II. Fields & Order #
Problems on the field and order axioms, the least upper bound property, and the structures (like \(\mathbb{Q}[\sqrt{2}]\)) that satisfy them.
Problem 1 #
Define the set of real numbers
$$\mathbb{Q}[\sqrt{2}] = \{\, a + b \sqrt{2} \mid a, b \in \mathbb{Q} \,\}.$$Together with the usual addition and multiplication, \((\mathbb{Q}[\sqrt{2}], +, \cdot)\) forms a field. One could verify this fact directly by checking all the axioms of a field hold, but it turns out that is not necessary to work quite that hard. The key observation is that the set of real numbers \((\mathbb{R},+,\cdot)\) is itself a field and every element of \(\mathbb{Q}[\sqrt{2}]\) is a real number.
Question: Identify which field axioms must be explicitly checked to verify that \(\mathbb{Q}[\sqrt{2}]\) is indeed a field, and which axioms already hold by viewing \(\mathbb{Q}[\sqrt{2}]\) as a subset of \(\mathbb{R}\). It is not necessary to carry out the proofs.
Solutions: Hint · Solution · Long solution
Problem 2 #
Suppose that \((F, +, \cdot)\) is a field and \(x, y,\) and \(z\) are elements of \(F\). Show that
(a) if \(x \neq 0\) and \(xy = xz\), then \(y = z\);
(b) if \(x \neq 0\) and \(xy = x\), then \(y = 1\).
Solutions: Hint · Solution · Long solution
Problem 3 #
Suppose that \((F, +, \cdot)\) is an ordered field with \(x, y \in F\). Show that
(a) if \(x < y\) then \(-y < -x\);
(b) \(1 > 0\).
Solutions: Hint · Solution · Long solution
Problem 4 #
Can the set of complex numbers \(\mathbb{C}\) form an ordered field?
Hint: Consider the relationship between \(i\) and \(0\).
Solutions: Hint · Solution · Long solution
Problem 5 #
Consider a set \(S \subset \mathbb{R}\) and suppose that it has a lower bound. Define the set \(-S\) as \(\{\, -r \mid r \in S \,\}\).
(a) Define precisely the notion of a greatest lower bound for a set \(S\) and explain why one exists if \(S\) has a lower bound. The greatest lower bound for a set \(S\) is often written as \(\inf S\), the infimum of \(S\).
(b) Show that \(\inf S = -\sup(-S)\).
Solutions: Hint · Solution · Long solution
Problem 6 #
A map \(f:\mathbb{Q} \to \mathbb{Q}\) is called an additive homomorphism if it satisfies
$$f(a+b)=f(a)+f(b).$$What are all the additive homomorphisms from \(\mathbb{Q}\) to \(\mathbb{Q}\)?
Solutions: Hint · Solution · Long solution
Problem 8 #
Recall the field \(\mathbb{Q}[\sqrt{2}]\) from above. Since all of its elements are real numbers, it inherits the usual ordering from \(\mathbb{R}\) and it is easy to verify that \(\mathbb{Q}[\sqrt{2}]\) is also an ordered field. Find another way of ordering the elements of \(\mathbb{Q}[\sqrt{2}]\), with the new order denoted by \(\prec\), so that \((\mathbb{Q}[\sqrt{2}], \prec)\) is still an ordered field.
Solutions: Hint · Solution · Long solution