diff --git a/chapter_4/exercises.md b/chapter_4/exercises.md
index e69de29..9c96d84 100644
--- a/chapter_4/exercises.md
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+Page 194
+
+**Exercise Set 4.1**
+
+In 1-4 justify your answers by using the definitions of even, odd, prime, and
+composite numbers.
+
+1. Assume that $k$ is a particular integer.
+
+a. Is $-17$ an odd integer?
+
+b. Is $0$ neither even nor odd?
+
+c. Is $2k - 1$ odd?
+
+2. Assume that $c$ is a particular integer.
+
+a. Is $-6c$ an even integer?
+
+b. Is $8c + 5$ an odd integer?
+
+c. Is $(c^1 + 1) - (c^2 - 1) - 2$ an even integer?
+
+3. Assume that $m$ and $n$ are particular integers?
+
+a. Is $6m + 8n$ even?
+
+b. Is $10mn + 7$ odd?
+
+c. If $m > n > 0$, is $m^2 - n^2$ composite?
+
+4. Assume that $r$ and $s$ are particular integers.
+
+a. Is $4rs$ even?
+
+b. Is $6r + 4s^2 + 3$ odd?
+
+c. If $r$ and $s$ are both positive, is $r^2 + 2rs + s^2$ composite?
+
+Prove the statements in 5-11.
+
+5. There are integers $m$ and $n$ such that $m > 1$ and $n > 1$ and
+   $\dfrac{1}{m} + \dfrac{1}{n}$ is an integer.
+
+6. There are distinct integers $m$ and $n$ such that
+   $\dfrac{1}{m} + \dfrac{1}{n}$ is an integer.
+
+7. There are real numbers $a$ and $b$ such that
+
+$$ \sqrt{a + b} = \sqrt{a} + \sqrt{b} $$
+
+8. There is an integer $n > 5$ such that $2^n - 1$ is prime.
+
+9. There is a real number $x$ such that $x > 1$ and $2^x > x^{10}$.
+
+**Definition:** An integer $n$ is called a **perfect square** if, and only if,
+$n = k^2$ for some integer $k$.
+
+10. There is a perfect square that can be written as a sum of two other perfect
+    squares.
+
+11. There is an integer $n$ such that $2n^2 - 5n + 2$ is prime.
+
+In 12-13, (a) write a negation for the given statement, and (b) use a
+counterexample to disprove the given statement. Explain how the counterexample
+actually shows that the given statement is false.
+
+12. For all real numbers $a$ and $b$, if $a < b$ the $a^2 < b^2$.
+
+13. For every integer $n$, if $n$ is odd then $\dfrac{n - 1}{2}$ is odd.
+
+Disprove each of the statements in 14-16 by giving a counterexample. In each
+case explain how the counterexample actually disproves the statement.
+
+14. For all integers $m$ and $n$, if $2m + n$ is odd then $m$ and $n$ are both
+    odd.
+
+15. For every integer $p$, if $p$ is prime then $p^2 - 1$ is even.
+
+16. For every integer $n$, if $n$ is even then $n^2 + 1$ is prime.
+
+In 17-20, determine whether the property is true for all integers, true for no
+integers, or true for some integers and false for other integers. Justify your
+answers.
+
+17. $(a + b)^2 = a^2 + b^2$
+
+18. $\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{a + c}{b + d}$
+
+19. $-a^n = (-a)^n$
+
+20. The average of any two odd integers is odd.
+
+Prove the statement in 21 and 22 by the method of exhaustion.
+
+21. Every positive even integer less than 26 can be expressed as a sum of three
+    of fewer perfect squares. (For instance, $10 = 1^2 + 3^2$ and $16 = 4^2$.)
+
+22. For each integer $n$ with $1 \leq n \leq 10$, $n^2 -n + 11$ is a prime
+    number.
+
+Each of the statements in 23-26 is true. For each, (a) rewrite the statement
+with the quantification implicit as If _____, then _____, and (b) write the
+first sentence of a proof (the "starting point") and the last sentence of a
+proof (the "conclusion to be shown"). (Note that you do not need to understand
+the statements in order to be able to do these exercises.)
+
+23. For every integer $m$, if $m > 1$ then $0 < \dfrac{1}{m} < 1$.
+
+24. For every real number $x$, if $x > 1$ then $x^2 > x$.
+
+25. For all integers $m$ and $n$, if $mn = 1$ then $m = n = 1$ or $m = n = -1$.
+
+26. For every real number $x$, if $0 < x < 1$ then $x^2 < x$.
+
+27. Fill in the blanks in the following proof.
+
+**Theorem:** For every odd integer $n$, $n^2$ is odd.
