# 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

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

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

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

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

## v

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

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

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

$= x^n - y^n\,$

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

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

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

## 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 $ac < bd$.

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 $a^2 < b^2$.