# Question 2.1.1Edit

## 1Edit

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

## 2Edit

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 $l$.

## 3Edit

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

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

## 4Edit

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

# Question 2.2.2Edit

## 1Edit

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

## 2Edit

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

## 3Edit

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 2.2.3Edit

## 1Edit

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

Let $j=3k$.

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

## 2Edit

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

Let $j=3k+1$.

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

## 3Edit

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 2.2.6Edit

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).

# Question 2.2.7Edit

$a|b$ 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 2.3.3Edit

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 $a|bc$