User:Tetrahedron93/Continuum

Let $$f : {Z_+}^n -> R$$ $$f[ (a,b,c...)] = \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}+...$$ As $$1/x$$ is defined for all $$x \in Z_+$$, and $$a+b$$ is defined for all $$a,b \in R$$, it follows that f is defined for all input. It therefore maps all $${Z_+}^n$$ to some member of R. It therefore follows that $$|{Z_+}^n| >= |R|$$

Also, let $$g(R) : R -> {Z_+}^n$$. g(R) is defined as the set produced when each 0 contained in the decimal expansion of R is treated as separating the integers members. Although g(R) is not defined for recurring decimals that do not contain a 0, there is still a real number that constructs ever g(R). Hence, $$|R| >= |{Z_+}^n|$$

From the two conclusions above, $$|{Z_+}^n|=|R|$$