# Chapter 1
## Section 2
### 1

$(a+b)+(c+d)=(a+d)+(b+c)\,$
| $a+b+c+d=(a+d)+(b+c)\,$ | by associativity |
| $a+b+d+c=(a+d)+(b+c)\,$ | by commutativity |
| $a+d+b+c=(a+d)+(b+c)\,$ | by commutativity |
| $(a+d)+(b+c)=(a+d)+(b+c)\,$ | by associativity |

#### 1 (alternative)

$(a+b)+(c+d)=(a+d)+(b+c)\,$
| $=a+(b+(c+d))\,$ | by associativity |
| $=a+(b+(d+c))\,$ | by commutativity |
| $=a+(d+(b+c))\,$ | by associativity |
| $=a+(d+(b+c))\,$ | by associativity |

### 2

$(a+b)+(c+d)=(a+c)+(b+d)\,$
| $a+b+c+d=(a+c)+(b+d)\,$ | by associativity |
| $a+c+b+d=(a+c)+(b+d)\,$ | by commutativity |
| $(a+c)+(b+d)=(a+c)+(b+d)\,$ | by associativity |

#### 2 (alternative)

$(a+b)+(c+d)=(a+c)+(b+d)\,$
| $a+(b+(c+d))=(a+c)+(b+d)\,$ | by associativity |
| $a+((b+c)+d)=(a+c)+(b+d)\,$ | by associativity |
| $a+((c+b)+d)=(a+c)+(b+d)\,$ | by commutativity |
| $a+(c+(b+d))=(a+c)+(b+d)\,$ | by associativity |
| $(a+c)+(b+d)=(a+c)+(b+d)\,$ | by associativity |

### 3

$(a-b)+(c-d)=(a+c)+(-b-d)\,$
| $a-b+c-d=(a+c)+(-b-d)\,$ | by associativity |
| $a+c-b-d=(a+c)+(-b-d)\,$ | by commutativity |
| $(a+c)+(-b-d)=(a+c)+(-b-d)\,$ | by associativity |

### 14

$-2+x=4\,$
$x=4+2\,$
$x=6\,$

### 15

| $2-x=5\,$ | |
| $-x=3\,$ | |
| $x=-3\,$ | multiply both sides by -1. |

## Section 3
### 30, 31, 32, 33

Although these can be done by hand the exercises suggest that you derive a general formula for finding the final population figures. The following one will suffice for the stated problems, where $p\,$ is the initial population, $x\,$ is the scaling factor (doubles, triples, etc.), $y_{1}\,$ and $y_{2}\,$ are the first year and end year respectively, and $n\,$ is the number of years taken for the population to go up by $x\,$:
$p\cdot x^{({\frac {y_{2}-y_{1}}{n}})}$
For example, 32a:
$200000\cdot 3^{({\frac {2215-1915}{50}})}=200000\cdot 3^{6}=145800000$

## Section 4
### 8-15

Write out the powers of 2 not exceeding the largest number in the problem set. Systematically check to see if each power divides the given number, starting with the largest.

### 16-23

Write out the powers of 3 not exceeding the largest number in the problem set. Systematically check to see if each power divides the given number, starting with the largest.

### 24
#### a)

Examining the definitions first, if $a\equiv b{\pmod {5}}$ and $x\equiv y{\pmod {5}}$ then $a-b=5n\,$ and $x-y=5m\,$.
$a+x\equiv b+y{\pmod {5}}$ means that $(a+x)-(b+y)=5k\,$.
$(a+x)-(b+y)=(a-b)+(x-y)\,$
$(a+x)-(b+y)=5n+5m\,$
$(a+x)-(b+y)=5(n+m)\,$
Let $k=(n+m)\,$
$(a+x)-(b+y)=5k\,$

#### b)

Examine the definitions in part a) again. We have $a-b=5n\,$ and $x-y=5m\,$.
Solving for $a\,$ and $x\,$:
$a=5n+b\,$ and $x=5m+y\,$
$ax=(5n+b)(5m+y)\,$
$ax=25nm+5ny+5mb+by\,$
$ax=5(5nm+ny+mb)+by\,$
Let $t=5nm+ny+mb\,$.
$ax=5t+by\,$
$ax-by=5t+by-by\,$
$ax-by=5t\,$.

## Section 5
### 7
#### a)

120, 720, 5040 and 40320.

#### c)

${\binom {m}{n}}={\binom {m}{m-n}}$
${\frac {m!}{n!(m-n)!}}={\frac {m!}{(m-n)!(m-(m-n))!}}$,
${\frac {m!}{n!(m-n)!}}={\frac {m!}{(m-n)!(m-m+n)!}}$,
${\frac {m!}{n!(m-n)!}}={\frac {m!}{(m-n)!n!}}$

#### d)

${\binom {m}{n}}+{\binom {m}{n-1}}={\binom {m+1}{n}}$
${\frac {m!}{n!(m-n)!}}+{\frac {m!}{(n-1)!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}$
Multiply both sides of the equation by ${\frac {n}{n}}$ and ${\frac {(m-n+1)}{(m-n+1)}}$, cancelling unnecessary factors to 1. We do this in order to achieve the denominator $n!(m-n+1)!$:
$1\cdot {\frac {(m-n+1)}{(m-n+1)}}\cdot {\frac {m!}{n!(m-n)!}}+{\frac {n}{n}}\cdot 1\cdot {\frac {m!}{(n-1)!(m-n+1)!}}=1\cdot 1\cdot {\frac {(m+1)!}{n!(m-n+1)!}}$
Use the fact that $a!(a+1)=(a+1)!$ and $a(a-1)!=a!$ to achieve:
${\frac {m!(m-n+1)}{n!(m-n+1)!}}+{\frac {m!n}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}$
${\frac {m!(m-n+1)+m!n}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}$
${\frac {m!((m-n+1)+n)}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}$
${\frac {m!(m+1)}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}$
${\frac {(m+1)!}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}$
Hence,
${\binom {m}{n}}+{\binom {m}{n-1}}={\binom {m+1}{n}}$

## Section 6
### 1
#### a)

${\frac {2x-1}{3x+2}}=7\,$
$2x-1=7(3x+2)\,$
$2x-1=21x+14\,$
$-19x=15\,$
$x=-{\frac {15}{19}}\,$

### 2
#### a)

${\frac {1}{x+y}}-{\frac {1}{x-y}}={\frac {-2y}{x^{2}-y^{2}}}$
${\frac {(x-y)-(x+y)}{(x+y)(x-y)}}=...$
${\frac {x+(-x)+(-y)+(-y)}{x^{2}+yx-yx-y^{2}}}=...$
${\frac {-2y}{x^{2}-y^{2}}}={\frac {-2y}{x^{2}-y^{2}}}$

#### b)

${\frac {x^{3}-1}{x-1}}=1+x+x^{2}$
Recall from the chapter that if ${\frac {a}{b}}={\frac {c}{d}}$, then $ad=bc$.
$1\cdot (x^{3}-1)=(x-1)(1+x+x^{2})$
$x^{3}-1=x+x^{2}+x^{3}-1-x-x^{2}\,$
$x^{3}-1=x^{3}-1\,$