Discrete Maths - Function Watch

Maths94
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Report Thread starter 5 years ago
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I'm currently studying CS, and as i didn't do maths A level i'm finding the module particularly difficult. We've now changed topics and lecturer, going onto discrete maths; and i'm refusing to fall behind :P. So, i'm going to post regularly/daily questions, just to make sure i have an understanding.

Hopefully some of you guys have the time to give detailed answers to give me some sort of foundation

Question - Range of a Functon

Let X = {a,b,c,d} and Y = {1,2,3,4,5} and define f: X>(that is meant to be an arrow) Y by f(a) =1, f(b) =2, f(c) = 5, f (d) = 2.

Find the domain, codomain and range. If someone could explain this question in detail so i can do some revision on it, i'd be grateful. Thanks
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Maths94
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Resolved my first question, from the help of another forum. If anyone's interested :

"By definition the domain of f is the set of all inputs for which f is defined; in this case that’s {a,b,c,d}=X. This is actually implicit in the notation f:X→Y, which almost always implies that that the domain of f is X. (I say almost because in some areas of mathematics one deals with so-called partial functions from X to Y, whose domains may not be all of X. I would not worry about this: it should not come up in what you’re doing.)

The codomain can also be read straight from the notation f:X→Y: it’s the target set Y, which here is {1,2,3,4,5}. The range is always a subset of the codomain: it’s the set of values that the function actually assumes (or if you prefer — and in CS you might! — outputs). For your function f those values are 1,2, and 5, so the range of f is the set {1,2,5}.

That’s all there is to it."
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