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as if $f(x)=1/x$ we would get a constant value each time or may be other weird function may satify this inversive property making it constant each time but idk )I think the problem with $f(x)=\frac{1}{x}$ is that its domain is not $\mathbb{R}$, it is $\mathbb{R} \backslash \{0\}$. But the function should have a domain of the whole set of real numbers. This is what I think, but I may be wrong too.Asif Hossain wrote: ↑Wed May 05, 2021 2:04 pmDoes $f(xf(x))=xf(x) \Rightarrow f(x)=x$ $\forall x \in \mathbb{R}$??(I don't think so
Doesn't $x=0 \Rightarrow f(0)=0$? then the domain $\mathbb{R} \backslash \{0\}$ shouldn't mess around.Mehrab4226 wrote: ↑Wed May 05, 2021 9:09 pmI think the problem with $f(x)=\frac{1}{x}$ is that its domain is not $\mathbb{R}$, it is $\mathbb{R} \backslash \{0\}$. But the function should have a domain of the whole set of real numbers. This is what I think, but I may be wrong too.Asif Hossain wrote: ↑Wed May 05, 2021 2:04 pmDoes $f(xf(x))=xf(x) \Rightarrow f(x)=x$ $\forall x \in \mathbb{R}$??(I don't think so
SO I guess your function is $f(x)=\frac{1}{x}$ if $x \neq 0$, and $f(0)=0$ if $x=0$. I guess then it works.Asif Hossain wrote: ↑Wed May 05, 2021 10:19 pmDoesn't $x=0 \Rightarrow f(0)=0$? then the domain $\mathbb{R} \backslash \{0\}$ shouldn't mess around.Mehrab4226 wrote: ↑Wed May 05, 2021 9:09 pmI think the problem with $f(x)=\frac{1}{x}$ is that its domain is not $\mathbb{R}$, it is $\mathbb{R} \backslash \{0\}$. But the function should have a domain of the whole set of real numbers. This is what I think, but I may be wrong too.Asif Hossain wrote: ↑Wed May 05, 2021 2:04 pmDoes $f(xf(x))=xf(x) \Rightarrow f(x)=x$ $\forall x \in \mathbb{R}$??(I don't think so
Actually the main question is does $f(xf(x))=xf(x) \Rightarrow f(x)=x\ \ \forall x \in \mathbb{R}$?Mehrab4226 wrote: ↑Thu May 06, 2021 4:02 amSO I guess your function is $f(x)=\frac{1}{x}$ if $x \neq 0$, and $f(0)=0$ if $x=0$. I guess then it works.Asif Hossain wrote: ↑Wed May 05, 2021 10:19 pmDoesn't $x=0 \Rightarrow f(0)=0$? then the domain $\mathbb{R} \backslash \{0\}$ shouldn't mess around.Mehrab4226 wrote: ↑Wed May 05, 2021 9:09 pm
I think the problem with $f(x)=\frac{1}{x}$ is that its domain is not $\mathbb{R}$, it is $\mathbb{R} \backslash \{0\}$. But the function should have a domain of the whole set of real numbers. This is what I think, but I may be wrong too.
Probably notAsif Hossain wrote: ↑Thu May 06, 2021 12:37 pmActually the main question is does $f(xf(x))=xf(x) \Rightarrow f(x)=x\ \ \forall x \in \mathbb{R}$?Mehrab4226 wrote: ↑Thu May 06, 2021 4:02 amSO I guess your function is $f(x)=\frac{1}{x}$ if $x \neq 0$, and $f(0)=0$ if $x=0$. I guess then it works.Asif Hossain wrote: ↑Wed May 05, 2021 10:19 pm
Doesn't $x=0 \Rightarrow f(0)=0$? then the domain $\mathbb{R} \backslash \{0\}$ shouldn't mess around.
why? i have just shown example of such functions but can it proven?(except this existential contradiction)Mehrab4226 wrote: ↑Thu May 06, 2021 2:05 pmProbably notAsif Hossain wrote: ↑Thu May 06, 2021 12:37 pmActually the main question is does $f(xf(x))=xf(x) \Rightarrow f(x)=x\ \ \forall x \in \mathbb{R}$?Mehrab4226 wrote: ↑Thu May 06, 2021 4:02 am
SO I guess your function is $f(x)=\frac{1}{x}$ if $x \neq 0$, and $f(0)=0$ if $x=0$. I guess then it works.