Examples of Additive-Only and Multiplicative-Only Functions

A function can preserve products without preserving sums, or preserve sums without preserving products. These are different properties: additivity means $f(x+y)=f(x)+f(y)$, while multiplicativity means $f(xy)=f(x)f(y)$. Neither condition by itself says that a function is linear.

Here are examples in both directions. One caveat matters: an additive function from $\mathbb{R}$ to $\mathbb{R}$ that is not real-linear cannot be an ordinary continuous function. A genuinely nonlinear example in that setting requires a Hamel basis.

Multiplicative but not additive: squaring

Define $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=x^2$. For all real $x,y$,

$$f(xy)=(xy)^2=x^2y^2=f(x)f(y),$$

so $f$ is multiplicative. But it does not preserve sums. For example,

$$f(1+1)=4,\qquad f(1)+f(1)=2.$$

Thus $f$ is multiplicative but not additive. It is also nonlinear as a map of real vector spaces: it fails the scalar rule $f(cx)=cf(x)$. For instance, $f(2\cdot1)=4$, whereas $2f(1)=2$.

The absolute-value function is another example: $f(x)=|x|$ satisfies $|xy|=|x||y|$, but $|1+(-1)|=0$ while $|1|+|-1|=2$.

Additive but not multiplicative: a Hamel-basis example

For an additive map $g:\mathbb{R}\to\mathbb{R}$ that is not real-linear, use a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$. A Hamel basis is a set of real numbers such that every real number has a unique expression as a finite rational linear combination of basis elements. Its existence uses the axiom of choice.

Choose a basis containing $1$ and $\sqrt{2}$. Define $g$ on the basis by

• $g(\sqrt{2})=0$;
• $g(b)=b$ for every other basis element $b$.

Extend the definition to every real number by rational linearity. In other words, if $x=\sum_i q_i b_i$ with rational $q_i$ and basis elements $b_i$, set $g(x)=\sum_i q_i g(b_i)$.

This construction makes $g$ additive. If $x$ and $y$ have finite rational-coordinate expansions in the basis, the coordinates of $x+y$ are their coordinate-wise sums, so $g(x+y)=g(x)+g(y)$.

It is not multiplicative. Taking $x=\sqrt{2}$ gives

$$g(x^2)=g(2)=2g(1)=2,\qquad g(x)g(x)=0.$$

It is not real-linear either: $g(\sqrt{2})=0$, but $\sqrt{2}\,g(1)=\sqrt{2}$. This is a genuine additive-but-not-multiplicative nonlinear map over $\mathbb{R}$, but it is highly discontinuous and cannot be described by a familiar elementary formula.

Why simple additive examples can be misleading

A tempting example is $g(x)=2x$. It is additive, and it is not multiplicative, since $g(1\cdot1)=2$ but $g(1)g(1)=4$. However, $g$ is still a linear map over $\mathbb{R}$: it satisfies both additivity and real scalar homogeneity. So it works if you only need a function with one property but not the other; it does not work if “nonlinear” means “not a linear transformation.”

There is also a short complex-valued example if linearity is judged over $\mathbb{C}$. Set $h(z)=z+\overline{z}$ for $z\in\mathbb{C}$. It is additive, but not multiplicative: $h(i)=0$ while $h(i^2)=h(-1)=-2$. It is not complex-linear, although it is linear when $\mathbb{C}$ is treated as a vector space over $\mathbb{R}$.

Common mistakes

• Confusing multiplicativity with scalar homogeneity. Multiplicativity concerns products of two inputs, $f(xy)=f(x)f(y)$. Linearity’s scalar rule is $T(cv)=cT(v)$; these are not the same condition.
• Calling every additive nonmultiplicative map nonlinear. A map such as $g(x)=2x$ is additive and nonmultiplicative, but it is still real-linear.
• Assuming an additive real function has an ordinary formula. If an additive $g:\mathbb{R}\to\mathbb{R}$ is continuous (or satisfies other regularity conditions such as measurability), then $g(x)=cx$ for some constant $c$. The discontinuous Hamel-basis construction avoids those regularity conditions.
• Leaving the scalar field unstated. A map can be linear over $\mathbb{R}$ but not over $\mathbb{C}$, as the example $h(z)=z+\overline z$ shows.

When choosing an example, state the domain, codomain, and what “nonlinear” means. For a simple multiplicative-only function, use $x^2$. For an additive-only function that is genuinely nonlinear over $\mathbb{R}$, use the Hamel-basis construction and mention its nonconstructive nature.