# Exercise 2.1.1

## 1

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

## 2

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 ${\displaystyle l}$.

## 3

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

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

## 4

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 2.2.2

## 1

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

## 2

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

## 3

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 2.2.3

## 1

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

Let ${\displaystyle j=3k}$.

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

## 2

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

Let ${\displaystyle j=3k+1}$.

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

## 3

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 2.2.6

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 2.2.7

${\displaystyle a|b}$ 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 2.2.8

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 2.3.3

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 2.3.4

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 2.3.5

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?