1.2: Rational and Irrational Numbers
( \newcommand{\kernel}{\mathrm{null}\,}\)
We start with a few results we need in the remainder of this subsection.
Theorem 1.7: Well-ordering Principle
Any non-empty set S in N∪{0} has a smallest element.
- Proof
-
Suppose this is false. Pick s1∈S. Then there is another natural number s2 in S such that s2≤s1−1. After a finite number of steps, we pass zero, implying that S has elements less than 0 in it. This is a contradiction.
Note that any non-empty set S of integers with a lower bound can be transformed by addition of a integer b∈N0 into a non-empty S+b in N0. Then S+b has a lower bound, and therefore so does S. Furthermore, a non-empty set S of integers with a upper bound can also be transformed into a non-empty −S+b in N0. Here, −S stands for the collection of elements of S multiplied by −1. Thus we have the following corollary of the well-ordering principle.
Corollary 1.8
Let be a non-empty set S in Z with a lower (upper) bound. Then S has a smallest (largest) element.
Definition 1.9
An element x∈R is called rational if it satisfies qx−p=0 where p and q≠0 are integers. Otherwise it is called an irrational number. The set of rational numbers is denoted by Q.
The usual way of expressing this, is that a rational number can be written as pq. The advantage of expressing a rational number as the solution of a degree 1 polynomial, however, is that it naturally leads to Definition 1.12.
Theorem 1.10
Any interval in R contains an element of Q. We say that Q is dense in R.
- Proof
-
Let I=(a,b) with b>a any interval in R. From Corollary 1.8 we see that there is an n such that n>1b−a. Indeed, if that weren’t the case, then N would be bounded from above, and thus it would have a largest element n0. But if n0∈N, then so is n0+1. This gives a contradiction and so the above inequality must hold.
It follows that nb−na>1. Thus the interval (na,nb) contains an integer, say, p. So we have that na<p<nb. The theorem follows upon dividing by n.
The crux of the following proof is that we take an interval and scale it up until we know there is an integer in it, and then scale it back down.
Theorem 1.11
√2 is irrational.
- Proof
-
Suppose √2 can be expressed as the quotient of integers rs. We may assume that gcd(r,s)=1 (otherwise just divide out the common factor). After squaring, we get
2s2=r2
The right hand side is even, therefore the left hand side is even. But the square of an odd number is odd, so r is even. But then r2 is a multiple of 4. Thus s must be even. This contradicts the assumption that gcd(r,s)=1.
It is pretty clear who the rational numbers are. But who or where are the others? We just saw that √2 is irrational. It is not hard to see that the sum of any rational number plus √2 is also irrational. Or that any rational non-zero multiple of √2 is irrational. The same holds for √2,√3,√5, etcetera. We look at this in exercise 1.7. From there, is it not hard to see that the irrational numbers are also dense (exercise 1.8). In exercise 1.15, we prove that the number e is irrational. The proof that π is irrational is a little harder and can be found in [1][section 11.17]. In Chapter 2, we will use the fundamental theorem of arithmetic, Theorem 2.14, to construct other irrational numbers. In conclusion, whereas rationality is seen at face value, irrationality of a number may take some effort to prove, even though they are much more numerous as we will see in Section 1.4.