Proving Strange Function is Neither Even nor Odd

[roam]

Here's a problem

Let k: R\{-1} → R be given by k(x) = \frac{2x-1}{x+1}

Prove that k is neither even nor odd.

That is strange! But to prove it we go back to the definition;
A function, f: (-a,a) \rightarrow R is said to be even if for all x \in (-a,a) => f(x) = f(-x)
And it is odd if f(x) = -f(-x).

So: if k is even then k(-x) = k(x) for any x
and k(x) = -k(-x) if it is odd.

I get the feeling that my proof doesn't seem right and somewhat unimplemented... (1) is there a better way to prove and conclude that k is neither even nor odd?
And (2) if it is neither even nor odd, then what could it be?

 

[Defennder]

Well, as you have stated it, all you need to do to prove that is to show that k(x) doesn't satisfy either the odd or even criterion. And there's nothing strange about a function being neither odd nor even since a whole lot of functions are neither odd nor even, such as e^x, ln(x) etc.

 

[HallsofIvy]

Here's a problem

Let k: R\{-1} → R be given by k(x) = \frac{2x-1}{x+1}

Prove that k is neither even nor odd.

That is strange! But to prove it we go back to the definition;
A function, f: (-a,a) \rightarrow R is said to be even if for all x \in (-a,a) => f(x) = f(-x)
And it is odd if f(x) = -f(-x).

So: if k is even then k(-x) = k(x) for any x
and k(x) = -k(-x) if it is odd.

I get the feeling that my proof doesn't seem right and somewhat unimplemented... (1) is there a better way to prove and conclude that k is neither even nor odd?
And (2) if it is neither even nor odd, then what could it be?

I'm not sure why you think that function is "strange". It's a simple rational function. If it is because it is neither even nor odd, well most functions are neither even nor odd.

To be even, a function must satisfy, as you say, f(x)= f(-x) for all x.
To be odd, a function must satisfy, as you say, f(x)= -f(-x) for all x.

To prove something is NOT "true for all x", you only need to find a counterexample.

If f(x)= (2x-1)/(x+1), what is f(2)? What is f(-2)?

That's all you need.

 

[roam]

I'm not sure why you think that function is "strange". It's a simple rational function. If it is because it is neither even nor odd, well most functions are neither even nor odd.

Hall, that's because I thought since even functions are their own reflection in the y-axis. Even functions are symmetric about the y-axis.
Whereas other functions which are not even (odd functions) aren't like that.
Yeah, I understood the error I made.

Yes, I get it. Hi@Defennder

To be even, a function must satisfy, as you say, f(x)= f(-x) for all x.
To be odd, a function must satisfy, as you say, f(x)= -f(-x) for all x.

To prove something is NOT "true for all x", you only need to find a counterexample.

If f(x)= (2x-1)/(x+1), what is f(2)? What is f(-2)?

\frac{2 \times 2-1}{2+1} = \frac{2 \times -2-1}{-2+1}
1 ≠ -5
But they don't equal, thus they don't satisfy neither of the conditions f(x) = f(-x) or f(x) = -f(-x).


^ is that a valid proof?

 

[HallsofIvy]

Yes, that is a perfectly valid proof: to show that a general statement is NOT true, it is sufficient to give a 'counter example': one example where it is not true.

Yes, it is true that an even function is symmetric about the x-axis. It is NOT true that if a function is not symmetric about the x-axis- that is that it is not even- it is necessarily "odd". An odd function is symmetric through the origin: if (x,y) is on the graph then extending a line from (x,y) through (0,0) an equal distance on the other side of (0,0), that is to (-x,-y), you are again on the graph: f(-x)= -f(x).

 

[vkroom]

If one maps the excluded point (here -1) to the origin such the (new) domain is R\{0}, things fall back to the old definition of f(x) and f(-x). Which in this case would be \frac{2x-3}{x}. This is pretty simple now :)