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Every function f: Z-> R is injective Answer: False. Why tho? Doesn't all the integer hit...

Every function f: Z-> R is injective Answer: False. Why tho? Doesn't all the integer hit real number?

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Answer #1

NO! Recall the definition of injective maps - A function is injective if it takes distinct elements of the domain to distinct elements of the co-domain set, or in other words, it is a map in which no two distinct elements have the same image. Consider any constant function f from Z to R, for instance consider the map f which takes all integers to the real number 0, (say.) Clearly this map is not injective, as the two distinct integers 1 and 2 have the same image 0. (There's no speciality about 1 and 2, you can consider any two different integers!)

I guess why you asked this question. Maybe it is because of the fact that Z is a subset of R. But note that, if A is a subset of B, you can always get atleast one injective function from A to B, which is the inclusion map itself. But in general not EVERY function is injective! (Unless A is a singleton)

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