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+# 1.1.6 Additional Exercises
+
+1.
+
+Q: Suppose $P$ and $Q$ are the statements: $P$: Jack passed math. $Q$: Jill
+passed math.
+
+(a) Translate "Jack and Jill both passed math" into symbols.
+
+(b) Translate "If Jack passed math, then Jill did not" into symbols.
+
+\(c\) Translate "$P \vee Q$" into English
+
+(d) Translate "$\neg(P \wedge Q) \to Q$" into English.
+
+(e) Suppose you know that if Jack passed math, then so did Jill. What can you
+conclude if you know that:
+
+i. Jill passed math?
+
+ii. Jill did not pass math?
+
+A:
+
+(a) Translate "Jack and Jill both passed math" into symbols.
+
+$$ P \wedge Q $$
+
+(b) Translate "If Jack passed math, then Jill did not" into symbols.
+
+$$ P \to \neg Q $$
+
+\(c\) Translate "$P \vee Q$" into English
+
+"Either Jack passed math or Jill passed math or they both passed math."
+
+(d) Translate "$\neg(P \wedge Q) \to Q$" into English.
+
+"If it is true that Jack did not pass math or Jill did not pass math or both did
+not pass math, then Jill passed math."
+
+(e) Suppose you know that if Jack passed math, then so did Jill. What can you
+conclude if you know that:
+
+i. Jill passed math?
+
+$$ P \to Q $$
+
+| $P$ | $Q$ | $P \to Q$ |
+| --- | --- | --------- |
+| T   | T   | T         |
+| T   | F   | F         |
+| F   | T   | T         |
+| F   | F   | T         |
+
+While it is true that Jack passed math:
+
+| $P$ | $Q$ | $P \to Q$ |
+| --- | --- | --------- |
+| T   | T   | T         |
+
+It is also possible that Jack did not pass math:
+
+| $P$ | $Q$ | $P \to Q$ |
+| --- | --- | --------- |
+| F   | T   | T         |
+
+Therefore we can say that while it is true that "If Jack passed math, then Jill
+passed math." And it is also true that "Jill passed math.", we cannot know if
+"Jack passed math" is a true or false statement.
+
+ii. Jill did not pass math?
+
+$$ P \to Q $$
+
+| $P$ | $Q$ | $P \to Q$ |
+| --- | --- | --------- |
+| T   | F   | F         |
+| F   | F   | T         |
+
+So given our first assumption, which is that if "Jack passed math, then Jill
+passed math", $P \to Q$, is a true statement. We are also told by part ii that
+Jill did not pass math. Given this, and given that we know that $P \to Q$ is
+true, we can only therefore take the last row:
+
+| $P$ | $Q$ | $P \to Q$ |
+| --- | --- | --------- |
+| F   | F   | T         |
+
+So we know that Jack did not pass math.
+
+2.
+
+Q: Translate into symbols. Use $E(x)$ for "$x$ is even" and $O(x)$ for "$x$ is
+odd."
+
+(a) No number is both even and odd.
+
+(b) One more than any even number is an odd number.
+
+\(c\) There is a prime number that is even.
+
+(d) Between any two numbers there is a third number.
+
+(e) There is no number between a number and one more than that number.
+
+A:
+
+(a) No number is both even and odd.
+
+$$ \forall x \neg (E(x) \wedge O(x)) $$
+
+Also:
+
+$$ \neg \exists x (E(x) \wedge O(x)) $$
+
+(b) One more than any even number is an odd number.
+
+$$ \forall x (E(x) \to O(x + 1)) $$
+
+\(c\) There is a prime number that is even.
+
+Say that $P(x)$ symbolizes "$x$ is prime." then:
+
+$$ \exists x (P(x) \wedge E(x)) $$
+
+(d) Between any two numbers there is a third number.
+
+This one goes outside the scope of the section a bit, but you can say:
+
+$$ \forall x \forall z \left(x < z \to \exists y(x < y < z)\right) $$
+
+(e) There is no number between a number and one more than that number.
+
+$$ \forall x \neg \exists y (x < y < x + 1) $$
+
+3.
+
+Q: For each of the statements below, give a domain of discourse for which the
+statement is true, and a domain for which the statement is false.
+
+(a) $\forall x \exists y \left(y^2 = x\right)$.
+
+(b) $\forall x \forall y \left(x < y \to \exists z(x < z < y)\right)$.
+
+\(c\) $\exists x \forall y \forall z(y < z \to y \leq x \leq z)$.
+
+A:
+
+(a) $\forall x \exists y \left(y^2 = x\right)$.
+
+The domain of discourse for the statement is true for all real numbers greater
+than or equal to $0$.
+
+$$ \text{DOMAIN WHERE TRUE: } \forall x \exists y \left(y^2 = x\right), x \in \mathbb{R_{\geq 0}} $$
+
+The domain of discourse for this statement is false for all real numbers (or
+integers) less than $0$
+
+$$ \text{DOMAIN WHERE FALSE: } \forall x \exists y \left(y^2 = x\right), x \in \mathbb{R_{<0}} $$
+
+(b) $\forall x \forall y \left(x < y \to \exists z(x < z < y)\right)$.
+
+This is true for the domain of all real numbers:
+
+$$ x, y, z \in \mathbb{R} $$
+
+This is false for the domain of all whole integers, since there is no number
+between, say $2$ and $3$ in $\mathbb{Z}$:
+
+$$ x, y, z \in \mathbb{Z} $$
+
+\(c\) $\exists x \forall y \forall z(y < z \to y \leq x \leq z)$.
+
+This statement is true in any domain with at most one element, and false in any
+ordered set with at least two distinct elements.
+
+Examples:
+
+True in: $\{0\}$ False in: $\mathbb{Z}, \mathbb{R}$