Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.3

 3. Prove that both the sum and the product of two rational numbers are rational.
Solution:

We assume that x and y are rational numbers. Then these rational numbers will be of form x=\frac{a}{b} and y=\frac{c}{d}. Where a,b,c,d are integers with b\neq0 and d\neq0.

Now,  xy=\left(\begin{array}{c}\frac{a}{b}\end{array}\right)\cdot\left(\begin{array}{c}\frac{c}{d}\end{array}\right)

xy=\frac{ac}{bd}

And x+y=\frac{a}{b}+\frac{c}{d}

x+y=\frac{(ad+bc)}{bd}

Where bd\neq0 Also both x+y and xy are ratios of integrs.Therefore they are both rational numbers. Hence our proof is complete.

Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.2

 2. Show that if a and b are positive integers, then there is a smallest positive integer of the form a – bk, k \in Z.

Solution:

Let a and b be positive integers and let S = { n : n is a positive integer and n = a – bk for some k \in \mathbb{Z}. Now S is non empty since a + b = a – b( – 1) is in S . By the Well Ordering Principle S has a least element.

Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.1

1.  Determine whether each of the following sets is well ordered. Either give a proof using the well-ordering property of the set of positive integers, or give an example of a subset of the set that has no smallest element.
a) the set of integers greater than 3
b) the set of even positive integers
c) the set of positive rational numbers
d) the set of positive rational numbers that can be written in the form a/2, where a is a positive integer
e) the set of nonnegative rational numbers

Solution:

We begin by stating the Well Ordering Principle. It states that every non-empty set of positive integers has a smallest element.

(a) The set of Integers greater than 3 is well ordered as it is non-empty set of positive Integers and the smallest element in this set is 3.

(b) The set of even positive Integers is well ordered as it is non-empty set of positive Integers and the smallest element in this set is 2.

(c) The set of positive rational numbers is not well ordered. Let us take a subset of this set as

\left\{ 1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\right\}

We can see that this non-empty positive subset has no least element. Instead it has a greatest element 1.

(d) The set of positive rational numbers of form \frac{a}{2} where a , is a positive integer is well Ordered.

The smallest positive Integer is 1 so the smallest positive rational number in this set is \frac{1}{2}

(e) The set of non-negative rational numbers is not well ordered. We take a subset of this set as the set of positive rational numbers which has a least element as can been seen from part (c).