Is {R-Z} a Subring of the Reals?

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Let S = {R-Z}, the set of all reals that are not integers. Is S a subring of R? I think not because 1/2 is in S but 1/2-1/2=0 so S is not closed under subtraction so is not a subring.

is that right?
 
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what do you think? and why do you think it? if i say you are wrong, would you believe me? why or why not?
 
i think I am right, I am asking because some of the kids in my class said zero is not an integer and they said i should of picked two distinct elements to show its not closed but i said it didnt matter.
 
In which class are you learning about rings where the students don't believe that 0 is an integer and believe that you have to pick two distinct elements to show it's not closed?
 

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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