Please help me! how to proof that the rational numbers are dense in the real numbers? Thank you?

Question

how do I prove that the rational numbers are dense in the real numbers?

Answer 1

Well, there are many ways to prove this, depending on what axioms you have. In all cases below, I'll assume the definition of "X dense in Y" as "every open set in Y contains a point of X." This is equivalent to any other definition you choose.

Here's one proof:
Let x be any real number, and ε>0; we need to find a rational number in (x-ε,x+ε). By the archimedian principle, for any real number M there is a natural number J greater than M. Let n > 1/ε; then 0 < 1/n < ε. Now, consider the rational numbers {j/n: j an integer}. I claim that one of these must lie between (x-ε) and (x+ε), since (x+ε) - (x-ε) = 2ε while (j+1)/n - j/n < ε. Therefore, some j/n is in (x-ε,x+ε).
§

A different proof is to go back and look at how you construct the real numbers; it's pretty immediate from Dedekind cuts, for example.
§§

A third proof is simply to look at decimal representations of real numbers. If x is a real number and ε>0, let d_i be the first digit at which the decimal representation for x+ε differs from that for x. Then a rational number between x and x+ε is given by taking just the first i digits of x+ε. (For example, if x=√2 and ε=0.001, then q=1.415 is a rational between x and x+ε.)
§§§

Answer 2

Rational Numbers Dense

Answer 3

By dense, I think you mean that the closure of the rationals is the set of the real numbers, which is the same as saying that every open interval of R intersects Q. We can do this by means of the decimal representation of a rational number, but I think it's better to take a different approach. Let's show that for every real numbers x < y there is a rational r in (x, y).

To make things simpler, first suppose 0 <= x < y, with y - x >1. Then, there exists a postive integer n - so a rational number - satisfying x < n < y, which proves the assertion for this 1st case.

Now, suppose 0 <= x < y with y - x <=1. By the Archmedian property of the natural numbers, there exists a positive integer n such that n(y - x) = ny - nx > 1. By the previous conclusion, there exists a positive integer m such that

nx < m < ny => x < m/n < y. So, m/n is a rational in (x, y).

Finally, suppose x < y are any real numbers. There exists
a rational r1 such that 0 < r1 + x < r1 + y. From what we have seen, there exists a rational r2 such that

r1 + x < r2 < r1 + y => x < r2 - r1 < y. So, r2 - r1 is a rational in (x, y).

The proof is now complete.

This can also be done based on the construction of the set of real numbers from the rationals. One approach is based on Cauchy sequences. In this approach, a real number is defined as the limit of a Cauchy sequence, which makes the conclusuion automatic.

Anoter approach is by means of Dedekind's cuts.

Answer 4

I'm going to have to guess what you mean by dense. All rational numbers are real, but not all real numbers are rational. Rational numbers are ratios of integers.

Take two rational numbers: 1/4 and 2/7.
Find their average: (1/2)(1/4 + 2/7) = 15/56
15/56 is a ratio of integers, so it's a rational number
You can keep finding the average between the previous average and either of the latest bounds, subdividing intervals without end.

But I'm not really sure what you mean by dense.

Answer 5

The only imaginary numbers are things like the square root of a negative number. It's not possible, as squaring times a number by itself, and positive x positive = positive, negative x negative = positive, so no number can be squared to reach a negative number. Rational numbers are real numbers, becuase, quite frankly, you can write them. If they can be written, they probably are real.

Check here: http://en.wikipedia.org/wiki/Real_number
And here: http://en.wikipedia.org/wiki/Rational_numbers

Answer 6

which type of numbers is a subset of the other ?
perhaps its a clue

Answer 7

I'm sorry the persons you are trying to reach do not understand broken English. Please hang up and try again.

Answer 8

where is the question???