrt{2}$). Then $\sqrt{2}\cdot \sqrt{2}=2\in \mathbb{Q}$, and $\sqrt{2}\cdot 3\sqrt{2}=6\in \mathbb{Q}$. Both factors irrational, product rational. Lesson: the irrationals fail closure under both $+$ and $\times$ — reinforcing that they do not form a field.

$\mathbb{Q}$ is countable (zig-zag list of $\mathbb{N}\times \mathbb{N}$, see Cardinality and Infinity), so $|\mathbb{Q}|={\aleph }_0$. But $|\mathbb{R}|=2^{{\aleph }_0}>{\aleph }_0$ (via Cantor's Diagonal Argument). If the irrationals $\mathbb{R}\setminus \mathbb{Q}$ were also countable, then $\mathbb{R}=\mathbb{Q}\cup (\mathbb{R}\setminus \mathbb{Q})$ would be a union of two countable sets, hence countable — contradiction. So the irrationals are uncountable. Precise meaning of "almost every": you cannot list all reals, yet you can list all rationals; in the sense of cardinality the rationals are a vanishingly thin slice (${\aleph }_0$) of the uncountable continuum ($2^{{\aleph }_0}$). "Almost every real is irrational" = the exceptions (rationals) form a countable set.

($\Leftarrow$) $n$ not a perfect square $\Rightarrow$ $\sqrt{n}$ irrational. Suppose $\sqrt{n}=\frac{p}{q}$ in lowest terms, $\gcd (p,q)=1$. Then $n{q}^2=p^2$. If $q>1$, then by the Fundamental Theorem of Arithmetic (every integer $>1$ has a prime divisor — cited in the definition box) pick a prime $r$ with $r\mid q$. From $n{q}^2=p^2$ we get $r\mid p^2$, and since $r$ is prime, $r\mid p$ (Euclid's lemma) — contradicting $\gcd (p,q)=1$. Hence $q=1$, giving $n=p^2$, a perfect square. Contrapositive: if $n$ is not a perfect square, no such $p/q$ exists, so $\sqrt{n}$ is irrational. ($\Rightarrow$) contrapositive: if $n=k^2$ is a perfect square, then $\sqrt{n}=k\in \mathbb{Z}\subset \mathbb{Q}$, so it is rational. Combining both directions gives the "iff". □

Consider $t={\sqrt{2}}^{\sqrt{2}}$. Either $t$ is rational or it is irrational.

• Case 1: $t$ is rational. Take $a=b=\sqrt{2}$ (both irrational). Done — $a^b=t$ is rational.
• Case 2: $t$ is irrational. Take $a=t={\sqrt{2}}^{\sqrt{2}}$ (irrational by assumption) and $b=\sqrt{2}$. Then $a^b={\left({\sqrt{2}}^{\sqrt{2}}\right)}^{\sqrt{2}}={\sqrt{2}}^{\sqrt{2}\cdot \sqrt{2}}={\sqrt{2}}^2=2\in \mathbb{Q}.$ In either case a valid pair $(a,b)$ exists. The beauty: we never had to decide which case is true! (This concerns Algebraic vs. Transcendental Numbers deeply, but the logic needs no such classification.)

First, two distinct ideas (do not confuse them):

• The Archimedean property: for any real $x$ there is a natural number $n$ with $n>x$ — "no real is bigger than every natural". This is what we use here.
• The Completeness Axiom (a different statement): $\mathbb{R}$ has no gaps — every bounded set has a least upper bound. It implies the Archimedean property but is not the same thing. We only need the Archimedean property below.

Recall $\lfloor x\rfloor$ = greatest integer $\le x$ (from the definition box). Step 1 — pick a fine enough denominator. Since $b-a>0$, by the Archimedean property choose $n\in \mathbb{N}$ with $n>\frac{1}{b-a}$, so $\frac{1}{n}<b-a$. Why: we want a grid of spacing $<$ the gap, so a grid point is forced inside (see the figure: the ruler ticks are closer together than the gap). Step 2 — pick the numerator. Let $m=\lfloor na\rfloor +1$, the first integer with $m>na$. Then $na<m\le na+1\Rightarrow \frac{m}{n}>a\text{and}\frac{m}{n}\le a+\frac{1}{n}<a+(b-a)=b.$ Step 3 — conclude. So $a<\frac{m}{n}<b$, and $\frac{m}{n}\in \mathbb{Q}$. Since $a,b$ were any reals (including irrationals), a rational sits between any two reals. □

One-line takeaway per level ::: L1 simplify before you classify; L2 run midpoint & $10^k$-shift procedures; L3 pick the right prime-divisibility / cancellation tool; L4 combine density + counting; L5 prove the general theorems. Does density of $\mathbb{Q}$ imply $\mathbb{Q}=\mathbb{R}$? ::: No — dense but countable; $\mathbb{R}$ is uncountable, so infinitely many irrational gaps remain.