Solutions To Mathematics Textbooks/Proofs and Fundamentals/Chapter 2

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Exercise 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...


Exercise 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\,.


Exercise 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.

n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 = 2j + 1, so 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 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 2.2.7Edit

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\,.


Exercise 2.2.8Edit

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 2.3.3Edit

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 2.3.4Edit

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 equatily 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 2.3.5Edit

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.


Exercise 2.3.6Edit

Anyone?