Mixing
Why is the inverse function of a monotonous function $f:mathbb{R}rightarrow mathbb{R}$ the reflection on the...
$begingroup$
Every monotonous function $f:mathbb{R}rightarrowmathbb{R}$ is injective, therefore $f:mathbb{R}rightarrow f(mathbb{R})$ is bijective. I.e. it is invertible. The function $bar{f}:f(mathbb{R})rightarrowmathbb{R}$ is the inversefunction. Where is the Connection with the reflection on the diagonal intersecting $0$ of the coordinate System, or also the $id$ function.
real-analysis
edited Jan 12 at 13:49
mwt
943416
asked Jan 12 at 13:20
RM777RM777
41112
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add a comment |
$begingroup$
Every monotonous function $f:mathbb{R}rightarrowmathbb{R}$ is injective, therefore $f:mathbb{R}rightarrow f(mathbb{R})$ is bijective. I.e. it is invertible. The function $bar{f}:f(mathbb{R})rightarrowmathbb{R}$ is the inversefunction. Where is the Connection with the reflection on the diagonal intersecting $0$ of the coordinate System, or also the $id$ function.
real-analysis
edited Jan 12 at 13:49
mwt
943416
asked Jan 12 at 13:20
RM777RM777
41112
$endgroup$
$begingroup$
Because if $(x,y) in Gr_{f}$ then $(y,x) in Gr_{f^{-1}}$ and vice-versa, it's not too difficult to show this.
$endgroup$
– rtybase
Jan 12 at 13:31
add a comment |
$begingroup$
Every monotonous function $f:mathbb{R}rightarrowmathbb{R}$ is injective, therefore $f:mathbb{R}rightarrow f(mathbb{R})$ is bijective. I.e. it is invertible. The function $bar{f}:f(mathbb{R})rightarrowmathbb{R}$ is the inversefunction. Where is the Connection with the reflection on the diagonal intersecting $0$ of the coordinate System, or also the $id$ function.
real-analysis
edited Jan 12 at 13:49
mwt
943416
asked Jan 12 at 13:20
RM777RM777
41112
$endgroup$
Every monotonous function $f:mathbb{R}rightarrowmathbb{R}$ is injective, therefore $f:mathbb{R}rightarrow f(mathbb{R})$ is bijective. I.e. it is invertible. The function $bar{f}:f(mathbb{R})rightarrowmathbb{R}$ is the inversefunction. Where is the Connection with the reflection on the diagonal intersecting $0$ of the coordinate System, or also the $id$ function.
real-analysis
real-analysis
edited Jan 12 at 13:49
mwt
943416
asked Jan 12 at 13:20
RM777RM777
41112
edited Jan 12 at 13:49
mwt
943416
asked Jan 12 at 13:20
RM777RM777
41112
edited Jan 12 at 13:49
mwt
943416
edited Jan 12 at 13:49
mwt
943416
edited Jan 12 at 13:49
mwt
943416
943416
asked Jan 12 at 13:20
RM777RM777
41112
asked Jan 12 at 13:20
RM777RM777
41112
asked Jan 12 at 13:20
RM777RM777
41112
41112
$begingroup$
Because if $(x,y) in Gr_{f}$ then $(y,x) in Gr_{f^{-1}}$ and vice-versa, it's not too difficult to show this.
$endgroup$
– rtybase
Jan 12 at 13:31
add a comment |
$begingroup$
Because if $(x,y) in Gr_{f}$ then $(y,x) in Gr_{f^{-1}}$ and vice-versa, it's not too difficult to show this.
$endgroup$
– rtybase
Jan 12 at 13:31
$begingroup$
Because if $(x,y) in Gr_{f}$ then $(y,x) in Gr_{f^{-1}}$ and vice-versa, it's not too difficult to show this.
$endgroup$
– rtybase
Jan 12 at 13:31
$begingroup$
Because if $(x,y) in Gr_{f}$ then $(y,x) in Gr_{f^{-1}}$ and vice-versa, it's not too difficult to show this.
