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Functions  is this proof satisfactory? 
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#1
Mar1612, 02:42 AM

P: 272

1. The problem statement, all variables and given/known data
Let f be a function from A to B and g a function from B to C. Show that if the composite function g°f is oneone, then f is oneone. 2. Relevant equations 3. The attempt at a solution By definition of a function every value of A has to be maped to a value in B. Likewise every value of B has to be maped to a value in C. So by definition of 11... since g°f satisfies this rule and g°f is a composite function... Therefore f must be 11. Does this make a valid argument? and justify the answer? 


#2
Mar1612, 03:28 AM

P: 85

Its not so clear to me.
If f(x) = f(y) for some x,y in A you have to show x=y. 


#3
Mar1612, 03:35 AM

P: 272

hmm I think I dont understand.



#4
Mar1612, 03:43 AM

P: 297

Functions  is this proof satisfactory?
You have to prove that if f(x) is not one one
then function g(f(x) is not one one. Suppose that f(x) is not one one and for two values of x lets say x1 and x2(both lying in the domain) give a value y1. Now this y1 lies on domain of function g. What happens when you calculate g(y1)? What does it tell us about our original assumption the f(x) is not one one. p.s Your proof is not right.(even if you understand why f should be one one, the proof is not satisfactory) 


#5
Mar1612, 03:58 AM

P: 272

But from the assumption say I have 3 sets A,B,C the same as the original question. Now x1 & x2 lie in A. when we plug in the values x1 & x2 into f wwe have they both = y1 which lies in B. So from this picture alone the function f :A→B is not oneone. But we have to prove the opposite! And to this statement. "What happens when you calculate g(y1)?" if the value of f at x1 and x2 = y1.... then the above statement is just saying g(f(x)). I'm confused? 


#6
Mar1612, 04:11 AM

P: 297

You are given g(f(x)) is one one.now you are supposed to prove f(x) is one one.
To prove this we say let f(x) be manyone(i.e not one one). In that case we observe that g(f(x)) comes out to be many one and not one one(How?).However we are told that g(f(x)) is one one. This contradiction signifies that our assumption of f(x) being many one is wrong and therefore f(x) is one one. This is the way you can proceed.Solving the indivudual bits is now upto you 


#7
Mar1612, 08:16 AM

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#8
Mar1612, 05:57 PM

P: 272

ie, f(x1) & f(x2) cannot both = y1. 


#9
Mar1612, 06:35 PM

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#10
Mar1612, 07:37 PM

P: 272

Im not so good putting what I want to say in words. 


#11
Mar1612, 07:57 PM

P: 297

Hello charmed beauty.
The most elegant propf will be by proving that f(x) cant be many one(which i mentioned a while ago). Drawing VennDiagrams is also a v useful approach. However, as the no. of functions forming the composite increase, proving using Ven Diagrams will get v big. But they are good way to start as they convey a lot of information 


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