How does denseness in Q imply a sequence exists converging to x in R?

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Denseness of the rationals (Q) in the reals (R) implies that between any two real numbers, there exists a rational number, which allows for the construction of a sequence of rationals converging to any real number x. This does not require x to be irrational, as sequences can also converge to rational numbers. For every irrational number, a sequence of rational numbers can indeed be formed that converges to it. The process involves iteratively selecting rationals closer to x, demonstrating convergence through decreasing distances. The discussion clarifies the concept of denseness and its implications for sequences in real analysis.
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In this link:

http://math.berkeley.edu/~scanez/courses/math104/fall11/homework/hw7-solns.pdf

In number 3, the text says "By the denseness of Q in R, there exists a sequence (r_n) of rationals converging to x."

I have several questions about this:

1) Why/how does denseness imply that there exists a sequence? because Q is dense in R, we know that any two real numbers have a rational number between them. But how does that tell us that there is a "sequence" of rational numbers?

2) Do we have to assume that x is an irrational number?

3) Does this statement mean that for every irrational number, there is a sequence of rational number converging to it?

Thanks in advance...
 
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Artusartos said:
1) Why/how does denseness imply that there exists a sequence? because Q is dense in R, we know that any two real numbers have a rational number between them.
Well, that's not all that denseness tells you. Between any two even integers is an odd integer, but that doesn't mean that the odd integers are dense in the even integers. What is the definition of denseness?

2) Do we have to assume that x is an irrational number?
The first paragraph proves that f(x) = x for any rational x. The second paragraph uses that fact, but you don't have to assume that x is irrational in the second paragraph. The argument also works for rational x, because given a rational number, you can certainly find a sequence of rationals converging to that number.

3) Does this statement mean that for every irrational number, there is a sequence of rational number converging to it?
I'm not sure what "this statement" refers to, but it's true that for every irrational number, there is a sequence of rational numbers converging to it.

If x is irrational, start by picking any rational, say r_1. Since x \neq r_1, the distance between these numbers is positive: |x - r_1| > 0. Choose a rational r_2 such that |x - r_2| < (1/2)|x - r_1|, i.e. r_2 is less than half the distance from x as r_1.

Then just keep repeating this process. At the n'th step, you will choose a rational r_n such that |x - r_n| < (1/2)^n |x - r_1|. Clearly the sequence r_n converges to x.

Another, more concrete (but less rigorous) way to see this is to consider the decimal expansion of x. Any real number has a decimal expansion. You can form a sequence of rationals converging to the real number by simply taking more and more digits, e.g.

x_1 = 3
x_2 = 3.1
x_3 = 3.14
x_4 = 3.141
x_5 = 3.1415
x_6 = 3.14159
and so on
 
jbunniii said:
Well, that's not all that denseness tells you. Between any two even integers is an odd integer, but that doesn't mean that the odd integers are dense in the even integers. What is the definition of denseness?

This is the definition that my textbook gives us "If a,b in R and a < b, then there is a rational r in Q such that a < r < b".

But I think I understand why sequences exist, from the answer you gave to my 3rd question. Thanks a lot. :)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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