anonymous
  • anonymous
what is a function? just a simple question. who can answer?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
a function is defined as for any one input; there is only one output that can be produced
amistre64
  • amistre64
a relation exists between any given sets of data; but a function is useful in that we can use it to predict events
anonymous
  • anonymous
ahaa tnx

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anonymous
  • anonymous
Let A and B be sets of numbers. A function f is a rule that assigns each element of A to a unique element of B.
anonymous
  • anonymous
what if B has infinitely many element from A ., is that can be called as a function?
amistre64
  • amistre64
If A results in many B's then a relation exists; but not a function
amistre64
  • amistre64
If B results from multiple A's, then a function can be determined to best match the results
anonymous
  • anonymous
*an
anonymous
  • anonymous
how is it done?
anonymous
  • anonymous
give all the functions. A={1,2,3} B={d,f}
anonymous
  • anonymous
There are infinitely many.
anonymous
  • anonymous
prove it.
amistre64
  • amistre64
are you mapping A into B?
amistre64
  • amistre64
f:A -> B
anonymous
  • anonymous
f:A->B
anonymous
  • anonymous
Let A={x: x=1,2 or 3} and B={d or f}. Then\[f:A \rightarrow B \] where f can be any function.
anonymous
  • anonymous
Can you show it for me all the functions?
anonymous
  • anonymous
No. And here are some corrections: B={g(x): g(x)=d or f} and g: A -> B.
anonymous
  • anonymous
Suppose that the set C containing all g such that g: x -> g(x) (where x is in A and g(x) is in B) is finite. Then there exists a h in C such that h: x -> h(x), where h(x) is in B. But h1= h(x)+1-1 is also in C, and so is h2=h1+1-1, and so is h3=h2+1-1, . . . Therefore, C is infinite. This is a contradiction. Therefore, C must be infinite.
anonymous
  • anonymous
And therefore there are infinitely many g.
anonymous
  • anonymous
No: wait. I have made an error. Two function are the same if their domains (A in this case), codomains (B), and effects are the same. Thus h=h1=h2= . . . However, I still cannot list all functions g by the same argument as in the above comment. Can you see why?

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