# Question 1

## i

${\displaystyle ax=a\,}$

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

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

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

${\displaystyle 1\cdot x=1}$

${\displaystyle x=1\,}$

## ii

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

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

${\displaystyle =x^{2}-xy+xy-y^{2}\,}$

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

## iii

${\displaystyle x^{2}=y^{2}\,}$, then

${\displaystyle x^{2}-y^{2}=(x+y)(x-y)=0\,}$

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

## iv

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

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

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

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

## v

${\displaystyle x^{n}-y^{n}=(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})\,}$

${\displaystyle =x(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})-y(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})\,}$

${\displaystyle =(x^{n}+x^{n-1}y+...+x^{2}y^{n-2}+xy^{n-1})-(x^{n-1}y+x^{n-2}y^{2}+...+xy^{n-1}+y^{n})\,}$

${\displaystyle =x^{n}-y^{n}\,}$

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

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

# Question 4

## ii

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

## iii

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

Thus, ${\displaystyle x>{\sqrt {7}}\,}$, or ${\displaystyle x<-{\sqrt {7}}\,}$.

## v

${\displaystyle x^{2}-2x+2}$ cannot be factored in its current form, so we first turn a part of the expression into a perfect square:

${\displaystyle (x^{2}-2x+1)+1>0\,}$

We then complete the square on ${\displaystyle x^{2}-2x+1=(x-1)^{2}\,}$, so now we have the expression:

${\displaystyle (x-1)^{2}+1>0\,}$, which is positive for all values of x.

## vi

If ${\displaystyle x^{2}+x+1>2\,}$, then ${\displaystyle x^{2}+x-1>0\,}$.

${\displaystyle x^{2}+x-1=\left(x+{\frac {1+{\sqrt {5}}}{2}}\right)\left(x-{\frac {{\sqrt {5}}-1}{2}}\right)}$

Thus ${\displaystyle x<-{\frac {1+{\sqrt {5}}}{2}}}$, or ${\displaystyle x>{\frac {{\sqrt {5}}-1}{2}}}$.

## viii

Complete the square:

${\displaystyle x^{2}+x+1=x^{2}+x+{\frac {1}{4}}+{\frac {3}{4}}=\left(x+{\frac {1}{2}}\right)^{2}+{\frac {3}{4}}}$

Thus, ${\displaystyle \left(x+{\frac {1}{2}}\right)^{2}+{\frac {3}{4}}>0}$.

## ix

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

## x

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

## xi

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

${\displaystyle log_{2}2^{x}

## xii

${\displaystyle x<1\,}$

## xiii

NOTE: The answer in the 3rd Edition provides ${\displaystyle 0 or ${\displaystyle x>1}$. Plugging in values ${\displaystyle x>1}$, e.g. 10, gives us ${\displaystyle 1/10-1/9<0}$, so I think this is a misprint or an incorrect answer.

If ${\displaystyle {\frac {1}{x}}+{\frac {1}{1-x}}>0}$, then ${\displaystyle {\frac {1}{x-x^{2}}}>0}$, which is only positive when ${\displaystyle 0 since ${\displaystyle x^{2}>x}$.

# Question 5

## i

${\displaystyle a+c

${\displaystyle =(b+d)-(a+c)>0\,}$

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

## ii

${\displaystyle b-a>0\,}$

${\displaystyle =-a+b>0\,}$

${\displaystyle =-a-(-b)>0\,}$

${\displaystyle =-b<-a\,}$

## iv

${\displaystyle (b-a),c>0\,}$, therefore ${\displaystyle c(b-a)>0\,}$.

${\displaystyle c(b-a)=bc-ac\,}$, therefore ${\displaystyle ac.

## v

${\displaystyle a>1\,}$, therefore ${\displaystyle 1-a>0\,}$.

${\displaystyle a(1-a)>0\,}$

${\displaystyle =a^{2}-a>0\,}$

${\displaystyle =a^{2}>a\,}$

## viii

If ${\displaystyle a}$ or ${\displaystyle c}$ are 0, then by definition ${\displaystyle ac.

Otherwise, we have already proved that ${\displaystyle ac for ${\displaystyle a,c>0}$, therefore if ${\displaystyle c, ${\displaystyle ac is true.

## ix

If ${\displaystyle a=0}$, then ${\displaystyle a^{2} since ${\displaystyle b^{2}>0}$.

If ${\displaystyle 0, then ${\displaystyle a^{2}. Since ${\displaystyle ab, we have ${\displaystyle a^{2}.