35) Let 6 > 0. There exists an N E N so that 0

35) Let e > 0. There exists an N e N so that 0 < < e, because a) b) c) d) the Archimedean Property shows that 3 is not an upper bound for N. the Archimedean Property shows that c- is not an upper bound for N. 1 the Archimedean Property shows that 3 is not an upper bound for N. the Archimedean Property shows that -r is not an upper bound for N.