# Irrational Numbers

1. Sep 26, 2008

### anantchowdhary

How do we exactly define irrational numbers..
but id like to know about any infinite series,
if any which is used to define irrational numbers...
and how can one prove properties of basic operations for irrational numbers

Thanks

2. Sep 26, 2008

### Mentallic

An irrational number is one that the decimal places do not terminate or recur. There are no perceivable patterns.

I'm sure you already knew this and want further detail on the subject though.

3. Sep 26, 2008

### HallsofIvy

Staff Emeritus
"Irrational" numbers are simply defined as real numbers that are not rational. But I think you are really asking how real numbers are defined.

One method is the "Dedekind cut" (see "Baby Rudin"). A Dedekind cut is defined as a set of rational numbers such that:
1) It is not empty- there exist at least one rational number in the set.
2) It is not all rational numbers- there exist a least one rational number NOT in the set
3) It has no largest member.
4) if a is a rational number in the set and b< a, then b is also in the set.

You might argue that we must have the rational numbers a subset of the real numbers, and rational numbers are NOT "sets of rational numbers"! But we can identify each rational number with such a set.

For example, if a is a rational number then {x| x is rational and x< a} is a cut- called a "rational cut" and is identified with the rational number a. It is non-empty: it contains a-1. It is not all rational numbers: it does not contain a. It has no largest member: if b is any member, then (a+b)/2 is a rational number in the set and is large than b. If b is in the set and c< b, then c< b< a so c is in the set.

The set {x| x is rational and either x2< 2 or x< 0} is also a Dedekind cut. It is non-empty: it includes 0. It is not all rational numbers: it does not include 2. It is harder to show that it has no largest member, but basically, if there were a largest member, its square would have to be 2- and there is no such rational number. Finally, suppose a is in the set and c< a. If a$\le$0 then c< 0. If 0< c< a, then c2< a2< 2. This Dedekind cut would be identified with the irrational number $\sqrt{2}$.

The "Dedekind cut" definition has the nice property that it becomes very easy to prove one of the "fundamental" properties of the real numbers: that every non-empty, bounded set of real numbers has a least upper bound.

But since you ask specifically about infinite sequences here are two other ways of defining the real numbers:

Consider the collection of all increasing, bounded, sequences of rational numbers. Say that two such sequences, {an} and {bn} are "equivalent" if and only if the sequence {an- bn} converges to 0. That is an "equivalence relation" and so partitions the collection into "equivalence classes"- sets of sequences so that any two sequences in the same set are equivalent, two sequences in different sets are not. Now we identify the real numbers with these equivalence classes. Again, we want to have the rational numbers a subset of this and rational numbers aren't sequences of rational numbers! But given any rational number, a, we can identify it with the sequence of rational numbers {a, a, a, a, ...} and it is not too difficult to show that a sequence of rational numbers is "equivalent" to that sequence if and only if it converges to a.

Of course, there exist increasing, bounded, sequences of rational numbers that do NOT converge to a rational number- the "monotone convergence property" is not true for the rational numbers. In fact, the "naive" way of thinking about real numbers is that a real number is any number that can be "written as a decimal": of the form a.a1a2.... We can identify that with the sequence {a, a.a1, a.a1a2, ...} and then with the class of all sequeces equivalent to that. In particular, "$\pi$" is identified with the class of sequences containing {3, 3.1, 3.14, 3.141, 3.1415, ...} and "$\sqrt{2}$" is identified with the class of sequences containing {2, 2.1, 2.14, 2.141, 2.1414, ...}

That definition makes it easy to prove the "monotone convergence property" for the real numbers.

I promised two ways of defining real numbers in terms of sequences. The second is almost the same. Consider the collection of all Cauchy sequences. That is, the collection of all sequences {an} such that |an- am| goes to 0 as m and n go to infinity (independently). We again say that two such sequences {an} and {bn} are equivalent if and only if the sequence {an- bn} converges to 0 and identify the real numbers with the equivalence classes of such sequences as before.

This definition makes it easy to prove the "Cauchy Criterion"- that all Cauchy sequences converge.

4. Sep 26, 2008

### anantchowdhary

how do we prove the basic results for operations like multiplication and division of irrational numbers?

5. Sep 26, 2008

### Tac-Tics

You don't. You prove them for all the reals.

I am a little fuzzy on the details, but it goes something like this. Take cauchy sequences (a_n) and (b_n) representing two reals a and b. Define an operation + such that a+b is the sequence c_i = a_i + b_i. You then show that (c_n) is cauchy, and thus, c is real. Multiplication is defined similarly. The usual properties for arithmetic (distributivity, commutativity, associativity-type things) are all inherited in a straightfoward way from the underlying rational operations used.

6. Sep 27, 2008

### arildno

We must distinguish between two levels of analysis:

1. To specify a set of axioms that some (otherwise undefined) set of objects (numbers) are said to obey.

2. To CONSTRUCT, on basis of another axiom set and their associated "numbers", structures that can be proven (according to this latter, more "fundamental" set of axioms) to be obeying the set of axioms as specified under 1.

Halls' post shows you how to construct the real numbers out of the rational number set (which again can be constructed out of the natural number set and basic set theory).
In this approach, we need to prove that multiplication of the reals does, in fact, obey the rules arbitrarily set up in 1.

However, we might let the axioms under 1 be our fundamental axioms, and in that case, we do not prove that reals can be multiplied, rather, that they obey the multiplication rules are part of their definition.

The advantage of the more fundamental approach is mainly theoretical in that by proving the constructibility of the reals out of simpler number sets, we have also proven that there aren't any other logical problems lurking within the concept of the reals than those problems that might hide within the concept of the naturals.

7. Sep 27, 2008

### anantchowdhary

is there any particular infinite series whic may converge to a given irrational number...if there is could someone please provide a derivation

thanks...

8. Sep 27, 2008

### HallsofIvy

Staff Emeritus
I wrote a paper that essentially proved the properties of the real numbers from the Dedekind Cut definition but don't have a URL for it available right now.

Of course, the first thing you would do is DEFINE addition and multiplication.

For example, using the Dedekind Cut definition, you would define "x+ y" to be the set of rational numbers of the form {a+ b| a is in x, b is in y}. You would need to show that satifies the requirements for a Dedekind Cut itself. If I remember correctly, defining multiplication is quite a lot harder- you have to do the case x, y both positive first.

Using the "equivalence classes of increasing, bounded sequences of rational numbers" definition, you would select one "representative" of each of x and y (one sequence in each of the classes defining x and y), say {xn} and {yn} and then show that the sequence {xn+ yn} is also an "increasing, bounded sequence of rational numbers" and so is in some equivalence class. That equivalence class IS "x+ y". Basically, then, you could show the properties of addtion or real numbers, commutativity, associativity, additive identity, additive inverse, from the corresponding properties of rational numbers, applied to the numbers in the sequences.

9. Sep 29, 2008

### anantchowdhary

Is ther any general infinite series which lets us express an irrational number in terms of an irrational number...ALl proofs in calculus are pretty much incomplete without basic proofs of operations on irrational numbers!

10. Sep 29, 2008

### Tac-Tics

Do you mean "in terms of *rational* numbers"?

Operations on irrational numbers are fully defined and proofs of their properties are very simple once you are comfortable working with sequences of rational numbers.

11. Sep 29, 2008

### HallsofIvy

Staff Emeritus
Why do you keep asking the same question over and over again? I just explained- not a complete explanation, that would take a whole book- how we can define the real numbers and show the basic properties of them.