I am studying differential forms from Rudin's PMA. Here's the definition of $k$-form 
Also he proves the anticommutative relation: $dx_1 \land dx_2=-dx_2\land dx_1$
Does the following expressions make sense: $dx_1\land cdx_2$ where $c\in \mathbb{R}^1$? I guess NO since the definition 10.11 different from what I wrote.
Does the following is true $dx_1\land cdx_2=cdx_1\land dx_2$? If yes how to prove it?
EDIT: First of all, I am going to prove that $d(cx_1)\land dx_2=cdx_1\land dx_2$. Let $\omega_1=d(cx_1)\land dx_2$ and $\omega_2=cdx_1\land dx_2$. Since $$\omega_1(\Phi)=\int \limits_{\Phi}\omega_1=\int \limits_{D}\dfrac{\partial(c\phi_1,\phi_2)}{\partial(u_1,u_2)}d\mathbb {u}=\int \limits_{D}c\dfrac{\partial(\phi_1,\phi_2)}{\partial(u_1,u_2)}d\mathbb {u}=\int \limits_{\Phi}\omega_2=\omega_2(\Phi).$$ Hence $\omega_1=\omega_2$ then $$d(cx_1)\land dx_2=cdx_1\land dx_2$$ $$dx_1\land d(cx_2)=cdx_1\land dx_2.$$ How to conclude that $dx_1\land cdx_2=cdx_1\land dx_2$ from above relations?
The expression $dx_1 \wedge cdx_2$ does make sense. It's not in the form given in definition 10.11, but if you skip ahead and take a look at section 10.17 where the (exterior/wedge) product of differential forms is defined, you'll notice that for the $1$-forms $\omega = dx_1$ and $\lambda = cdx_2$, both of which are in the canonical form of definition 10.11, we have
$$dx_1 \wedge cdx_2 = \omega \wedge \lambda,$$
and that is defined as
$$c\,dx_1 \wedge dx_2.$$