Problem 1.1. If is rational () and is irrational, prove that and are irrational.
Solution. For the sake of contradiction, suppose that . Then, as is a field, we have by (A5) and hence by (M1). But then , contradicting the fact that is irrational. Analogously, suppose that . As , there exists a multiplicative inverse . So by (M1), but , so . This is, once more, a contradiction, so we conclude that both and are irrational, as required.
Problem 1.2. Prove that there is no rational number whose square is .
Solution. Suppose, for the sake of contradiction, that there exists satisfying . Then write for and . Dividing the numerator and denominator by if necessary, we can assume that and are relatively prime. Then and hence . So divides . As is prime, Euclid’s lemma then implies that divides . So for some . So we find that and hence . So divides . But then, as does not divide , we must have that divides and hence divides by Euclid’s lemma. But then and share a common factor of , contradicting the fact that they were chosen to be relatively prime.
Problem 1.3. Prove Proposition 1.15.
Solution. (a) Let with , and suppose that . By (M5), there exists a multiplicative inverse of . We then find that (b) Applying part (a) with gives the result.
(c) Applying part (a) with gives the result.
(d) We apply part (c), replacing with (which is likewise non-zero) and with . We then find that .
Problem 1.4. Let be a nonempty subset of an ordered set; suppose is a lower bound of and is an upper bound of . Prove that .
Solution. As is nonempty, we may choose . Then since is a lower bound of , and since is an upper bound of . This reduces to four cases. If and , then . If and , then by the ordered-field axioms. If and , then substituting yields . Finally, if and , then substituting yields . So we conclude that , as required.
Problem 1.8. Prove that no order can be defined in the complex field that turns it into an ordered field. Hint : is a square.
Solution. Suppose, seeking a contradiction, that there exists an order on that turns it into an ordered field. Then , so by Proposition 1.18(d), we have . As by Proposition 2.28, we have . The ordered-field axioms then imply that and hence . This contradicts the trichotomy of order, as is a square and hence strictly greater than by Proposition 1.18(d).
Problem 1.12. If are complex, prove that Solution. We proceed by induction on . When , the statement reduces to . The case is given by Theorem 1.33(e). Suppose inductively that we have for a fixed . It then follows that which closes the induction.
Problem 1.13. If are complex, prove that Solution. Let . By the triangle inequality (Theorem 1.33(e)), we find that hence Analogously, one has hence from which it follows that We therefore find that so it follows that as required.
Problem 1.17. Prove that if and . Interpret this geometrically, as a statement of parallelograms.
Solution. Given , we obtain Geometrically, this is a statement about the -parallelogram in spanned by and . This parallelogram has diagonals and , two sides of length , and two sides of length . This statement asserts that the sum of the squared lengths of the diagonals is equal to the sum of the squared lengths of the four sides of the parallelogram.
Problem 1.18. If and , prove that there exists such that but . Is this also true if ?
Solution. If , then for any non-zero , one has . Suppose that . Then there exists such that . As dot products are invariant under permutation of indices, we can assume that . Now define . Then , but The result is no longer true if . Indeed, in , the dot product coincides exactly with multiplication of real scalars. But is a field and hence lacks zero divisors, so if and only if or . If , then again any will suffice. If , then only if .