SOLUTION: show that the mapping f:N->Q defined by f(n) = 1/n, where n is a natural number is one-one but not onto

Question 1132672: show that the mapping f:N->Q defined by f(n) = 1/n, where n is a natural number is one-one but not onto
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
A mapping(f) from a set X to a set Y is onto if and only if for every member y of the set Y there is some x in the other set X for which f(x)=y.
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N is the natural numbers, 1, 2, 3, 4, 5, ...
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Q is the rational numbers, x/y such that x is an element of Z and y is an element of N
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Z is the integers, ..., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ...
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f(n) = 1/n where n is an element of N is 1-1 since each element in N is associated with only 1 element in Q
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f(n) is not onto since f(n) is always positive and Q has negative elements
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