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Exercise 1.3 · Q4

Q.State whether the following relations are functions or not. If it is a function check for one-to-oneness and ontoness. If it is not a function, state why?

(i) If A={a,b,c}A=\{a,b,c\} and f={(a,c),(b,c),(c,b)}f=\{(a,c),(b,c),(c,b)\}; (f:A→Af:A\to A).
(ii) If X={x,y,z}X=\{x,y,z\} and f={(x,y),(x,z),(z,x)}f=\{(x,y),(x,z),(z,x)\}; (f:X→Xf:X\to X).
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Concept understanding — Types of Functions

Definition. A relation f⊆A×Bf\subseteq A\times B is a function f:A→Bf:A\to B if (i) every a∈Aa\in A has some image b∈Bb\in B with (a,b)∈f(a,b)\in f, and (ii) that image is unique: (a,b),(a,c)∈f⇒b=c(a,b),(a,c)\in f\Rightarrow b=c. Then f(a)=bf(a)=b; bb is the image of aa, aa is a pre-image of bb. The range {b:(a,b)∈f for some a}\{b:(a,b)\in f\text{ for some }a\} is always ⊆\subseteq co-domain BB. Only the domain side is required to be fully, uniquely covered -- how many pre-images a co-domain point has, or whether it has any, are separate questions (injectivity/surjectivity below).

Representing a function: tabularly (a list of argument/value pairs), graphically (plot with the Vertical Line Test: a curve is a function's graph iff every vertical line meets it at exactly one point), or analytically (a formula, whose natural domain is wherever that formula is actually defined -- found by excluding zero denominators, requiring even-root radicands ≥0\ge0, etc., often via a sign-chart over intervals cut out by the critical points). Functions may also be piecewise (different formula on different sub-intervals).

Named elementary functions: identity (f(x)=xf(x)=x), constant (and the zero function as its special case), modulus ∣x∣|x|, signum x/∣x∣x/|x| (with 0↦00\mapsto0), floor ⌊x⌋\lfloor x\rfloor (always rounds down, even for negatives) and ceiling ⌈x⌉\lceil x\rceil (always rounds up) -- the last two are "step functions". …

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