$endgroup$
– rtybase
Jan 12 at 13:31
add a comment |
1 Answer
1
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$begingroup$
As per my comment, let's define $Gr_f =left{(x,y)in mathbb{R}^2 mid f(x)=yright}$ (from Graphic).
Now, if $(x,y)in Gr_f iff f(x)=y iff f^{-1}(y)=x iff (y,x) in Gr_{f^{-1}}$, $f^{-1}$ is the notation for the $f$'s inverse. As a result
$(x,y) in Gr_f iff (y,x) in Gr_{f^{-1}}$
But $(x,y)$ and $(y,x)$ are symmetric to each other with respect to $id_{mathbb{R}}$
answered Jan 12 at 14:14
rtybasertybase
10.9k21533
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add a comment |
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$begingroup$
As per my comment, let's define $Gr_f =left{(x,y)in mathbb{R}^2 mid f(x)=yright}$ (from Graphic).
Now, if $(x,y)in Gr_f iff f(x)=y iff f^{-1}(y)=x iff (y,x) in Gr_{f^{-1}}$, $f^{-1}$ is the notation for the $f$'s inverse. As a result
$(x,y) in Gr_f iff (y,x) in Gr_{f^{-1}}$
But $(x,y)$ and $(y,x)$ are symmetric to each other with respect to $id_{mathbb{R}}$
answered Jan 12 at 14:14
rtybasertybase
10.9k21533
$endgroup$
add a comment |
$begingroup$
As per my comment, let's define $Gr_f =left{(x,y)in mathbb{R}^2 mid f(x)=yright}$ (from Graphic).
Now, if $(x,y)in Gr_f iff f(x)=y iff f^{-1}(y)=x iff (y,x) in Gr_{f^{-1}}$, $f^{-1}$ is the notation for the $f$'s inverse. As a result
$(x,y) in Gr_f iff (y,x) in Gr_{f^{-1}}$
But $(x,y)$ and $(y,x)$ are symmetric to each other with respect to $id_{mathbb{R}}$
answered Jan 12 at 14:14
rtybasertybase
10.9k21533
$endgroup$
add a comment |
$begingroup$
As per my comment, let's define $Gr_f =left{(x,y)in mathbb{R}^2 mid f(x)=yright}$ (from Graphic).
Now, if $(x,y)in Gr_f iff f(x)=y iff f^{-1}(y)=x iff (y,x) in Gr_{f^{-1}}$, $f^{-1}$ is the notation for the $f$'s inverse. As a result
$(x,y) in Gr_f iff (y,x) in Gr_{f^{-1}}$
But $(x,y)$ and $(y,x)$ are symmetric to each other with respect to $id_{mathbb{R}}$
answered Jan 12 at 14:14
rtybasertybase
10.9k21533
$endgroup$
As per my comment, let's define $Gr_f =left{(x,y)in mathbb{R}^2 mid f(x)=yright}$ (from Graphic).
Now, if $(x,y)in Gr_f iff f(x)=y iff f^{-1}(y)=x iff (y,x) in Gr_{f^{-1}}$, $f^{-1}$ is the notation for the $f$'s inverse. As a result
$(x,y) in Gr_f iff (y,x) in Gr_{f^{-1}}$
But $(x,y)$ and $(y,x)$ are symmetric to each other with respect to $id_{mathbb{R}}$
answered Jan 12 at 14:14
rtybasertybase
10.9k21533
answered Jan 12 at 14:14
rtybasertybase
10.9k21533
answered Jan 12 at 14:14
rtybasertybase
10.9k21533
answered Jan 12 at 14:14
rtybasertybase
10.9k21533
10.9k21533
add a comment |
add a comment |
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$begingroup$
Because if $(x,y) in Gr_{f}$ then $(y,x) in Gr_{f^{-1}}$ and vice-versa, it's not too difficult to show this.
$endgroup$
– rtybase
Jan 12 at 13:31