11. A function f: X → Y is bijective, where X and Y are finite sets. The number of possible relations from X to Y is 512. If X and Y have exactly one element in common, then what is the number of elements in X ∪ Y

Question

11.	A function f: X → Y is bijective, where X and Y are finite sets. The number of possible relations from X to Y is 512. If X and Y have exactly one element in common, then what is the number of elements in X ∪ Y

experimentx · Answer

seriously, isn't the number of possible relation supposed to be the product of number of elements in those two sets??

anonymous · Answer

square root of 512 is 22.6. isnt the answer a interger?? thats what baffles me

anonymous · Answer

@experimentX

experimentx · Answer

i thought the same ..

anonymous · Answer

since its bijective, then means that f is surjective (for every element in Y, there is an element in X which corresponds to it ) and injective(its one to one, meaning for every element in Y, there could only be one element in X corresponding to it and the same goes for X). This will look something like this:
|dw:1338839137033:dw|
so that means, that if I have a set A and set B with 3 and 3 elements respectively, and that with f:A->B, where f is bijective, then I will only be able to generate 3 relations, if I have more than that, it will not be injective which contradicts that f is bijective, and if we have less, it cannot be surjective which also contradicts. and since A has 3 elements and B has 3 elements AUB will have 6 elements supposing they have no same elements. if they have same elements of course AUB will have 5 elements. Now that means also the same for our two sets. since with f:X->Y, we get 512, and since f is bijective, X must have 512 elements which f will map one by one to distinct 512 elements in Y. otherwise it will not be bijective as stated. So X has 512 elements and Y has 512 elements. now since they have 1 same element then the cardinality of the set XUY is 512+512-1=1023. any questions?