---
tags: 225-spr22, mth225
---

# Partial Answer Key for Learning Target Quiz 4

## Learning Targets 1--5

All of these can be checked using online tools except for Learning Target 3. The latter had no unsuccessful attempts on the latest quiz.

## Learning Target 6

1. False, False, True, False, Undetermined.
2. (a) $\forall x (\neg P(x))$
(b) Some even numbers are not prime. (Or: There exists an even number that is not prime.)

## Learning Target 7

1. (a) $\{0, 1, 8, 64\}$
(b) $\{4\}$

2. There is more than one correct answer.
(a) $\{ 10^x \, : \, x \in \mathbb{N} \}$
(b) $\{ 2n+4 \, : \, n \in \{0,1,2\}\}$

3. (a) Technically false since $C$, the set, is not an element of $A$. But I threw this one out because it's too tricky to distinguish $c$ the element from $C$ the set.
(b) True
(c) True
(d) True
(e) False

## Learning Target 8

1. 11
1. $\{0,1,2,...,10\}$
1. $\{1, 5, 9\}$
1. $\{(0,u), (0,v), (1,u), (1,v), (2,u), (2,v)\}$
1. 8
1. $\{3,4,...,10\}$

## Learning Target 9

1. It's a function - domain $\{1,2,3\}$, codomain $\mathbb{N}$, range $\{0,10,100\}$
1. Not a function because 4 doesn't map to anything
1. It's a function - domain is the set of all possible strings, codomain $\mathbb{Z}$ or $\mathbb{N}$, range $\mathbb{N}$

## Learning Target 10

1. Neither injective nor surjective. Make a table of inputs and outputs:

| $x$ | $1$ | $2$ | $3$ | $4$ | $5$|
| -------- | -------- | -------- | --| -- | -- |
| $f(x)$ | $1$ | $2$ | $2$ | $3$ | $3$|

We see that $2$ and $3$ map to $2$ (not injective) and $4$ in the codomain has nothing mapping to it (not surjective).

2. Surjective, but not injective because for example `1001` and `1100` both map to 2.

3. Bijective.

## Learning Target 11

1. (a) $\{2,4\}$
(b) $\{1,3\}$

2. All of these are "DNE".

## Learning Target 12

Can be checked with a calculator.

## Learning Target 13

1. There are $2^{12}$ binary strings of length 12 because every such string involves choosing 2 bits for 12 positions, then use the Multiplicative Principle.
2. Having `1` in the leftmost bit and `00` in the rightmost two bits locks in three of 12 bits and leaves the other 9 freely chooseable. As in part 1 that leads to $2^9$ possible bitstrings.
3. There are $2^9$ bitstrings with three `1` bits on the left. There are $2^{10}$ bitstrings with two `1` bits on the right. **And there are $2^7$ bitstrings that satisfy both conditions.** Therefore the Inclusion/Exclusion Principle says the total number that satisfy either condition is $2^9 + 2^{10} - 2^7$.