Sequence that has all rational numbers

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Homework Statement



Construct a sequence that has all rational numbers in it

Homework Equations



None.

The Attempt at a Solution



Here are my thoughts, though I have no solutions yet.

If I construct a sequence Sn= n*sin(n)-1/n, will it work?

Thanks guys!
 
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The sine term will often give irrational numbers, so that won't work. Try putting the rationals in an array and finding a path that goes through all of them.
 
thanks! can you elaborate a little bit? I'm trying to self-study real analysis, and I'm not really familiar with what you just mentioned...
how would i put them in an array and find a "path"?

thanks!
 
ask_questions said:

Homework Statement



Construct a sequence that has all rational numbers in it

Homework Equations



None.

The Attempt at a Solution



Here are my thoughts, though I have no solutions yet.

If I construct a sequence Sn= n*sin(n)-1/n, will it work?

Thanks guys!
Just drop the sine term.

OR

Do you really want a sequence with all of the rational numbers in it. -- maybe just all of the positive rationals?

Better yet: Please type the problem word fro word as it was presented to you.
 
Hi guys:

Thank you so much! Here's the problem as it was typed on the book:

Construct a sequence such that every real number is its limit point.

I know this is different from the question I typed above, but my reasoning is that if i can have a sequence that contains all rational numbers, then I can prove that every real number is its limit point. Does that make sense? How should I solve the original question if this does not?

Thank you!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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