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Needs help in function

Posted on 2011-09-08
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Last Modified: 2012-05-12
Define a function f: R-->R by the formula f(x) = 3x - 5.
a. Prove that f is one-to-one
b. Prove that f is onto.

Thanks.
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Question by:mustish1
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6 Comments
 
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Expert Comment

by:TommySzalapski
ID: 36507494
Prove it by contradiction. Assume that it is not one-to-one, then find a contradiction.
If it is not one-to-one, then there exist two values x1 and x2 such that f(x1) = f(x2) but x1 not= x2 or x1 = x2 and f(x1) not= f(x2)
Continue this until it fails miserably. The onto proof is very similar.

This is an academic question so to facilitate learning, please no one just jump in with a full solution.
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LVL 31

Accepted Solution

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GwynforWeb earned 250 total points
ID: 36507676
Keep it simple is the key to this question.

Suppose y= 3x-5 then

   x= (y+5)/3

then for

(1) For any y in R there is only one possible x hence 1-1.  ( given by x= (y+5)/3 )

(2) For any y in R there is an x such y =  3x-5.  ( given by x= (y+5)/3 )
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LVL 37

Assisted Solution

by:TommySzalapski
TommySzalapski earned 250 total points
ID: 36507707
To me that doesn't seem like the kind of rigorous proof that would be required for that type of question. If you do have rules that you can use though to show one-to-one, then use them of course.
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LVL 37

Expert Comment

by:TommySzalapski
ID: 36507728
The goal, of course, isn't to prove one-to-one and onto; it's to prove them given the constructs that have been provided to you. Since we have no idea what those are, we can only throw out suggestions. Gwen's solution is very intuative and anyone can understand it fairly easily. If it works for your class/self-learing/whatever, then it's much simpler than mine and it great. If you are in some kind of foundations or logic class where you need to use very specific rules, then proof by contradiction is almost always included.
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LVL 31

Expert Comment

by:GwynforWeb
ID: 36508265
... my guess is that is an introductory course on functions. I doubt proof by contradiction has been covered yet.
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LVL 37

Expert Comment

by:TommySzalapski
ID: 36510261
Could be.
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