Do equations like $y = \sqrt {x} \cdot 0$ have real solutions for $x<0$?

What I mean by the title, is that some equations, such as $$y=\frac1x$$ or $$y=\sqrt{x}$$ don't have any real solutions for some values of $$x$$. My question is:

Does multiplying such functions by zero (ie $$y=\sqrt{x}\cdot0$$) make it so they have no solutions for those values (in this case $$x<0$$), are undefined, or that all values of $$x$$ are had the solution $$y=0$$?

If it depends on the function that is being multiplied, how would one know which one is correct?

Sorry if it is a bit of a dumb question, but I could genuinly not find the answer (except for the case $$y=\frac00$$, which is undefined (y can have any value))

• When multiplying by $0$ you should do it in both members of the equation, so you get $0=0$ which doesn't add anything new – JoseSquare Feb 20 at 10:26
• I meant multiplying to right side of the equation to get a new equation :) – Treehee Feb 20 at 10:27
• In some contexts it is allowed to say $\langle undefined\rangle\cdot0=0$, but the context should be well delimited with very careful definitions. – egreg Feb 20 at 10:32
• could you explain what you mean specifically with "some contexts"? Could you give an example of such a context? – Treehee Feb 20 at 10:33
• Now I get your point, what I would say is no, the domain of $y=0\sqrt{x}$ is $\mathbb{R}_{\geq 0}$ so your function will really be $f:\Bbb{R}_{\geq 0} \rightarrow \Bbb{R}_{\geq 0}$ with $f(x)=0$ for all $x$ in its domain. A similar thing happens with $f(x)=\frac{x}{x}$ which is $1$ but it is not defined on $x=0$ (although you can define it there continuously). – JoseSquare Feb 20 at 10:33

When multiplying two functions, the new function domain will be the intersection of the domains of the two functions although the domain could be continuously extendable, but that would be a new function.

In you example as $$f: \Bbb{R}_{\geq 0} \rightarrow \Bbb{R}_{\geq 0}$$ where $$f(x)=\sqrt{x}$$ and $$g: \Bbb{R} \rightarrow \Bbb{R}$$ with $$g(x)=0$$ the product function will have $$\Bbb{R} \cap \Bbb{R}_{\geq 0}= \Bbb{R}_{\geq 0}$$ as domain.

$$y=0$$ is the only solution to the equation $$y(x) = \sqrt{x} \cdot 0 = 0$$ of unknown $$x$$. $$y$$ is a constant. However, I am not sure if you ask for the domain of $$\sqrt{x}$$. Well, it has a definition for negative reals, but I am not sure you are talking about that.

So if $$\sqrt{x}$$ is the one that goes from $$\mathbb{R}^+$$ to $$\mathbb{R}^+$$, then it's not defined. However wise, for $$\sqrt{x}$$ from $$\mathbb{C}$$ to $$\mathbb{C}$$, it is defined.

However, in general, if you don't mention the definition of squareroot, we will assume that it is the complex one if you are dealing with negative reals.

• Maybe $\sqrt{x}$ was a bad example as it does indeed have complex solutions for $x<0$. I'm wondering if the same applies to functions which do not have any solution for certain values (ie does multiplying a function by zero change its domain?) – Treehee Feb 20 at 10:31
• No, it doesn't change its domain because the initial equation is defined with an operator that doesn't have a part of the total domain. – PackSciences Feb 20 at 10:39
• I'm rather confused by that comment, could you rephrase? – Treehee Feb 20 at 10:41
• If $g(x) \cdot a$, it doesn't change its domain, even if $a$ is zero. – PackSciences Feb 20 at 10:42

actually the function you defined is only defined for x greater or equal to zero, and the value is 0. but if you define the function y=0, it yields the same values for x>=0 and also values for x<0. so you extended the function to a bigger domain, which is a common practice in mathematics.

I guess you mean, does an equation like $$\sqrt x\cdot0=0$$ for example, have real solutions different from the real solutions of $$\sqrt x=0.$$

The answer is that $$\sqrt x\cdot0=0$$ has infinitely many solutions, namely all nonnegative real $$x,$$ whereas the latter has only $$0$$ as solution. I assume all through that we never leave $$\mathbf R,$$ otherwise the first equation has imaginary solutions for all real numbers, even negative ones.

All this follows from the meaning of square root of $$x,$$ which is a number $$r$$ satisfying $$rr=r^2=x.$$

• No, that is not at all what I'm asking. I'm asking if $y=\sqrt{-1}*0$ has a real solution (assuming $\sqrt{-1}$ is undefined - only real numbers. – Treehee Feb 20 at 10:44
• @Treehee Since you don't want to work with $i,$ then an equation involving $i$ would be undefined, too. – Allawonder Feb 20 at 10:46

Indeed,

$$0\cdot\sqrt{x}$$ is undefined for negative $$x$$. Multiplication by $$0$$ "comes too late".