Table of Contents

- 1 Why the set of rational numbers is not complete?
- 2 Is set of rational numbers a complete set?
- 3 Why is Q dense in R?
- 4 Is Z dense in R?
- 5 Why is R both open and closed?
- 6 How can a set be neither open nor closed?
- 7 Why are the rational numbers not complete in the sense that?
- 8 Is there a lower bound of rationals greater than 1?

## Why the set of rational numbers is not complete?

Therefore the assumption that some x∈Q is equal to infnxn is paradoxical. So the non-empty subset {xn}n of Q has a lower bound in Q but no greatest lower bound in Q, so Q is not order-complete.

### Is set of rational numbers a complete set?

The set of rational numbers is denoted by Q. A real number that is not rational is termed irrational. 2 , −5 6 , 100, 567877 −1239 , 8 2 are all rational numbers. Exercise 1 1.

**Is set of rational numbers a complete ordered field?**

Example 13. The rational numbers Q are an ordered field, with the usual +, ·, 0 and 1, and with P = {q ∈ Q : q > 0}. The real numbers are a complete ordered field.

**Why is the set of rational numbers neither open nor closed?**

The set of rational numbers Q ⊂ R is neither open nor closed. It isn’t open because every neighborhood of a rational number contains irrational numbers, and its complement isn’t open because every neighborhood of an irrational number contains rational numbers.

## Why is Q dense in R?

Theorem (Q is dense in R). For every x, y ∈ R such that x, there exists a rational number r such that x

### Is Z dense in R?

(a) Z is dense in R . A counterexample would be any interval that doesn’t contain an integer, like (0 , 1). (b) The set of positive real numbers is dense in R .

**Is set of real number complete?**

Axiom of Completeness: The real number are complete. Theorem 1-14: If the least upper bound and greatest lower bound of a set of real numbers exist, they are unique.

**Does the set of rational number hold properties of completeness?**

The Completeness property makes into a complete ordered field. The following example shows that does not have the completeness property. The set of rational numbers is an ordered field but it is not complete.

## Why is R both open and closed?

R is open because any of its points have at least one neighborhood (in fact all) included in it; R is closed because any of its points have every neighborhood having non-empty intersection with R (equivalently punctured neighborhood instead of neighborhood).

### How can a set be neither open nor closed?

The interval (0,1) as a subset of R2, that is {(x,0)∈R2:x∈(0,1)} is neither open nor closed because none of its points are interior points and (1,0) is a limit point not in the set. The rational numbers Q are neither open nor closed. Therefore, the set of rationals is neither open nor closed.

**Why are the rationals dense?**

Informally, for every point in X, the point is either in A or arbitrarily “close” to a member of A — for instance, the rational numbers are a dense subset of the real numbers because every real number either is a rational number or has a rational number arbitrarily close to it (see Diophantine approximation).

**Are integers dense?**

The integers, for example, are not dense in the reals because one can find two reals with no integers between them.

## Why are the rational numbers not complete in the sense that?

The real numbers are complete in the sense that every set of reals which is bounded above has a least upper bound and every set bounded below has a greatest lower bound. The rationals do not have this property because there is a “gap” at every irrational number.

### Is there a lower bound of rationals greater than 1?

Similarly, there is no greatest lower bound of the rationals greater than although 1 is a perfectly good example of a lower bound. The easiest way to see this is to construct an increasing sequence of rationals, all of which are lower bounds of that set of rationals greater than .

**How can we prove that Q is not complete?**

We can prove that Q is not complete because the square root of 2 is not a rational number. If x is negative (i.e. smaller than 0 in the ordering) substitute -x in place for that. The key is that the square root of x (which would be the obvious supremum) is not a rational function.

**Why are rational numbers written as integer fractions?**

A number that can be written as an integer fraction. Integers are the whole numbers and their negative counter parts . So the rational numbers are . Now a fraction is fully simplefied when the greatest common divisior of and is . So is even, but if is even so is because an odd number squared would be odd again.