Solutions To Mathematics Textbooks/Proofs and Fundamentals/Chapter 2

Exercise 2.1.1

1

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

2

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 $l$.

3

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

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

4

If e is raised to the power of x+y, then it is equivalent to the product of e, each one rised to the power of x and y, respectively.

5

If $f$ is a continuous function on the interval [a, b] and $F$ is any function such that $F^{\prime }(x)=f(x)$, then the integral of f(x) on [a,b] equals F(b) - F(a).

Exercise 2.2.2

1

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

2

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

3

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 $m|-n$.

Exercise 2.2.3

1

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

Let $j=3k$.

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

2

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

Let $j=3k+1$.

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

Exercise 2.2.4

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 $j=2k^2+2k$.

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

$ak=a(qm+rn)=(aq)m+(ar)n=bm+cn$

Because $ak=bm+cn$, it follows $a|(bm+cn)$

Exercise 2.2.7

$a|b$ 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 $ac|bd$.

Exercise 2.2.8

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

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

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

Exercise 2.3.2

Proof by contrapositive:

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

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

Exercise 2.3.3

It is true that $a$ does not divide $bc$. 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 $a|bc$, Contradiction!

Exercise 2.3.4

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

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

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

Exercise 2.3.5

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

$ax+by=c$

$(dp)x+(dq)y=c$

$d(px+qy)=c$

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