# Math Help - analysis summation

1. ## analysis summation

Hey peeps - i just started my first year at university in Industrial Applied Mathematics. Another problem is i'm not studying in a language i fully understand so it's even more hard to understand. From what i understand from the assignment we were given:

Prove that the sum of the intervals comtaining all rational numbers is less than or equal to one.

Another variation: m/n is a rational number. There is an interval I(m,n).
Prove that (SIGMA) Σ I(m,n) < or = to 1.

The Professor also explained it as: If i give a different quantity of money to every rational number of people could the amount i gave out be less than or equal to \$1.

Now i have NO idea where to start. I know however that the answer is not 1 = 1/2 + 1/3 + 1/4 + 1/5 +...+1/n

2. The heart of this proof depends upon the fact that the set of rationals is countable.
Start by picking a rational, $r_1$. Define $I_1 = \left( {r_1 - \frac{1}{4},r_1 + \frac{1}{4}} \right)$, then pick a rational $r_2 \in \Re \backslash I_1$ and define $I_2 = \left( {r_2 - \frac{1}{{2^3 }},r_1 + \frac{1}{{2^3 }}} \right)$. For $n \ge 3$ pick a rational $r_n \in \Re \backslash \bigcup\limits_{k = 1}^{n - 1} {I_k }$ and define $I_n = \left( {r_n - \frac{1}{{2^{n+1} }},r_1 + \frac{1}{{2^{n+1} }}} \right)$.

What is the sum of the lengths of all the $I_n$?
Do these ‘cover’ the rationals?
In fact, we could make the cover as small as we wish by using $\frac{\varepsilon }{{2^{n + 1} }}$.

3. ## You're a genius....but there is one flaw...

I thought of a solution on the same line as yours - yours although is almost perfect.

In your formulae you are assuming that size of (N) and size of (Q) are the same....I need to prove that....does that also revolve around the fact that all rational numbers are countable...

also - the sum of all the lengths of I(n) has to be < or = to 1...and it has to cover ALL rational numbers...

i like it that you started at 1/(2^n+1) so that the intervals from [-oo; 0] dont double up and add to one...clever...

4. Originally Posted by ARozanski
...

In your formulae you are assuming that size of (N) and size of (Q) are the same....I need to prove that......
Hello,

that isn't necessary because this proof was done by Georg Cantor (in 1891). Google for Cantor and his famous diagonal argument. Maybe you find an example of his proof that the set Q is numerable infinite.

5. Actually there is a correctable flaw in the argument. It was done haste I was on my way to lecture after a frustrating morning of printers not working with my laptop. But anyway, it has to do with care one must take in picking the next rational. I was not careful enough. We really need to pick the first rational in a listing of rational not in the union. That way we can be sure to cover all rationals.

As to the rationals being denumerable, note that $\left( {Z^ + \times Z^ + } \right) \leftrightarrow \left\{ {1,2,3, \cdots } \right\}$ using the function $\phi (m,n) = 2^{n - 1} \left( {2m - 1} \right)$. One that is done it is fairly simple to show that the rational are denumerable.

BTW. earboth I just checked Dauben’s book on Cantor. The 1891 paper was using the diagonal argument to show the nondenumerablity of the real numbers.

6. ## Confusion

Plato - you seem like you should be in league with Einstein - but I have NO idea what you just said...please explain to me in further...

thanks for your help so far...im very close to finishing the project...

oh - and i used the proof to say that the size (N) = size (Q)...

7. I am not sure where to begin. It would seem that you are in need of a great of background material to even start this problem. As earboth told you there is really to worry about the denumerablity of the rationals. That is well known. In any case I gave you a way to proceed with that.

So we make a list of the rationals. We use the first as $r_1$. Then using the well ordering of the positive integers, $r_2$ is the first rational in our list not in $I_1$. Then we proceed to construct our open intervals. That way we can easily show that the union of the $I_n$ cover $Q$.

8. ## ...

Hmmm....

This is my formula ----->

$
Q_n = \left( {Q_n - \frac{1}{{10^{n} }},Q_n + \frac{1}{{10^{n} }}} \right)
$

then i went on to explaining it - AND i proved that size (N) = size (Q) by using the cardinal number rule and some other formula.

am i missing anything else??