Question

Prove that the set of rational numbers is not
complete.​

Answer 1

Step-by-step explanation:

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.

Answer 2

Concept

In mathematical analysis, a metric space M is called complete (or a Cauchy space) if every Cauchy sequence of points in M has a limit that is also in M. Intuitively, a space is complete if there are no missing "points". (inside or at the border)

Given

set of rational number is given

Find

We need to find whether the set of rational numbers is complete or not

Solution

First, note that the rational numbers form a metric space.

We need to show that Q is not complete.

Consider the sequence ⟨an⟩ given by:

an:=f(n+1) / f(n)

where ⟨fn⟩ is a sequence of Fibonacci numbers.

According to the ratio of consecutive Fibonacci numbers:

lim (n→∞)an=ϕ:=1+5–√2

But the square root of Prime is irrational.

This means that 1+5–√2 is also irrational.

Thus ⟨an⟩ is a Cauchy sequence of rational numbers that converges to a number not in Q.

Hence the set of rational number is not complete

#SPJ2