Last modified on 22 November 2011, at 13:52

Solutions To Mathematics Textbooks/Basic Mathematics/Chapter 1

Chapter 1Edit

Section 2Edit

1Edit

(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

14Edit

 -2 + x = 4 \,

 x = 4 + 2 \,

x = 6 \,

15Edit

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

Section 3Edit

30, 31, 32, 33Edit

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}\, is the inital 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 4Edit

8-15Edit

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.

16-23Edit

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.

24Edit

a)Edit

Examining the definitions first, if a \equiv b \pmod{5} and x \equiv y \pmod{5} then a-b = 5n \, and x-y = 5m \,.


a+x \equiv b+y \pmod{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)Edit

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 5Edit

7Edit

a)Edit

120, 720, 5040 and 40320.

c)Edit

\binom{m}{n} = \binom{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)Edit

\binom{m}{n} + \binom{m}{n-1} = \binom {m+1}{n}


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


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


\binom{m}{n} + \binom{m}{n-1} = \binom {m+1}{n}


Section 6Edit

1Edit

a)Edit

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


2x-1 = 7(3x+2) \,

2x-1 = 21x + 14 \,

-19x = 15 \,


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

2Edit

a)Edit

\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)Edit

\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 \,