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Prove : if a and b are any positive integers, then there exists a positive integer n such that n*a >= b
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assume the statment is not true, so for some a,b na>b for positive integer n then the set S={b-na , n positive integer} by the "well-ordering princple" S will posses a least element ,say b-ma. notice that b-(m+1)a also lies in S. becouse S contains all integer of the form , furthermore, we have b-(m+1)a=(b-ma)-a<b-ma contrary to the choice of b-ma as the smallest integer in S . this contradiction arose out of our original assumption that the Archimeadean property didnt hold; ence this property is proven true. got it or still confused ??
actually this one :P i have doubt if u got it xD @ganeshie8
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