A proof using the completeness axiom

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We prove the Archimedean principle doing a proof by contradiction using the completeness axiom
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We can contradict right at the second line after negated the statement. Suppose there is an x in R that is larger than any natural number n. Then the set of the natural number is bounded above, but N is not bounded above, a contradiction.

sitienlieng
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Archimedean principle use proof by contradiction please

yparraguirreromalynq.