# Question 1
## i

$ax=a\,$
$a\cdot (x\cdot a^{-1})=a\cdot a^{-1}=1$
$a\cdot (a^{-1}\cdot x)=1$
$(a\cdot a^{-1})\cdot x)=1$
$1\cdot x=1$
$x=1\,$

## ii

$(x-y)(x+y)=(x\cdot (x-y)+y\cdot (x-y))$
$=(x^{2}-xy)+(xy-y^{2})\,$
$=x^{2}-xy+xy-y^{2}\,$
$=x^{2}-y^{2}\,$

## iii

$x^{2}=y^{2}\,$, then
$x^{2}-y^{2}=(x+y)(x-y)=0\,$
Either $(x+y)=0\,$, which means $x=-y\,$ or $(x-y)=0\,$, which means that $x=y\,$.

## iv

$x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})\,$
$=(x-y)x^{2}+(x-y)xy+(x-y)y^{2}\,$
$=(x^{3}-x^{2}y+x^{2}y-xy^{2}+xy^{2}-y^{3}\,$
$=x^{3}-y^{3}\,$

## v

$x^{n}-y^{n}=(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})\,$
$=x(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})-y(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})\,$
$=(x^{n}+x^{n-1}y+...+x^{2}y^{n-2}+xy^{n-1})-(x^{n-1}y+x^{n-2}y^{2}+...+xy^{n-1}+y^{n})\,$
$=x^{n}-y^{n}\,$
Note that the two middle terms do no appear to match, but if you follow the logical development in each term, you will see that there is a matching additive inverse for each term on both sides.

## vi

Use the same method as iv, expand the expression and cancel.

# Question 2

$x^{2}=xy\,$ implies that $x=y\,$ and thus $x-y=0\,$. Step 4 requires division by $x-y=0\,$ and thus is an invalid step.

# Question 4
## ii

All values of x satisfy the inequality, since it can be rewritten as $-3<x^{2}\,$, and $x^{2}>=0\,$ for all x.

## iii

If $5-x^{2}<-2\,$, then $x^{2}>7\,$.
Thus, $x>{\sqrt {7}}\,$, or $x<-{\sqrt {7}}\,$.

## v

$x^{2}-2x+2$ cannot be factored in its current form, so we first turn a part of the expression into a perfect square:
    $(x^{2}-2x+1)+1>0\,$
We then complete the square on $x^{2}-2x+1=(x-1)^{2}\,$, so now we have the expression:
    $(x-1)^{2}+1>0\,$, which is positive for all values of x.

## vi

If $x^{2}+x+1>2\,$, then $x^{2}+x-1>0\,$.
$x^{2}+x-1=\left(x+{\frac {1+{\sqrt {5}}}{2}}\right)\left(x-{\frac {{\sqrt {5}}-1}{2}}\right)$
Thus $x<-{\frac {1+{\sqrt {5}}}{2}}$, or $x>{\frac {{\sqrt {5}}-1}{2}}$.

## viii

Complete the square:
$x^{2}+x+1=x^{2}+x+{\frac {1}{4}}+{\frac {3}{4}}=\left(x+{\frac {1}{2}}\right)^{2}+{\frac {3}{4}}$
Thus, $\left(x+{\frac {1}{2}}\right)^{2}+{\frac {3}{4}}>0$.

## ix

Solve for $(x-\pi )(x+5)>0\,$ first, then consider the third factor $(x-3)\,$ for both cases, giving us $-5<x<3\,$, or $x>\pi \,$

## x

$x>{\sqrt {2}}$, or $x<{\sqrt[{3}]{2}}$

## xi

If $2^{x}<8\,$, then taking the base 2 logarithm on both sides:
$log_{2}2^{x}<log_{2}8=x<3\,$

## xii

$x<1\,$

## xiii

NOTE: The answer in the 3rd Edition provides $0<x<1$ or $x>1$. Plugging in values $x>1$, e.g. 10, gives us $1/10-1/9<0$, so I think this is a misprint or an incorrect answer.
If ${\frac {1}{x}}+{\frac {1}{1-x}}>0$, then ${\frac {1}{x-x^{2}}}>0$, which is only positive when $0<x<1$ since $x^{2}>x$.

# Question 5
## i

$a+c<b+d\,$
$=(b+d)-(a+c)>0\,$
$=(b-a)+(d-c)>0\,$, which is true since $(b-a),(d-c)>0\,$

## ii

$b-a>0\,$
$=-a+b>0\,$
$=-a-(-b)>0\,$
$=-b<-a\,$

## iv

$(b-a),c>0\,$, therefore $c(b-a)>0\,$.
$c(b-a)=bc-ac\,$, therefore $ac<bc\,$.

## vi

$a>1\,$, therefore $1-a>0\,$.
$a(1-a)>0\,$
$=a^{2}-a>0\,$
$=a^{2}>a\,$

## viii

If $a$ or $c$ are 0, then by definition $ac<bd$.
Otherwise, we have already proved that $ac<bc$ for $a,c>0$, therefore if $c<d$, $ac<bd$ is true.

## ix

If $a=0$, then $a^{2}<b^{2}$ since $b^{2}>0$.
If $0<a<b$, then $a^{2}<ab$. Since $ab<b^{2}$, we have $a^{2}<b^{2}$.