Solutions To Mathematics Textbooks/Proofs and Fundamentals/Chapter 2

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Exercise 2.1.1Edit

1Edit

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

2Edit

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

3Edit

If   is a triangle with sides of length   and   then

 

4Edit

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


Exercise 2.2.2Edit

1Edit

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

2Edit

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

3Edit

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


Exercise 2.2.3Edit

1Edit

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

Let  .

Then  .

2Edit

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

Let  .

Then  .

3Edit

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 2.2.6Edit

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 2.2.7Edit

  implies that   for some integer, x.

  implies that   for some integer, y.


 

Therefore,

  for some integer, j.

 

Let  , hence  .


Exercise 2.2.8Edit

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

 

Because  , it follows  


Exercise 2.3.3Edit

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 2.3.4Edit

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 2.3.5Edit

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 2.3.6Edit

Anyone?