---
tags: mth225
---

# Key for Learning Target quiz 3, 2021-10-08

## CA.1, CA.2, and SF.5

You can check these using calculators:

- [Base conversion calculator](https://www.rapidtables.com/convert/number/base-converter.html)
- [Binary arithmetic calculator](https://www.calculator.net/binary-calculator.html)
- [Wolfram|Alpha](https://wolframalpha.com) for the rest

## L.1

| Statement: | $P \rightarrow (Q \wedge R)$ | If a figure is a triangle then it has three sides |
| --- | --- | --- |
| Hypothesis | $P$ | A figure is a triangle |
| Conclusion | $Q \wedge R$ | The figure has three sides |
| Negation | $P \wedge \neg(Q \wedge R)$ | A figure is a triangle but does not have three sides. |
| Converse | $(Q \wedge R) \rightarrow P$ | If a figure has three sides, then it is a triangle. |
| Contrapositive | $\neg (Q \wedge R) \rightarrow \neg P$ | If a figure does not have three sides, it is not a triangle. |

Note, "but" and "and" in the negation are interchangeable.

## L.2

1.
| $P$ | $Q$ | $P \vee Q$ | $\neg (P \vee Q)$ |
| --- | --- | ---------- | ----------------- |
| T | T | T | F |
| T | F | T | F |
| F | T | T | F |
| F | F | F | T |

| $P$ | $Q$ | $\neg P$ | $\neg Q$ | $(\neg P) \wedge (\neg Q)$ |
| --- | --- | ---------- | ----------------- | ---- |
| T | T | F | F | F
| T | F | F | T | F
| F | T | T | F | F
| F | F | T | T | T

So the two statements are logically equivalent.

2.
| $P$ | $Q$ | $R$ | $P \rightarrow Q$ | $Q \rightarrow R$ | $(P \rightarrow Q) \wedge (Q \rightarrow R)$ |
| :--: | :--: | :--: | :--: | :--: | :--: |
| T | T | T | T | T | T |
| T | F | T | F | T | F |
| F | T | T | T | T | T |
| F | F | T | T | T | T |
| T | T | F | T | F | F |
| T | F | F | F | T | F |
| F | T | F | T | F | F |
| F | F | F | T | T | T |

## L.3

1. True, False, True, True
2. (a) False because not all integers are even, for example $n = 3$
(b) True, for example $P(2)$ is true
(c) False, because for example $Q(5)$ is false
3. No mathematics courses are online. (Other formulations are possible.)

## SF.1

1. (a) $\{2, 9, 16, 23, 30, \dots\}$
(b) $\{0,1,2,3,4,5,6\}$ (No repetitions!)
(c) $\{2,3,4,5,6,7,8,9,10\}$

2. Two possible correct answers:
- $\{10(x+1) \, : \, x \in \mathbb{N}\}$
- $\{x \in \{1,2,3,4,5,\dots\} : x \ \% \ 10 = 0\}$
There are lots of others.

3. (a) False
(b) True
(c) False
(d) False
(e) True
(f) False
(g) True

## SF.2

1. $\{1,2,4,6,8,9\}$
2. $\{9\}$
3. $\{1,9\}$
4. $\{0,1,2,3,4,6,8,10\}$
5. $\{2,4,5,6,7,8,9\}$
6. $\{0,1,2,3,4,5,6,7,8,10\}$
7. $7$
8. $\{\emptyset, \{9\}\}$

## SF.3

1. This is a function with domain = $\{1,2,3,4\}$, codomain = $\{x,y,z,t\}$, and range $\{x,y,t\}$.
2. This is not a function because $2$ maps to two different outputs.
3. This is a function with domain = $\{1,2,3,4\}$ and both codomain and range = $\{x,y,z,t\}$.

## SF.4

1. This function is not injective because $f(1) = f(2)$. The function is not surjective because nothing maps to $z$. Therefore it's not bijective.
2. This function is injective, surjective, and bijective.
3. This function is surjective, but it is not injective because for example $h(1.3) = h(1.2)$. So it's not bijective.

## C.1

1. There is 1 choice for the letter with 3 options, followed by 4 choices for the digit with 10 options each. By the Multiplicative Principle the total number of ID's available is $3 \times 10^4 = 30,000$.
2. The Principle of Inclusion/Exclusion says the number of patients with either disease is the number of patients with pneumonia, plus the number of patients with bonchitis, *minus the number of patients with both*. So that's $25 + 30 - 10 = 45$.