# Chapter 1Edit

## Section 2Edit

### 1Edit

$(a + b) + (c + d) = (a + d) + (b + c) \,$

 $a + b + c + d = (a + d) + (b + c) \,$ 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

### 14Edit

$-2 + x = 4 \,$

$x = 4 + 2 \,$

$x = 6 \,$

### 15Edit

 $2 - x = 5 \,$ $-x = 3 \,$ $x = -3 \,$ 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 $p\,$ is the initial population, $x\,$ is the scaling factor (doubles, triples, etc.), $y_{1}\,$ and $y_{2}\,$ is the inital 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 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 $a \equiv b \pmod{5}$ 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)Edit

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 5Edit

### 7Edit

#### a)Edit

120, 720, 5040 and 40320.

#### c)Edit

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

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

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

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

#### d)Edit

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

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

Mutliply both sides of the equation by $\frac{n}{n}$ and $\frac{(m-n+1)}{(m-n+1)}$, cancelling unecessary 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 6Edit

### 1Edit

#### a)Edit

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

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

$2x-1 = 21x + 14 \,$

$-19x = 15 \,$

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

### 2Edit

#### a)Edit

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

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

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

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

#### b)Edit

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

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

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

$x^3-1 = x + x^2 + x^3 - 1 -x - x^2 \,$

$x^3-1 = x^3-1 \,$

Last modified on 22 November 2011, at 13:52