Prove Isomorphism: R x S & S x R

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Homework Statement



Show that for any rings R and S, R x S and S x R are isomorphic, and A x B is the cartesian product, or ordered pairs. So an element of R x S can be written as (r1, s1).

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The Attempt at a Solution



So I have to show that there is a bijection from R x S to S x R, and this bijection must preserve addition and multiplication. This is tough for me since the mapping from R x S to S x R could be anything! How can I even start if I don't have this function or mapping?
 
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hi stripes! :smile:
stripes said:
… the mapping from R x S to S x R could be anything!

yes it could

but you're in charge, and you can choose any mapping you like :smile:

go for the "natural" mapping …

what do you think it would be really neat for (r1, s1) to be mapped onto? :wink:​
 

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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