Countable vs Finite Rationals in (0,1)

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Homework Statement


Are there countably many rational numbers in the interval (0,1) or are there finitely many?


Homework Equations





The Attempt at a Solution


I am confused. There are countably many rational numbers in the interval (0,1). Does this mean I can list them all in such a way that I can theoretically stop at the very last element and therefore say that there are finitely many rational numbers in (0,1).
 
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lmedin02 said:

Homework Statement


Are there countably many rational numbers in the interval (0,1) or are there finitely many?


Homework Equations





The Attempt at a Solution


I am confused. There are countably many rational numbers in the interval (0,1). Does this mean I can list them all in such a way that I can theoretically stop at the very last element and therefore say that there are finitely many rational numbers in (0,1).

No. Countably many does not mean you can count them like you would a bag of marbles 1,2,3,...25, for example, and be done with it. That would be finitely many. It means there is a 1-1 correspondence between the rationals on (0,1) and the positive integers, and, as you know, there isn't a last one of those.
 
thank you
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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