Solutions To Mathematics Textbooks/Basic Mathematics/Chapter 1

=Chapter 1=

1
$$(a + b) + (c + d) = (a + d) + (b + c) \,$$

1 (alternative)
$$(a + b) + (c + d) = (a + d) + (b + c) \,$$

2
$$(a + b) + (c + d) = (a + c) + (b + d) \,$$

2 (alternative)
$$(a + b) + (c + d) = (a + c) + (b + d) \,$$

3
$$(a - b) + (c - d) = (a + c) + (-b - d) \,$$

14
$$ -2 + x = 4 \,$$

$$ x = 4 + 2 \,$$

$$x = 6 \,$$

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:

$$200000 \cdot 3^{(\frac{2215-1915}{50})} = 200000 \cdot 3^6 = 145800000$$

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.

a)
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)
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 $$t = 5nm + ny + mb \,$$.

$$ax = 5t + by \,$$

$$ax - by = 5t + by - by \,$$

$$ax - by = 5t \,$$.

a)
120, 720, 5040 and 40320.

c)
$$\binom{m}{n} = \binom{m}{m-n}$$

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

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

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

d)
$$\binom{m}{n} + \binom{m}{n-1} = \binom {m+1}{n} $$

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

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,

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

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

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

$$2x-1 = 21x + 14 \,$$

$$-19x = 15 \,$$

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

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 $$ad = bc$$.

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

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

$$x^3-1 = x^3-1 \,$$