# Determining when a function is neither even nor odd

Suppose that $$f(x) = \dfrac{x}{x+1}$$. I need to determine if it is even, odd or neither.

Now $$f(-x) = \dfrac{-x}{-x+1} = \dfrac{x}{1-x} \text{ and } -f(x) = \dfrac{-x}{x+1}$$. I can "see" that $$f(-x) \neq -f(x)$$ and $$f(-x) \neq f(x)$$.

Is this sufficient to show that it is neither an even nor odd function? Wouldn't it be more correct to find a counterexample, i.e, an $$x$$ in the domain of $$f$$ such that $$f(-x) \neq -f(x)$$ and $$f(-x) \neq f(x)$$, because this will be the negation of the definitions of an even and odd function?

• Counterexample, is, of course, enough as negation requires existence. Jun 24, 2020 at 10:05
• You have to write a specific value $x$ for which $f(x) \neq f(-x)$ and a specific value $x$ for which $f(x) \neq -f(-x)$. Jun 24, 2020 at 10:06
• You recieved 5 answers to your question. Is any of them what you needed? If so, consider accepting the best answer and upvoting all useful answers you got. That's how the site works.
– 5xum
Jun 29, 2020 at 7:39

Since $$f(2)=\frac23$$ and $$f(-2)=2\ne\pm\frac23$$, $$f$$ is neither odd nor even. That's all you need.

I can "see" that $$f(-x) \neq -f(x)$$ and $$f(-x) \neq f(x)$$.

How can you "see" it? What do you mean by "see"?

Is this sufficient to show that it is neither an even nor odd function?

No, as my question demonstrates. Mathematics is rigorous, and it has rules. One of those rules is that you can only claim something is true if you know how to prove said something. "seeing" is not proof. A proof is a sequence of true claims where each claim follows logically from the previous claims, and the sequence conclusion is the statement being proven.

A special kind of proof pertains to disproving a statement. A statement can be disproven if the negation of the same statement is proven.

Wouldn't it be more correct to find a counterexample

YES! The statement "$$f$$ is odd" is in fact the statement "For all $$x$$ in the domain of $$f$$, $$f(x)=-f(-x)$$". This statement can be disproven by proving its negation, which is

There exists some $$x$$ such that $$f(x)\neq -f(-x)$$

in other words, you can disprove the statement by finding a counterexample.

It suffices to find a single counterexample showing $$|f(x)|\ne|f(-x)|$$ such as $$\frac2{1+2}\ne\frac{-2}{1-2}.$$

Also note that $$f(0)\ne0$$ with $$f$$ continuous at $$0$$ disqualifies an odd function.

But for $$x>1$$

$$\frac x{x+1}=\frac{-x}{-x+1}$$ implies

$$x+1=x-1,$$ which is not possible. This approach might be helpful if you were not able (or willing) to find a counterexample.

• @JamesA: stupid me, only for continuous functions ! Thanks for pointing that.
– user65203
Jun 24, 2020 at 10:20
• @YvesDaoust: I think it would be more precise to put the condition "if $0$ is in the domain". (Of course, continuous at $0$ would imply that.) Jun 24, 2020 at 10:48
• @AryamanMaithani: no, $0$ in the domain does not imply continuity.
– user65203
Jun 24, 2020 at 10:58
• @YvesDaoust: I'm aware but you don't need continuity to conclude that $f(0)=0$ for odd functions. Only $0$ being in the domain is sufficient is what I meant. Jun 24, 2020 at 10:59
• @AryamanMaithani: no. $x<0\to-1,x\ge0\to1$ has $f(0)$ in the domain and is not odd.
– user65203
Jun 24, 2020 at 11:03

Yes, it would be "more" correct. In fact, that is the correct way.

The fact that they don't "look" the same is not good enough.

As a more "convincing" example - take the function $$f:\Bbb R\to \Bbb R$$ given by $$f(x) = \log(x + \sqrt{1 + x^2})$$.

Upon replacing $$x$$ with $$-x$$, it might not look odd at first glance, but it certainly is.

EDIT: Based on the comments, I might not done a clear enough job of expressing what I meant.

I want to make the point that "see"ing that a function is neither odd nor even is not correct!

The way to prove that a function is neither even nor odd is by constructing a counterexample, i.e., finding an $$x$$ such that $$f(x) \neq -f(-x)$$ and a $$y$$ such that $$f(y) \neq f(-y)$$.

Of course, the above can be done by simply showing the existence of such an $$x$$ and $$y$$. For example, by solving $$|f(x)| = |f(-x)|$$ and showing that that has only finitely many solutions while your domain has infinitely many elements.

• Why "more" correct with a counterexample ? Solving the equation $f(x)=\pm f(\pm x)$ can return an infinity of counterexamples !
– user65203
Jun 24, 2020 at 10:15
• @YvesDaoust: I'm not sure what you mean. By a counterexample, I meant finding an $x$ such that $f(x) {\color{red} \neq} -f(-x)$ and similarly finding a $y$ such that $f(y) {\color{red} \neq} f(-y)$. Jun 24, 2020 at 10:18
• I should have said "solving the equation and showing it does not work for some (many) $x$" like I did in my answer.
– user65203
Jun 24, 2020 at 10:24
• The reason I wrote ""more"" was because OP has written "more correct" in their question. Jun 24, 2020 at 10:25
• How do you justify that "more" and "the" correct way ?
– user65203
Jun 24, 2020 at 10:26

Well, in this case, you have two rational functions, i.e. functions that are defined by quotients of polynomials. In your case, it is enough to check that the formal objects $$\frac{X}{1-X}$$ and $$\frac{-X}{1+X}$$ are not equal. This can be proved using the fact that the formal polynomials $$X(1+X)$$ and $$-X(1-X)$$ are not equal. The main argument here is that two polynomials that agree on infinitely many points are equal. In other words, if two polynomial formulas give the same result on infinitely many points, then there is a formal manipulation from one to the other. This idea extends to quotients of polynomials - you have to avoid the points where the denominator is $$0$$, but there are only finitely many of them.

Or you could decompose them as sums of simple elements. There is a uniqueness argument that ensures they have to be different.

In our case, we have $$f(-x) = -1 + \frac{1}{1-x}$$ and $$-f(x) = -1 + \frac{1}{1+x}$$, for all but finitely many points. If these functions were equal, then the functions $$x \mapsto \frac{1}{1-x}$$ and $$x \mapsto \frac{1}{1+x}$$ would be equal as well.

So, this argument of "they do not look the same" actually works for rational functions or polynomials. Yet, as duly pointed out, it does not hold for more complicated functions like logarithms.