One to one function

mohabitar

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..

HallsofIvy

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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mohabitar	One to one function 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..

HallsofIvy Recognitions: Gold Member Science Advisor Staff Emeritus	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.