# Exercise 2.1.1Edit

## 1Edit

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

## 2Edit

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

## 3Edit

If ${\displaystyle ABC}$  is a triangle with sides of length ${\displaystyle a,b,}$  and ${\displaystyle c}$  then

${\displaystyle {\frac {a}{sinA}}={\frac {b}{sinB}}={\frac {C}{sinC}}}$

## 4Edit

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

# Exercise 2.2.2Edit

## 1Edit

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

## 2Edit

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

## 3Edit

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

# Exercise 2.2.3Edit

## 1Edit

If n is an even integer, then for some integer k, ${\displaystyle n=2k}$ .

Let ${\displaystyle j=3k}$ .

Then ${\displaystyle 3n=3(2k)=2(3k)=2j}$ .

## 2Edit

If n is an odd integer, then for some integer k, ${\displaystyle n=2k+1}$ .

Let ${\displaystyle j=3k+1}$ .

Then ${\displaystyle 3n=3(2k+1)=6k+3=6k+2+1=2(3k+1)+1=2j+1}$ .

## 3Edit

If n is even, then ${\displaystyle n=2k}$ . For integers j and k, let ${\displaystyle j=2k^{2}}$ .

${\displaystyle n^{2}=(2k)^{2}=4k^{2}=2(2k^{2})=2j}$ , so ${\displaystyle n^{2}}$  is even.

If n is odd, then ${\displaystyle n=2k+1}$ . For integers j and k, let ${\displaystyle j=2k^{2}+2k}$ .

${\displaystyle n^{2}=(2k+1)^{2}=4k^{2}+4k+1=2(2k^{2}+2k)+1=2j+1}$ , so ${\displaystyle n^{2}}$  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 ${\displaystyle a|b}$  and ${\displaystyle a|c}$ . Hence there are integers ${\displaystyle q}$  and ${\displaystyle r}$  such that ${\displaystyle aq=b}$  and ${\displaystyle ar=c}$ . Define the integer ${\displaystyle k}$  by ${\displaystyle k=qm+rn}$ . Then

${\displaystyle ak=a(qm+rn)=(aq)m+(ar)n=bm+cn}$

Because ${\displaystyle ak=bm+cn}$ , it follows ${\displaystyle a|(bm+cn)}$

# Exercise 2.2.7Edit

${\displaystyle a|b}$  implies that ${\displaystyle ax=b}$  for some integer, x.

${\displaystyle c|d}$  implies that ${\displaystyle cy=d}$  for some integer, y.

${\displaystyle ac|bd=ac|ax\cdot cy\,}$

Therefore,

${\displaystyle acj=ax\cdot cy}$  for some integer, j.

${\displaystyle acj=ac(xy)\,}$

Let ${\displaystyle j=xy\,}$ , hence ${\displaystyle ac|bd\,}$ .

# Exercise 2.2.8Edit

Suppose that ${\displaystyle a|b}$ . Hence there is an integer ${\displaystyle q}$  such that ${\displaystyle aq=b}$ . if ${\displaystyle n}$  is a positive integer, define the integer ${\displaystyle k}$  by ${\displaystyle k=q^{n}}$ . Then

${\displaystyle a^{n}k=a^{n}(q^{n})=(aq)^{n}=b^{n}}$

Because ${\displaystyle a^{n}k=b^{n}}$ , it follows ${\displaystyle a^{n}|b^{n}}$

# Exercise 2.3.2Edit

Proof by contraposition:

Assume ${\displaystyle n}$  is not even, then ${\displaystyle n=2k+1}$  and ${\displaystyle n^{2}=(2k+1)^{2}=(2k+1)(2k+1)=(4k^{2}+2k)+1=2(2k^{2}+k)+1}$ .

Let ${\displaystyle i=(2k^{2}+k)}$  then ${\displaystyle n^{2}=2i+1}$ , it follows that if ${\displaystyle n}$  is not even then ${\displaystyle n^{2}}$  is not even.

# Exercise 2.3.3Edit

It is true that ${\displaystyle a}$  does not divide ${\displaystyle bc}$ . Suppose that ${\displaystyle a|b}$ . This means there is an integer ${\displaystyle n}$  such that ${\displaystyle b=an}$ . Then, we have:

${\displaystyle bc=(an)c=a(nc)}$

We may consider the integer ${\displaystyle k=nc}$ . Therefore, we have that ${\displaystyle bc=ak}$ . Then ${\displaystyle a|bc}$ , Contradiction!

# Exercise 2.3.4Edit

Let ${\displaystyle a}$  a non-zero rational number, hence there are integers ${\displaystyle m}$  and ${\displaystyle n}$  both different from zero, such that ${\displaystyle a={\frac {m}{n}}}$ . Let ${\displaystyle b}$  an irrational number.

Suppose that the product ${\displaystyle ab}$  is a rational number, hence there are integers ${\displaystyle p}$  and ${\displaystyle q}$  different from zero such that ${\displaystyle ab={\frac {p}{q}}}$ , this is ${\displaystyle {\frac {m}{n}}b={\frac {p}{q}}}$ , it follows that ${\displaystyle b={\frac {np}{mq}}}$ .

The last equatily means that ${\displaystyle b}$  is a rational number, which is a contradiction because we supposed that ${\displaystyle b}$  was irrational. By contradiction, it follows that the product ${\displaystyle ab}$  must be irrational.

# Exercise 2.3.5Edit

Suppose that ${\displaystyle d|a}$  and ${\displaystyle d|b}$ , but ${\displaystyle d}$  does not divide ${\displaystyle c}$ . Hence there are integers ${\displaystyle p}$  and ${\displaystyle q}$  such that ${\displaystyle dp=a}$  and ${\displaystyle dq=b}$ . Suppose that the equation ${\displaystyle ax+by=c}$  has a solution such that ${\displaystyle x}$  and ${\displaystyle y}$  are integers, then

${\displaystyle ax+by=c}$

${\displaystyle (dp)x+(dq)y=c}$

${\displaystyle d(px+qy)=c}$

Let ${\displaystyle k=px+qy}$ , then ${\displaystyle dk=c}$ , it follows that ${\displaystyle d|c}$ , which is a contradiction.

Anyone?