# Solutions To Mathematics Textbooks/Proofs and Fundamentals/Chapter 2

## Contents

# Exercise 2.1.1[edit]

## 1[edit]

If is a real number, then the area of a circle of radius is .

## 2[edit]

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

## 3[edit]

If is a triangle with sides of length and then

## 4[edit]

If is a continuous function on [a, b] and is an function such that , then...

# Exercise 2.2.2[edit]

## 1[edit]

If , then there is an integer q such that . Let q = n.

## 2[edit]

If , then there is an integer q such that . Let q = 1.

## 3[edit]

If , then there is an integer q such that . This implies , and so , and thus .

# Exercise 2.2.3[edit]

## 1[edit]

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

Let .

Then .

## 2[edit]

If n is an odd integer, then for some integer k, .

Let .

Then .

## 3[edit]

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

, so is even.

If n is odd, then . For integers j and k, let .

, so is odd.

# Exercise 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).

Another proof: Suppose that and . Hence there are integers and such that and . Define the integer by . Then

Because , it follows

# Exercise 2.2.7[edit]

implies that for some integer, x.

implies that **Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle cy=d}**
for some integer, y.

Therefore,

for some integer, j.

Let , hence .

# Exercise 2.2.8[edit]

Suppose that . Hence there is an integer such that . if is a positive integer, define the integer by . Then

Because , it follows

# Exercise 2.3.3[edit]

It is true that does not divide . Suppose that . This means there is an integer such that . Then, we have:

We may consider the integer . Therefore, we have that . Then , Contradiction!

# Exercise 2.3.4[edit]

Let a non-zero rational number, hence there are integers and both different from zero, such that . Let an irrational number.

Suppose that the product is a rational number, hence there are integers and different from zero such that , this is , it follows that .

The last equatily means that is a rational number, which is a contradiction because we supposed that was irrational. By contradiction, it follows that the product must be irrational.

# Exercise 2.3.5[edit]

Suppose that and , but does not divide . Hence there are integers and such that and . Suppose that the equation has a solution such that and are integers, then

Let , then , it follows that , which is a contradiction.

# Exercise 2.3.6[edit]

Anyone?