1. **Show that the set of rational numbers $\mathbb{Q}$ is a Field.** A field is a set equipped with two operations (addition and multiplication) satisfying certain axioms: closure, associativity, commutativity, distributivity, existence of identity elements, and existence of inverses (except for zero in multiplication). - **Closure:** For any $a, b \in \mathbb{Q}$, both $a+b$ and $a\cdot b$ are rational. - **Associativity:** Addition and multiplication are associative. - **Commutativity:** Addition and multiplication are commutative. - **Distributivity:** Multiplication distributes over addition. - **Identity elements:** $0$ is the additive identity, $1$ is the multiplicative identity. - **Inverses:** For any $a \in \mathbb{Q}$, there exists $-a$ such that $a + (-a) = 0$; for any $a \neq 0$, there exists $a^{-1}$ such that $a \cdot a^{-1} = 1$. All these properties hold for rational numbers by their definition and arithmetic properties. 2. **If $0 < a < b$, prove that $\frac{1}{b} < \frac{1}{a}$ using only the field and order axioms.** - Since $0 < a < b$, subtracting $a$ gives $0 < b - a$. - Multiply both sides by $\frac{1}{ab} > 0$ (since $a,b>0$), preserving inequality: $$0 < (b - a) \cdot \frac{1}{ab} = \frac{b}{ab} - \frac{a}{ab} = \frac{1}{a} - \frac{1}{b}$$ - Thus, $\frac{1}{a} - \frac{1}{b} > 0 \implies \frac{1}{b} < \frac{1}{a}$. 3. **If $a < b$ and $0 < a < b$, show that $a^2 < ab < b^2$.** - Since $0 < a < b$, multiply $a < b$ by $a > 0$: $$a \cdot a < a \cdot b \implies a^2 < ab$$ - Multiply $a < b$ by $b > 0$: $$a \cdot b < b \cdot b \implies ab < b^2$$ - Combining: $$a^2 < ab < b^2$$ 4. **Let $D = \{x \in \mathbb{R} \mid x = 2 - \frac{1}{n}, n \in \mathbb{N}\}$. Find:** (a) Maximum: - As $n$ increases, $\frac{1}{n} \to 0$, so $x \to 2$ from below. - For $n=1$, $x = 2 - 1 = 1$ (smallest element). - The largest element is when $n=1$, which is $1$, but this is the smallest, so check carefully. - Actually, as $n$ increases, $x$ increases towards $2$ but never reaches $2$. - So the maximum is the largest element in the set, which is $2 - \frac{1}{1} = 1$? No, this is the smallest. - The set is increasing: $2 - 1, 2 - \frac{1}{2}, 2 - \frac{1}{3}, ...$ so the maximum is the limit $2$ but not attained. - So no maximum. (b) Infimum: - The smallest element is $1$ (when $n=1$), so infimum is $1$. (c) Supremum: - The supremum is the least upper bound, which is $2$. (d) Is the infimum attained? - Yes, since $1 \in D$ (for $n=1$). 5. **Let $E = \{(-1)^n + \frac{1}{n} \mid n \in \mathbb{N}\}$. Determine:** (a) sup $E$: - For even $n$, $(-1)^n = 1$, so terms are $1 + \frac{1}{n} \to 1$ from above. - For odd $n$, $(-1)^n = -1$, terms are $-1 + \frac{1}{n} \to -1$ from above. - The largest term is when $n=2$: $1 + \frac{1}{2} = 1.5$. - So sup $E = 1.5$. (b) inf $E$: - The smallest term is when $n=1$: $-1 + 1 = 0$. - But for odd $n$, terms approach $-1$ from above. - So infimum is $-1$. (c) Maximum or minimum: - Maximum is $1.5$ (attained at $n=2$). - Minimum is $0$ (attained at $n=1$). (d) Accumulation points: - The subsequence for even $n$ tends to $1$. - The subsequence for odd $n$ tends to $-1$. - So accumulation points are $\{1, -1\}$. 6. **Find a rational number between $\sqrt{10}$ and $\pi$.** - Since $\sqrt{10} \approx 3.162$ and $\pi \approx 3.1415$, note $\pi < \sqrt{10}$. - So between $\pi$ and $\sqrt{10}$ means between $3.1415$ and $3.162$. - By density of rationals, there exists a rational number, e.g., $\frac{157}{50} = 3.14$ (too small), try $\frac{316}{100} = 3.16$ (too big). - Choose $\frac{315}{100} = 3.15$ which lies between $3.1415$ and $3.162$. 7. **Let $S = \{\frac{n}{10} \mid n \in \mathbb{Z}\}$.** (a) Is $S$ dense in $\mathbb{R}$? - No, because the distance between consecutive elements is $0.1$, so there are gaps. (b) Is $S$ dense in any open interval (e.g., $(0,1)$)? - No, same reason: gaps of $0.1$ prevent density. (c) Find an element of $S$ in $(1.1, 1.2)$: - $\frac{12}{10} = 1.2$ is not in the open interval. - $\frac{11}{10} = 1.1$ not in open interval. - No element strictly inside $(1.1,1.2)$ because elements are spaced by $0.1$. - So no element of $S$ lies strictly inside $(1.1,1.2)$. 8. **Let $d(x,y) = \frac{|x-y|}{1+|x-y|}$. Prove $d$ is a metric on $\mathbb{R}$.** - **Non-negativity:** $d(x,y) \geq 0$ since numerator and denominator positive. - **Identity:** $d(x,y) = 0 \iff |x-y|=0 \iff x=y$. - **Symmetry:** $d(x,y) = d(y,x)$ since $|x-y|=|y-x|$. - **Triangle inequality:** $$d(x,z) = \frac{|x-z|}{1+|x-z|} \leq \frac{|x-y| + |y-z|}{1 + |x-y| + |y-z|} < \frac{|x-y|}{1+|x-y|} + \frac{|y-z|}{1+|y-z|} = d(x,y) + d(y,z)$$ This uses the fact that $f(t) = \frac{t}{1+t}$ is increasing and subadditive in this form. Hence, $d$ is a metric. Final answer for the first problem only as per instructions.