Is there a function f such that f^2=f and f is not equal to 0 or 1?

[robertjordan]

Homework Statement

Show there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$ s.t. $f^2=f$ but $f\neq{0,1}$.

Homework Equations

Here $f^2=f$ means for arbitrary $a\in{\mathbb{R}}, f(a)^2=f(a)$

The Attempt at a Solution

I came up with the function $f(a)= \begin{cases}<br /> 0, & \text{if }a\text{> 0 } \\<br /> 1, & \text{if }a \leq 0<br /> \end{cases}$What do you guys think? Is this right? I figured the only real numbers r for which r^2=r are r=0 and r=1 so the function f will have to only spit out those values or else there would be some input a for which f(a)^2=/=f(a)

 

[Mark44]

Homework Statement

Show there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$ s.t. $f^2=f$ but $f\neq{0,1}$.

Homework Equations

Here $f^2=f$ means for arbitrary $a\in{\mathbb{R}}, f(a)^2=f(a)$

The Attempt at a Solution

I came up with the function $f(a)= \begin{cases}<br /> 0, & \text{if }n\text{> 0 } \\<br /> 1, & \text{if }n \leq 0<br /> \end{cases}$

Make that:
$f(x)= \begin{cases}<br /> 0, & \text{if }x > 0 \\<br /> 1, & \text{if }x \leq 0<br /> \end{cases}$

What do you guys think? Is this right? I figured the only real numbers r for which r^2=r are r=0 and r=1 so the function f will have to only spit out those values or else there would be some input a for which f(a)^2=/=f(a)

What you have works, but I don't think it's what the write of the problem had in mind. Instead, I think they had a kind of identity function in mind - one whose output is the same as its input.

 

[robertjordan]

Make that:
$f(x)= \begin{cases}<br /> 0, & \text{if }x > 0 \\<br /> 1, & \text{if }x \leq 0<br /> \end{cases}$

Thanks. Fixed it.

What you have works, but I don't think it's what the write of the problem had in mind. Instead, I think they had a kind of identity function in mind - one whose output is the same as its input.

But we need $f(a)^2=f(a)$ for all real numbers a. If we make $f(a)=a$, then in order for $f(a)^2=f(a)$, we would need $a^2=a$ which is clearly not true in general...

Can you elaborate some more on what you mean? I think I missed it...

 

[Stimpon]

What you have works, but I don't think it's what the write of the problem had in mind. Instead, I think they had a kind of identity function in mind - one whose output is the same as its input.

Assuming that robertjordan hasn't completely misunderstood the question, they definitely aren't looking for an identity function.

But we need $f(a)^2=f(a)$ for all real numbers a. If we make $f(a)=a$, then in order for $f(a)^2=f(a)$, we would need $a^2=a$ which is clearly not true in general...

Can you elaborate some more on what you mean? I think I missed it...

Normally $f^{2}=f \circ f$, so I think Mark44 thought that that's what the question writer meant by $f^{2}$. Certainly if the question writer did mean $f \circ f$ then a (the) real identity function would be correct, but you clarified that the writer meant $f \cdot f$ so Mark44 is wrong.

 

[Mark44]

The notation used here is confusing to me, and appears to be in contradiction to itself.

Show there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$ s.t. $f^2=f$ but $f\neq{0,1}$.

f² as used above normally indicates function composition, as in f(f(x)).

Here $f^2=f$ means for arbitrary $a\in{\mathbb{R}}, f(a)^2=f(a)$

To my mind, the notation used immediately above contradicts the meaning at the top of this page. Even if f² denotes multiplication, it should be written as [f(a)]² to be clear.

 

[Stimpon]

As an aside, if $f^{2}$ was being used to mean $f \circ f$, then the function given wouldn't work because we would have $f^{2}(x)=f(0)=1$ for $x>0$.

ETA: As well as $f^{2}(x)=f(1)=0$ for $x \leq 0$.