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## Main Question or Discussion Point

Any number c in the real numbers has the form [tex]x.{c_1}{c_2}...{c_n}[/tex], in which x is an integer and [tex]0 \le {c_n} \le 9[/tex] is a natural number. From the way that we have enumerated the decimal places, clearly number of decimal places is countable. Then there is a bijection from the indexes of the decimal places onto the set of the first n natural numbers. Consider the irrational numbers, namely [tex]\mathop {\lim }\limits_{n \to \infty } (x.{c_1}{c_2}...{c_n})[/tex]. The infinity in the enumeration of decimal places must then be the same infinity as that of the natural numbers. Then, [tex]\mathop {\lim }\limits_{n \to \infty } [\frac{{{{10}^n}}}{{{{10}^n}}}(x.{c_1}{c_2}...{c_n})] = \frac{{\mathop {\lim }\limits_{n \to \infty } ({{10}^n} \times x.{c_1}{c_2}...{c_n})}}{{\mathop {\lim }\limits_{n \to \infty } {{10}^n}}} = {\lim }\limits_{n \to \infty } (x.{c_1}{c_2}...{c_n})[/tex]. Is a rational expression in which the numerator and denominator are both integers.

If we consider Euclid's argument and acknowledge that the factor in the numerator is not unique in yielding an integer multiple then we can understand Euclid's flaw. The numerators and denominators will have infinite prime factorizations and therefore it would be impossible to simplify the fraction.

If we consider Euclid's argument and acknowledge that the factor in the numerator is not unique in yielding an integer multiple then we can understand Euclid's flaw. The numerators and denominators will have infinite prime factorizations and therefore it would be impossible to simplify the fraction.

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