Show that the elements of the lattice (N,≤), where N is the set of positive intergers and a≤b if and only if a divides b, satisfy the distributive property. Since in distributive lattice , atmost one complement exist for each element. So , if we get 2 complements for an element then we can say given lattice is not distributive. So, to prove above statement , I assumed 2 complements 'b' and 'c' for an element 'a' now I am unable to check these complements exist or not for a given relation "x divides y" because we can't say about upper bound of the lattice because given set is set of positive integers. So , how to prove it. Please help!
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This question was also answered here – amrsa May 12 '18 at 11:16
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It looks as if you're making the logical error of trying to prove distributivity by proving one of its consequences, namely uniqueness of complements. – Andreas Blass May 12 '18 at 12:12
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Thank you !@amrsa @Andreas – ankit May 12 '18 at 14:31
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The meet of two elements is the greatest common denominator
(gcd), the join is the least common multiple (lcm).
Show distributivity directly
a gcd (b lcm c) = (a gcd b) lcm (a gcd c).
Use the prime factorisation theorem to obtain
expressions for the gcf and lcm to prove this.
Clearly this lattice does not provide complements.
Thus you cannot use the unique complement property.
William Elliot
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hello sir, could you please explain why this equation a gcd (b lcm c) = (a gcd b) lcm (a gcd c). is always true for all a,b,c of the set N – ankit May 12 '18 at 09:07
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Use the prime factorisation theorem to obtain expressions for gcf and lcm. – William Elliot May 12 '18 at 21:50
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