MHB Right inverse clarification needed

[dana1]

f: (R*R)->R
f(x,y)=x+y

if I'm asked to write 2 right inversed fanctions of f.
can I say that:

f1: R-> (R*R)
f1(x)= (x-1, 1)

f2: R-> (R*R)
f1(x)= (x-2, 2)

because: f(f1(x))= f(x-1,1)=x-1+1=x

well this does matches the definition of right inverse function but what bothers me
I guess is that there is no more y on f1.
but it seems more of a problem to define:
f3(x)= (x-y, y)
because then we will receive a hole range of functions. (Whew)

please can someone clarify this point out? thanks so much in advance!(Blush)

 

[I like Serena]

f: (R*R)->R
f(x,y)=x+y

if I'm asked to write 2 right inversed fanctions of f.
can I say that:

f1: R-> (R*R)
f1(x)= (x-1, 1)

f2: R-> (R*R)
f1(x)= (x-2, 2)

because: f(f1(x))= f(x-1,1)=x-1+1=x

well this does matches the definition of right inverse function but what bothers me
I guess is that there is no more y on f1.
but it seems more of a problem to define:
f3(x)= (x-y, y)
because then we will receive a hole range of functions. (Whew)

please can someone clarify this point out? thanks so much in advance!(Blush)

Hi dana! Welcome to MHB!

That all seems fine to me. ;)

To make the matter less confusing, I would suggest to use a different letter, say t.
So you would have for instance:
$$f_1(t) = (t-1,t)$$
That is be cause this $t$ is in no way connected to the $x$ and $y$ that you have in $f$.

The $y$ that you have in $f_3$ is also not connected to the $y$ in $f$.
It is just an arbitrary constant.
So it would be less confusing to write:
$$f_3(t) = (t - c, c)$$

 

[chisigma]

f: (R*R)->R
f(x,y)=x+y

if I'm asked to write 2 right inversed fanctions of f.
can I say that:

f1: R-> (R*R)
f1(x)= (x-1, 1)

f2: R-> (R*R)
f1(x)= (x-2, 2)

because: f(f1(x))= f(x-1,1)=x-1+1=x

well this does matches the definition of right inverse function but what bothers me
I guess is that there is no more y on f1.
but it seems more of a problem to define:
f3(x)= (x-y, y)
because then we will receive a hole range of functions. (Whew)

please can someone clarify this point out? thanks so much in advance!(Blush)

You can define an inverse function $\displaystyle x = f^{- 1} (y)$ if it exists a function $\displaystyle y = f(x)$ in which for any value of x there is one and only one value of y. Similarly if You have a two variable function z= f(x,y) You can define two inverse two variable functions $\displaystyle x= f^{-1} (y,z)$ and $\displaystyle y = f^{-1}(x,z)$ if the same conditions are satisfied. In Your case is $\displaystyle z = f(x,y) = x + y$ , so that the two inverse functions are $\displaystyle x = f^{-1} (y,z) = z - y$ and $\displaystyle y = f^{-1} (x,z) = z - x$...

Kind regards$\chi$ $\sigma$