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, &amp; \text{if }a\text{&gt; 0 } \\<br /> 1, &amp; \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, &amp; \text{if }n\text{&gt; 0 } \\<br /> 1, &amp; \text{if }n \leq 0<br /> \end{cases}

Make that:
f(x)= \begin{cases}<br /> 0, &amp; \text{if }x &gt; 0 \\<br /> 1, &amp; \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, &amp; \text{if }x &gt; 0 \\<br /> 1, &amp; \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&gt;0.

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