- due February 5:
Rudin Chapter 1, Problems 1,3,5.
- [Exercise 1] If and are rational then so are and (since the rationals form a field). So if is rational and is real, then rational implies is rational. An irrational number is just a non-rational real number, so conversely if is irrational then must be irrational. Similarly if is rational then so is thus if is irrational then so is
- [Exercise 3]
- (a)
- If then exists and if then
- (b)
- Is (a) with
- (c)
- Multiply by so using associativity and definition of inverse.
- (d)
- The identity for gives by commutativity which means by the uniqueness of inverses.

- [Exercise 5] If is a set of real numbers which is bounded below
then is by definition a lower bound, i.e.
for all
and if
for any other lower bound We already
know that if it exists it is unique. Now if is bounded below then

is bounded above. Indeed if for all then for all which means for all Now, if is the least upper bound of it follows that is a lower bound for since

As noted above, if is any lower bound for then is an upper bound for so and This is the definition of so

- due February 11: Rudin Chapter 1, Problems 8, 9, 10 and Chapter 2 Problems 2,3,4.
- due February 18: Rudin Chapter 2, Problems 9a, 9b, 9c, 11.
- due February 25: Rudin Chapter 2, Problems 10, 12, 16, 22, 23, 25.
- due March 2: Rudin Chapter 2, Problems 19, 20, 21, 24, 26, 29.
- due March 9: Rudin Chapter 3, Problems 2, 7, 12, 16, 20, 21 (you should add the missing assumption that for all
- due March 30. Rudin Chapter 4, Problems 1, 4, 15.
- due April 6. Rudin Chapter 4, Problems 14, 18, 21. Chapter 5, Problems 1, 3, 4.
- due April 13. Rudin Chapter 5, Problems 15, 16, 22. Chapter 6, Problems 1, 5, 7.
- Nothing due April 20.
- due May 4. This is the last homework: Rudin Chapter 7, Problem 6, 9, 10, 16, 20.
- Nothing due May 11. The last homework was going to be a little project in
doing a piece of mathematics but it is too late given the fact
that there will be a final exam. To test yourself and see if you have
gained in understanding from the course, try to write all this out as
clearly as you can after you figure it out.
Read Rudin Chapter 9, ``The contraction principle'' on your own. Let be continuous and Lipschitz continuous in the second variable, meaning that there is a constant such that

Suppose that for some is a differentiable function, with satisfying the differential equation

- Show that is continuously differentiable on
and that it
satisfies the integral equation

- Conversely show that if is a continuous function which satisfies (4) then it is differentiable and satisfies (3).
- Fix some real number
and
show that

is closed with respect to the supremum metric. - For
and as above show that the map

defines a map from into and that this map satisfies

where is the constant in (2). - Conclude that for sufficiently small (depending only on and
but not there is a unique differentiable function on
satisfying

- No big hints for the last part! Show that, given
and under the same conditions as above there is a unque differentiable
function on satisfying the intial value problem

- Show that is continuously differentiable on
and that it
satisfies the integral equation

Richard B. Melrose 2004-05-18