How can one prove that a function $f(x,t)$ cannot be expressed by the form $X(x)\cdot T(t)$? I was reading about ODE variable separation solving and the book says that assuming that a function can be expressed as the product of two single-variable functions loses generality, which I understand. I cannot, however, prove that it does. For example: the function $\sqrt{x+t}$ cannot be expressed as $X(x)\cdot T(t)$. I tried proving it by contradiction and got to the result $X_1(x)\cdot T_1(t) = (x+it)\cdot(x-it)$ but I'm not sure if this is sufficient. That is, what guarantees that $(x+it)\cdot(x-it)$ cannot be rearranged in some way so that it only has the variable $x$ in one of the factors and only $t$ in the other?
 A: If $f(x,t) = \sqrt{x+t} = X(x) \cdot Y(t)$ for all $x, t \ge 0$ then
$$X(x)\cdot Y(0) = \sqrt{x} \text{ and } X(0) \cdot Y(t) = \sqrt t$$
But $X(0) \cdot Y(0) = \sqrt{0+0} = 0$ and therefore the contradiction $X(0) = 0$ or $Y(0) = 0$.
A: Note that $f(x,y)=X(x)T(t)$ implies things like
$$\tag1f(x_1,t_1)f(x_2,t_2)=f(x_1,t_2)f(x_2,t_1)$$
for all $x_1,x_2,t_1,t_2$.
In the concrete example $f(x,t)=\sqrt{x+t}$, nearly all choices show that $(1)$ does not hold, e.g.
$$f(1,1)f(2,2)=\sqrt2\sqrt 4=2\sqrt 2\ne 3=\sqrt3\sqrt 3 =f(1,2)f(2,1) $$
or even simpler
$$f(0,0)f(1,1)=\sqrt 0\sqrt 2=0\ne 1=\sqrt 1\sqrt 1 =f(0,1)f(1,0).$$
A: To prove that $\sqrt{x+t}=X(x)T(t)$ is not valid in general it is suffisant to show a conter-example.
$$\text{Suppose that}\quad  \sqrt{x+t}=X(x)T(t)\quad 
\text{is true any }(x,t) \tag 1$$
$$\text{then}\quad \sqrt{2x+t}=X(2x)T(t)\quad
\text{which implies}$$
$$\frac{\sqrt{2x+t}}{\sqrt{x+t}}=\frac{X(2x)}{X(x]}$$
The term on the right is a function of $x$ only while the term on the left is a function involving $t$. This is impossible. Thus the supposition $(1)$ is false.
A: I think $\log$ and partial derivative can do it.
$\frac{\partial}{\partial x}\log(\sqrt{x+t})=\frac{1}{2}\frac{\partial}{\partial x}\log(x+t)=\frac{1}{2(x+t)}$ is still related to $t$.
If $f(x,t)=h(x)g(t)$, $\frac{\partial}{\partial x}\log(f(x,t))=\frac{\partial}{\partial x}\log(h(x)g(t))=\frac{\partial}{\partial x}\log(h(x))+\frac{\partial}{\partial x}\log(g(t))=\frac{h'(x)}{h(x)}$ is unrelated to $t$.
