**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, Euclidâ€™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 Euclidâ€™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
.