About: http://webisa.webdatacommons.org/424565221

Property	Value
http://www.w3.org/1999/02/22-rdf-syntax-ns#type	http://www.w3.org/ns/prov#Entity
http://www.w3.org/ns/prov#value	Let a=kb, where k is an integer, such that k cannot be 1, since in that case a=b which is not allowedHence, \(kb=b^k\)\(b^k-bk=0\)\(b(b^{k-1}-k)=0\)Either b=0 OR \(b=k^{\frac{1}{k-1}}\)b=0 implies a=kb=0.
http://www.w3.org/ns/prov#wasQuotedFrom	gmatclub.com