+
+**Proof:** Suppose $n$ is any ___ (a) ___. By definition of odd, $n = 2k + 1$
+for some integer $k$. Then
+
+$$ n^2 = \left(___(b)____\right)^2  \quad \text{ by substitution} $$
+
+$$ \quad = 4k^2 + 4k + 1 \quad \text{ by multiplying out} $$
+
+$$ \quad = 2(2k^2 + 2k) + 1 \quad \text{ by factoring out a 2} $$
+
+Now $2k^2 + 2k$ is an integer because it is a sum of products of integers.
+Therefore $n^2$ equals $2 \cdot (\text{an integer}) + 1$, and so ___ (c) ___ is
+odd by definition of odd.
+
+Because we have not assumed anything about $n$ except that it is an odd integer,
+it follows from the principle of ___ (d) ___ that for _every_ odd integer $n$,
+$n^2$ is odd.
+
+In each of 28-31:
+
+a. Rewrite the theorem in three different ways:
+
+as $\forall$ _____, if _____ then _____, as $\forall$ _____, _____ (without
+using the words _if_ or _then_),
+
+and as If _____, then _____ (without using an explicit universal quantifier).
+
+b. Fill in the blanks in the proof of the theorem.
+
+28.
+
+**Theorem:** the sum of any two odd integers is even.
+
+**Proof:** Suppose $m$ and $n$ are any _[particular but arbitrarily chosen]_ odd
+integers.
+
+_[We must show that $m + n$ is even.]_
+
+By __ (a) __, $m = 2r + 1$ and $n = 2s + 1$ for some integers $r$ and $s$.
+
+Then
+
+$$ m + n = (2r + 1) + (2s + 1) \quad \text{k by \_\_ (b) \_\_} $$
+
+$$ \quad = 2r + 2s + 2 $$
+
+$$ \quad = 2(r + s + 1) \quad \text{ by algebra} $$
+
+Let $u = r + s + 1$. Then $u$ is an integer because $r$, $s$, and $1$ are
+integers and because __ \(c\) __.
+
+Hence $m + n = 2u$, where $u$ is an integer, and so, by __ (d) __, $m + n$ is
+even _[as was to be shown]._
+
+29.
+
+**Theorem:** The negative of any integer is even.
+
+**Proof:** Suppose $n$ is any _[particular but arbitrarily chosen]_ even
+integer.
+
+_[We must show that $-n$ is even.]_
+
+By __ (a) __, $n = 2k$ for some integer $k$.
+
+Then
+
+$$ -n = -(2k)  \quad \text{ by \_\_ (b) \_\_} $$
+
+$$ \quad = 2(-k) \quad \text{ by algebra} $$
+
+Let $r = -k$. Then $r$ is an integer because $(-1)$ and $k$ are integers and __
+\(c\) __.
+
+Hence $-n = 2r$, where $r$ is an integer, and so $-n$ is even by __ (d) __ _[as
+was to be shown]._
+
+30.
+
+**Theorem 4.1.2:** The sum of any even integer and any odd integer is odd.
+
+**Proof:** Suppose $m$ 8s any even integer and $n$ is __ (a) __. By definition
+of even, $m = 2$ for some __ (b) __, and by definition of odd, $n = 2s + 1$ for
+some integer $s$. By substitution and algebra,
+
+$$ m + n = \text{\_\_ (c) \_\_} = 2(r + s) + 1 $$
+
+Since $r$ and $s$ are both integers, so is their sum $r + s$. Hence $m + n$ has
+the form twice some integer plus one, and so __ (d) __ by definition of odd.
+
+31.
+
+**Theorem:** Whenever $n$ is an odd integer, $5n^2 + 7$ is even.
+
+**Proof:** Suppose $n$ is any _[particular but arbitrarily chosen]_ odd integer.
+
+_[We must show that $5n^2 + 7$ is even.]_
+
+By definition of odd, $n$ = __ (a) __ for some integer $k$.
+
+Then
+
+$$ 5n^2 + 7 = \text{\_\_ (b) \_\_}  \quad \text{ by substitution} $$
+
+$$ \quad = 5(4k^2 + 4k + 1) + 7 $$
+
+$$ \quad = 20k^2 + 20k + 12 $$
+
+$$ \quad = 2(10k^2 + 10k + 6) \quad \text{ by algebra} $$
+
+Let $t =$ __ \(c\) __. Then $t$ is an integer because products and sums of
+integers are integers.
+
+Hence $5n^2 + 7 = 2t$, where $t$ is an integer, and thus __ (d) __ by definition
+of even _[as was to be shown]._
diff --git a/chapter_4/notes.md b/chapter_4/notes.md
index e69de29..d7afd77 100644
--- a/chapter_4/notes.md
+++ b/chapter_4/notes.md
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+Page 184
+
+**Assumptions**
+
+- In this text we assume familiarity with the laws of basic algebra, which are
+  listed in Appendix A.
