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 , 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 , 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} xii

$x<1\,$ xiii

NOTE: The answer in the 3rd Edition provides $0 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 since $x^{2}>x$ .

Question 5

i

$a+c $=(b+d)-(a+c)>0\,$ $=(b-a)+(d-c)>0\,$ , 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 .

v

$a>1\,$ , 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 .

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

ix

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

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