Solutions To Mathematics Textbooks/Proofs and Fundamentals/Chapter 2

=Exercise 2.2.1=

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


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

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 $$F'(x) = f(x)$$, 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

$$a^nk = a^n(q^n)=(aq)^n=b^n$$

Because $$a^nk=b^n$$, 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.