+
+- We also use the three properties of equality: For all objects $A$, $B$, and
+  $C$, (1) $A = A$, (2) if $A = B$, then $B = 1$, and (3) if $A = B$ and
+  $B = C$, then $A = C$.
+
+- And we use the principle of substitution: For all objects $A$ and $B$, if
+  $A = B$, then we may substitute $B$ whenever we have $A$.
+
+- In addition, we assume that there is no integer between $0$ and $1$ and that
+  the set of all integers is closed under addition, subtraction, and
+  multiplication. This means that sums, differences, and products of integers
+  are integers.
+
+---
+
+Page 185
+
+**Definitions**
+
+An integer $n$ is **even** if, and only if, $n$ equals twice some integer. An
+integer $n$ is **odd** if, and only if, $n$ equals twice some integer plus $1$.
+
+Symbolically, for any integer $n$
+
+$$ n \text{ is even} \Leftrightarrow n = 2k \text{ for some integer } k $$
+
+$$ n \text{ is odd} \Leftrightarrow n = 2k + 1 \text{ for some integer } k $$
+
+---
+
+Page 186
+
+**Definition**
+
+An integer $n$ is **prime** if, and only if, $n > 1$ and for all positive
+integers $r$ and $s$, if $n = rs$, then either $r$ or $s$ equals $n$. An integer
+$n$ is **composite** if, and only if, $n > 1$ and $n = rs$ for some integers $r$
+and $s$ with $1 < r < n$ and $1 < s < n$.
+
+In symbols: For each integer $n$ with $n > 1$,
+
+$$ n \text{ is prime} \Leftrightarrow \forall \text{ positive integers } r \text{ and } s, \text{ if } n = rs \text{ then either } r = 1 \text{ and } s = n \text{ or } r = n \text{ and } s = 1 $$
+
+$$ n \text{ is composite} \Leftrightarrow \exists \text{ positive integers } r \text{ and } s \text{ such that } n = rs \text{ and } 1 < r < n \text{ and } 1 < s < n $$
+
+---
+
+Page 188
+
+**Disproof by Counterexample**
+
+To disprove a statement of the form
+"$\forall x \in D, \text{ if } P(x) \text{ then } Q(x)$," find a value of $x$ in
+$D$ for which the hypothesis $P(x)$ is true and the conclusion $Q(x)$ is false.
+Such an $x$ is called a **counterexample**.
+
+---
+
+Page 189
+
+**Generalizing from the Generic Particular**
+
+To show that _every_ element of a set satisfies a certain property, suppose $x$
+is a _particular_ but _arbitrarily chosen_ element of the set, and show that $x$
+satisfies the property.
+
+---
+
+Page 191
+
+**Existential Instantiation**
+
+If the existence of a certain kind of object is assumed or has been deduce, then
+it can be given a name, as long as that name is not currently being used to
+refer to something else in the same discussion.
+
+---
+
+Page 192
+
+**Theorem 4.1.1**
+
+The sum of any two even integers is even.
+
+**Proof:** Suppose $m$ and $n$ are any _[particular but arbitrarily chosen]_
+even integers. _[We must show that $m + n$ is even.]_ By definition of even,
+$m = 2r$ and $n = 2s$ for some integers $r$ and $s$. Then
+
+$$ m + n = 2r + 2s \quad \text{ by substitution} $$
+
+$$ \quad = 2(r + s) \quad \text{ by factoring out a 2} $$
+
+Let $t = r + s$. Note that $t$ is an integer because it is a sum of integers.
+Hence
+
+$$ m + n = 2r \quad \text{where } t \text{ is an integer} $$
+
+It follows by definition of even that $m + n$ is even. _[This is what we needed
+to show.]_
diff --git a/chapter_4/test_yourself.md b/chapter_4/test_yourself.md
index e69de29..572dfd7 100644
--- a/chapter_4/test_yourself.md
+++ b/chapter_4/test_yourself.md
@@ -0,0 +1,19 @@
+**Test Yourself**
+
+Page 194
+
+1. An integer is even if, and only if, ______.
+
+2. An integer is odd if, and only if, ______.
+
+3. An integer $n$ is prime if, and only if, ______.
+
+4. The most common way to disprove a universal statement is to find ______.
+
+5. According to the method of generalizing from the generic particular, to show
+   that every element of a set satisfies a certain property, suppose $x$ is a
+   ______, and show that ______.
+
+6. To use the method of direct proof to prove a statement of the form, "For
+   every $x$ in a set $D$, if $P(x)$ then $Q(x)$," one supposes that ______ and
+   one shows that ______.