find a 1-1 function f from J onto S where S is the set of all odd integers Solut

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Explanation / Answer

Let f J ->S be defined as: f(n) ={ n if n is odd ; -(n-1) if n is even With this function n 2 {1, 3, 5, 7 . . . } will be taken to themselves which are all the positive numbers in S. And if n 2 {2, 4, 6, 8 . . . } then we shift one back to an odd number and then take the opposite to ensure we hit all the negative odd numbers. So our function is definitely onto. As well this function is obviously 1-1, but to be more rigorous lets consider all the possibilities. If we had f(n1) = f(n2) then either n1 and n2 were both even, both odd, or one was even and one was odd. If n1, n2 both even then that means n1 = n2 which is trivial. If n1, n2 both odd then we get -(n1 - 1) = -(n2 - 1) =) n1 - 1 = n2 - 1 =) n1 = n2. Last, one was even and one was odd. Without loss of generality let n1 be even and n2 be odd. Then we have n1 = -(n2 - 1) but this means that n1 is a negative number which is obviously a contradiction if n1 belongs to J. This means it isn't even possible that one was even and one was odd. And the only other possibilities lead to f being 1-1. So we have an equivalence function from J -> S which means S is countable.

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