Solutions To Mathematics Textbooks/Proofs and Fundamentals/Chapter 2

From Wikibooks, open books for an open world
< Solutions To Mathematics Textbooks‎ | Proofs and Fundamentals
Jump to: navigation, search

Question 2.1.1[edit]

1[edit]

If r is a real number, then the area of a circle of radius r is \pi r^2.

2[edit]

If there is a line l and a point P not on l, then there is exactly one line m containing P that is parallel to l.

3[edit]

If ABC is a triangle with sides of length a, b, and c then

\frac{a}{sin A}=\frac{b}{sin B}=\frac{C}{sin C}

4[edit]

If f is a continuous function on [a, b] and F is an function such that F'(x) = f(x), then...


Question 2.2.2[edit]

1[edit]

If 1|n\,, then there is an integer q such that 1 \cdot q = n. Let q = n.

2[edit]

If n|n\,, then there is an integer q such that n \cdot q = n. Let q = 1.

3[edit]

If m|n\,, then there is an integer q such that m\cdot q = n\,. This implies -mq = -n\,, and so m\cdot -q = -n\,, and thus m|-n\,.


Question 2.2.3[edit]

1[edit]

If n is an even integer, then for some integer k, n = 2k.

Let j=3k.

Then 3n = 3(2k) = 2(3k) = 2j.

2[edit]

If n is an odd integer, then for some integer k, n = 2k+1.

Let j=3k+1.

Then 3n = 3(2k+1) = 6k+3 = 6k+2+1 = 2(3k+1)+1 = 2j+1.

3[edit]

If n is even, then n = 2k. For integers j and k, let j = 2k^2.

n^2 = (2k)^2 = 4k^2 = 2(2k^2) = 2j, so n^2 is even.

If n is odd, then n = 2k+1. For integers j and k, let j = 2k^2 + 2k+ 1.

n^2 = (2k+1)^2 = 4k^2 + 4k + 2 = 2(2k^2 + 2k + 1) + 1 = 2j, so n^2 is odd.

Question 2.2.6[edit]

If a|b, and b|bm then a|bm, implying aj = bm for some integer j.

Also, if a|c, and c|cn then a|cn, implying ai = cn for some integer i.

We let x = (j+i).

ax = aj+ai

ax = bm+cn

Which implies a|(bm+cn).

Question 2.2.7[edit]

a|b implies that  ax = b for some integer, x.

c|d implies that  cy = d for some integer, y.


ac|bd = ac|ax\cdot cy\,

Therefore,

acj = ax\cdot cy for some integer, j.

acj = ac(xy)\,

Let j=xy\,, hence ac|bd\,.

Question 2.3.3[edit]

Suppose that  a|b . This means there is an integer  n such that  b = an . Then, we have:

 bc = (an)c = a(nc)

We may consider the integer  k = nc . Therefore, we have that  bc = ak . Then  a|bc