# Why is absolute value function not a polynomial?

Why is absolute value function not a polynomial?

Why couldn't we consider absolute value function as a polynomial?

• You mean the function $x\mapsto |x|$. – hamam_Abdallah Jun 30 '17 at 22:11
• @Salahamam_ Fatima also some function like $|x^3|$ – user373141 Jun 30 '17 at 22:13
• You don't like my answer? It is most elemntary from the answers you got. – Aqua Jan 15 '20 at 19:08

All polynomials are differentiable, but the absolute value function $|x|$ is not (at $x=0).$

A polynomial is $by$ $definition$ a function which can be represented by

$$\sum_{i=1}^n a_ix^i$$ for all $x$ on the domain being considered, where $n$ is finite. Let's look at the simple example of $f(x)=|x|$. We would have to use combinations of polynomials and characteristic functions but then this wouldn't fit the definition.

Just quoting the definition of "polynomial" does not constitute a proof. Who knows, maybe there is a certain polynomial of degree $2017$ with particular coefficients that does the job. To be serious: We have to exhibit a property of ${\rm abs}$ that no polynomial can have. In this sense Reiner Martin's answer is fine.

Here is an argument not using differentiability: If $p$ is a polynomial of degree $\geq2$ then $x\mapsto{p(x)\over x}$ is unbounded when $x\to\infty$. If $p$ has degree $1$ then $p(x)p(-x)\to-\infty$ when $x\to\infty$.

Suppose $$|x|$$ is a polynomial $$p(x)$$, so $$|x|=p(x)\implies x^2=|x|^2= p(x)^2$$

So $$p(x)$$ is some linear polynomial, say $$p(x) = ax+b$$. Then we have $$x^2=a^2x^2+2abx+b^2\implies b=0\;\; \wedge\;\;a^2=1$$

If $$a=1$$ then we have $$|x|=x$$ for all $$x$$ which is not true and

if $$a=-1$$ we have $$|x|=-x$$ for all $$x$$ which is also not true.

Check out this link on polynomials to see how they are defined.

Polynomials are expressions which only consist of variables and coefficients, using only the operations of: addition, subtraction, multiplication and natural exponents of variables. The absolute value operation is therefore not included in the scope of available operations for polynomials.

As pointed out by Squirtle, polynomials are defined in a way which does not allow for functions such as $f(x)=|x^3|$ to be a polynomial, as it can not be written using the above conditions.

Sorry for the late answer, but here is one without differentiability or the definition:

Lemma 1: $$f$$ and $$g$$ and polynomials, then $$\deg (f * g) = \deg f + \deg g$$

Proof:

From the definition of the degree of a polynomial: $$\deg f=\deg f'$$ $$\deg g=\deg g'$$

where $$f'=a_nx^n$$, $$g'=b_mx^m$$ ($$n=\deg f$$, $$m=\deg g$$).

Then $$\deg(f' * g')=\deg(f * g)$$ because $$f'$$ and $$g'$$ are the summands with the largest power in their respective polynomials. But $$f'*g'=a_nb_mx^{m+n}$$. Thus $$\deg (f' * g')=\deg f'+\deg g'$$ and $$\deg (f*g)=\deg(f' * g')=\deg f'+\deg g'=\deg f+\deg g$$. $$\blacksquare$$

Note the special case of Lemma 1 for $$f=g$$, we have $$\deg(f^2)=2deg(f)$$.

Now, $$|x|^2$$ is equal to $$x^2$$. We will prove by contradiction that $$|x|$$ cannot be a polynomial. The proof is very simple: because $$\deg (|x|^2)=\deg(x^2)=2$$, $$\deg |x|$$ would be $$1$$. This would mean that $$|x|$$ is a linear function.

But we know that, given two points, there is exactly one line through them (Postulate III). Thus, if we have $$|0|=0$$ and $$|1|=1$$, there is exactly one line through these points, and we see $$y=x$$ matches the criterion and again it is the only one. Thus, $$|x|=x$$.

Now, recall that $$|-1|=1$$. But $$|-1|=-1$$, because $$|x|=x$$ as proven above. Thus $$1=-1$$, which is an obvious contradiction. Because $$|x|$$ being a polynomial leads to contradiction, $$|x|$$ is definitely not a polynomial. $$\blacksquare$$