One to one function

  • Thread starter mohabitar
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  • #1
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h(x,y)=x/(y+1)

I'm not understanding why this function is NOT one to one? How do I quickly see if this function is one to one? Im not getting the overall concept of this..
 

Answers and Replies

  • #2
HallsofIvy
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A function is "one to one" if and only if different values of the arguments give different values of the function.
A function of two variables, f(x, y) is "one to one" if and only if f(x, y)= f(x', y') implies that x'= x and y'= y. That is not the case here.

[tex]h(2, 1)= \frac{2}{2}= 1[/tex]
[tex]h(5}{4}= \frac{5}{5}= 1[/tex]

In fact, any point on the line x= c(x+ y) gives h(x, y)= c.
 

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