For any natural number a, we define aN={ax:x belongs to N}. If b, c, d belongs to N such that bN intersection cN = dN, then prove that d is the LCM of b and c.

Asked by mesamit | 30th Jul, 2010, 10:45: AM

Expert Answer:

Given:aN={ax/xN} ; bN={bx/xN}; cN={cx/xN}; dN={dx/xN}Also bNcN=dNAny element y in the set dN is element of both bN and cNHence y is a multiple of both b and  c.Therefore y must be the L.C.M of b and c.But y also belongs to dN , hence is a multiple of d.Therefore d is L.C.M of b and c.">

Answered by  | 30th Jul, 2010, 10:30: PM

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