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). **Step 1: Closure** - For any $a, b \in \mathbb{Q}$, $a + b$ and $a \times b$ are also in $\mathbb{Q}$ because sums and products of rational numbers are rational. **Step 2: Associativity and Commutativity** - Addition and multiplication of rational numbers are associative and commutative. **Step 3: Identity Elements** - Additive identity is $0 \in \mathbb{Q}$ since $a + 0 = a$. - Multiplicative identity is $1 \in \mathbb{Q}$ since $a \times 1 = a$. **Step 4: Inverses** - For every $a \in \mathbb{Q}$, there exists $-a \in \mathbb{Q}$ such that $a + (-a) = 0$. - For every $a \in \mathbb{Q} \setminus \{0\}$, there exists $a^{-1} = \frac{1}{a} \in \mathbb{Q}$ such that $a \times a^{-1} = 1$. **Step 5: Distributivity** - Multiplication distributes over addition: $a \times (b + c) = a \times b + a \times c$. Since all these axioms hold, $\mathbb{Q}$ is a field. 2. **If $0 < a < b$, prove that $\frac{1}{b} < \frac{1}{a}$ using only the field and order axioms.** **Step 1: Given $0 < a < b$, subtract $a$ from all parts:** $$0 < b - a$$ **Step 2: Since $a, b > 0$, multiply both sides of $a < b$ by $\frac{1}{ab}$ (positive, so inequality direction preserved):** $$\frac{a}{ab} < \frac{b}{ab}$$ **Step 3: Simplify:** $$\frac{1}{b} < \frac{1}{a}$$ This completes the proof. 3. **If $a < b$ and $0 < a < b$, show that $a^2 < ab < b^2$.** **Step 1: Since $0 < a < b$, multiply $a < b$ by $a > 0$:** $$a \times a < a \times b \Rightarrow a^2 < ab$$ **Step 2: Multiply $a < b$ by $b > 0$:** $$a \times b < b \times b \Rightarrow ab < b^2$$ **Step 3: Combine inequalities:** $$a^2 < ab < b^2$$ 4. **Let $D = \{x \in \mathbb{R} \mid x = 2 - \frac{1}{n}, n \in \mathbb{N}\}$. Find:** (a) Maximum - As $n \to 1$, $x = 2 - 1 = 1$. - As $n \to \infty$, $x \to 2$ from below. - So the maximum is $x = 1$ when $n=1$. (b) Infimum - The set approaches $2$ from below but never reaches it. - The infimum is the greatest lower bound, which is $1$. (c) Supremum - The supremum is the least upper bound, which is $2$. (d) Is the infimum attained? - Yes, the infimum $1$ is attained at $n=1$. 5. **Let $E = \{(-1)^n + \frac{1}{n} \mid n \in \mathbb{N}\}$. Determine:** (a) $\sup E$ - For even $n$, $(-1)^n = 1$, so $E$ contains values close to $1 + \frac{1}{n}$ approaching $1$ from above. - For odd $n$, $(-1)^n = -1$, so values approach $-1$ from above. - The supremum is $1 + 1/2 = 1.5$ (largest value at $n=2$). (b) $\inf E$ - The infimum is $-1$ (limit of odd terms as $n \to \infty$). (c) Maximum or minimum? - Maximum is $1.5$ at $n=2$. - No minimum because values approach $-1$ but never reach it. (d) Accumulation points - Accumulation points are $1$ and $-1$. 6. **Find a rational number between $\sqrt{10}$ and $\pi$.** - Since $\sqrt{10} \approx 3.162$ and $\pi \approx 3.1415$, note $\pi < \sqrt{10}$ is false. - Actually, $\pi < \sqrt{10}$ is false because $\pi \approx 3.1415 < 3.162$. - So $\pi < \sqrt{10}$. - By density of rationals, there exists $q \in \mathbb{Q}$ such that $\pi < q < \sqrt{10}$. - For example, $q = 3.15$ (which is $\frac{315}{100}$) lies between $\pi$ and $\sqrt{10}$. 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 as above. (c) Find an element of $S$ in $(1.1, 1.2)$ - $\frac{12}{10} = 1.2$ is at the boundary. - $\frac{11}{10} = 1.1$ is at the boundary. - No element strictly inside, but $1.15 = \frac{23}{20}$ is not in $S$. - So no element strictly inside, but $1.1$ and $1.2$ are in $S$. 8. **Let $d(x,y) = \frac{|x-y|}{1 + |x-y|}$. Prove $d$ is a metric on $\mathbb{R}$.** **Step 1: Non-negativity and identity of indiscernibles** - $d(x,y) \geq 0$ since numerator and denominator are positive. - $d(x,y) = 0 \iff |x-y|=0 \iff x=y$. **Step 2: Symmetry** - $d(x,y) = \frac{|x-y|}{1+|x-y|} = \frac{|y-x|}{1+|y-x|} = d(y,x)$. **Step 3: Triangle inequality** - Use $|x-z| \leq |x-y| + |y-z|$. - Then $$d(x,z) = \frac{|x-z|}{1+|x-z|} \leq \frac{|x-y| + |y-z|}{1 + |x-y| + |y-z|}.$$ - Since $\frac{a}{1+a}$ is increasing for $a \geq 0$, and $$\frac{|x-y|}{1+|x-y|} + \frac{|y-z|}{1+|y-z|} \geq \frac{|x-y| + |y-z|}{1 + |x-y| + |y-z|},$$ - So triangle inequality holds. Hence, $d$ is a metric.