Solutions To Mathematics Textbooks/Basic Mathematics/Chapter 1

Chapter 1

Section 2

1

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

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

1 (alternative)

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

$=a+(b+(c+d))$	by associativity
$=a+(b+(d+c))$	by commutativity
$=a+(d+(b+c))$	by associativity
$=a+(d+(b+c))$	by associativity

2

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

$a+b+c+d=(a+c)+(b+d)$		by associativity
$a+c+b+d=(a+c)+(b+d)$		by commutativity
$(a+c)+(b+d)=(a+c)+(b+d)$		by associativity

2 (alternative)

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

$a+(b+(c+d))=(a+c)+(b+d)$		by associativity
$a+((b+c)+d)=(a+c)+(b+d)$		by associativity
$a+((c+b)+d)=(a+c)+(b+d)$		by commutativity
$a+(c+(b+d))=(a+c)+(b+d)$		by associativity
$(a+c)+(b+d)=(a+c)+(b+d)$		by associativity

3

$(a-b)+(c-d)=(a+c)+(-b-d)$

$a-b+c-d=(a+c)+(-b-d)$		by associativity
$a+c-b-d=(a+c)+(-b-d)$		by commutativity
$(a+c)+(-b-d)=(a+c)+(-b-d)$		by associativity

14

$-2+x=4$

$x=4+2$

$x=6$

15

$2-x=5$
$-x=3$
$x=-3$		multiply both sides by -1.

Section 3

30, 31, 32, 33

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 $y_2$ are the first 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 4

8-15

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.

16-23

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.

24

a)

Examining the definitions first, if $a\equiv b(\mathrm{mod}5)$ and $x\equiv y(\mathrm{mod}5)$ then $a-b=5n$ and $x-y=5m$.

$a+x\equiv b+y(\mathrm{mod}5)$ 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)

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 5

7

a)

120, 720, 5040 and 40320.

c)

$(\genfrac{}{}{0ex}{}{m}{n})=(\genfrac{}{}{0ex}{}{m}{m-n})$

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

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

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

d)

$(\genfrac{}{}{0ex}{}{m}{n})+(\genfrac{}{}{0ex}{}{m}{n-1})=(\genfrac{}{}{0ex}{}{m+1}{n})$

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

Multiply both sides of the equation by $\frac{n}{n}$ and $\frac{(m-n+1)}{(m-n+1)}$, cancelling unnecessary 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,

$(\genfrac{}{}{0ex}{}{m}{n})+(\genfrac{}{}{0ex}{}{m}{n-1})=(\genfrac{}{}{0ex}{}{m+1}{n})$

Section 6

1

a)

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

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

$2x-1=21x+14$

$-19x=15$

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

2

a)

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

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

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

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

b)

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

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

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

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

$x^3-1=x^3-1$