# Chapter 1Edit

## Section 2Edit

### 1Edit

${\displaystyle (a+b)+(c+d)=(a+d)+(b+c)\,}$

 ${\displaystyle a+b+c+d=(a+d)+(b+c)\,}$ by associativity ${\displaystyle a+b+d+c=(a+d)+(b+c)\,}$ by commutativity ${\displaystyle a+d+b+c=(a+d)+(b+c)\,}$ by commutativity ${\displaystyle (a+d)+(b+c)=(a+d)+(b+c)\,}$ by associativity

### 14Edit

${\displaystyle -2+x=4\,}$

${\displaystyle x=4+2\,}$

${\displaystyle x=6\,}$

### 15Edit

 ${\displaystyle 2-x=5\,}$ ${\displaystyle -x=3\,}$ ${\displaystyle x=-3\,}$ multiply both sides by -1.

## Section 3Edit

### 30, 31, 32, 33Edit

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 ${\displaystyle p\,}$  is the initial population, ${\displaystyle x\,}$  is the scaling factor (doubles, triples, etc.), ${\displaystyle y_{1}\,}$  and ${\displaystyle y_{2}\,}$  is the inital year and end year respectively, and ${\displaystyle n\,}$  is the number of years taken for the population to go up by ${\displaystyle x\,}$ :

${\displaystyle p\cdot x^{({\frac {y_{2}-y_{1}}{n}})}}$

For example, 32a:

${\displaystyle 200000\cdot 3^{({\frac {2215-1915}{50}})}=200000\cdot 3^{6}=145800000}$

## Section 4Edit

### 8-15Edit

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.

### 16-23Edit

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.

### 24Edit

#### a)Edit

Examining the definitions first, if ${\displaystyle a\equiv b{\pmod {5}}}$  and ${\displaystyle x\equiv y{\pmod {5}}}$  then ${\displaystyle a-b=5n\,}$  and ${\displaystyle x-y=5m\,}$ .

${\displaystyle a+x\equiv b+y{\pmod {5}}}$  means that ${\displaystyle (a+x)-(b+y)=5k\,}$ .

${\displaystyle (a+x)-(b+y)=(a-b)+(x-y)\,}$

${\displaystyle (a+x)-(b+y)=5n+5m\,}$

${\displaystyle (a+x)-(b+y)=5(n+m)\,}$

Let ${\displaystyle k=(n+m)\,}$

${\displaystyle (a+x)-(b+y)=5k\,}$

#### b)Edit

Examine the definitions in part a) again. We have ${\displaystyle a-b=5n\,}$  and ${\displaystyle x-y=5m\,}$ .

Solving for ${\displaystyle a\,}$  and ${\displaystyle x\,}$ :

${\displaystyle a=5n+b\,}$  and ${\displaystyle x=5m+y\,}$

${\displaystyle ax=(5n+b)(5m+y)\,}$

${\displaystyle ax=25nm+5ny+5mb+by\,}$

${\displaystyle ax=5(5nm+ny+mb)+by\,}$

Let ${\displaystyle t=5nm+ny+mb\,}$ .

${\displaystyle ax=5t+by\,}$

${\displaystyle ax-by=5t+by-by\,}$

${\displaystyle ax-by=5t\,}$ .

## Section 5Edit

### 7Edit

#### a)Edit

120, 720, 5040 and 40320.

#### c)Edit

${\displaystyle {\binom {m}{n}}={\binom {m}{m-n}}}$

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

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

${\displaystyle {\frac {m!}{n!(m-n)!}}={\frac {m!}{(m-n)!n!}}}$

#### d)Edit

${\displaystyle {\binom {m}{n}}+{\binom {m}{n-1}}={\binom {m+1}{n}}}$

${\displaystyle {\frac {m!}{n!(m-n)!}}+{\frac {m!}{(n-1)!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}$

Mutliply both sides of the equation by ${\displaystyle {\frac {n}{n}}}$  and ${\displaystyle {\frac {(m-n+1)}{(m-n+1)}}}$ , cancelling unecessary factors to 1. We do this in order to achieve the denominator ${\displaystyle n!(m-n+1)!}$ :

${\displaystyle 1\cdot {\frac {(m-n+1)}{(m-n+1)}}\cdot {\frac {m!}{n!(m-n)!}}+{\frac {n}{n}}\cdot 1\cdot {\frac {m!}{(n-1)!(m-n+1)!}}=1\cdot 1\cdot {\frac {(m+1)!}{n!(m-n+1)!}}}$

Use the fact that ${\displaystyle a!(a+1)=(a+1)!}$  and ${\displaystyle a(a-1)!=a!}$  to achieve:

${\displaystyle {\frac {m!(m-n+1)}{n!(m-n+1)!}}+{\frac {m!n}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}$

${\displaystyle {\frac {m!(m-n+1)+m!n}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}$

${\displaystyle {\frac {m!((m-n+1)+n)}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}$

${\displaystyle {\frac {m!(m+1)}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}$

${\displaystyle {\frac {(m+1)!}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}$

Hence,

${\displaystyle {\binom {m}{n}}+{\binom {m}{n-1}}={\binom {m+1}{n}}}$

## Section 6Edit

### 1Edit

#### a)Edit

${\displaystyle {\frac {2x-1}{3x+2}}=7\,}$

${\displaystyle 2x-1=7(3x+2)\,}$

${\displaystyle 2x-1=21x+14\,}$

${\displaystyle -19x=15\,}$

${\displaystyle x=-{\frac {15}{19}}\,}$

### 2Edit

#### a)Edit

${\displaystyle {\frac {1}{x+y}}-{\frac {1}{x-y}}={\frac {-2y}{x^{2}-y^{2}}}}$

${\displaystyle {\frac {(x-y)-(x+y)}{(x+y)(x-y)}}=...}$

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

${\displaystyle {\frac {-2y}{x^{2}-y^{2}}}={\frac {-2y}{x^{2}-y^{2}}}}$

#### b)Edit

${\displaystyle {\frac {x^{3}-1}{x-1}}=1+x+x^{2}}$

Recall from the chapter that if ${\displaystyle {\frac {a}{b}}={\frac {c}{d}}}$ , then ${\displaystyle ad=bc}$ .

${\displaystyle 1\cdot (x^{3}-1)=(x-1)(1+x+x^{2})}$

${\displaystyle x^{3}-1=x+x^{2}+x^{3}-1-x-x^{2}\,}$

${\displaystyle x^{3}-1=x^{3}-1\,}$