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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))

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  • $\begingroup$ 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 $\endgroup$ – Ahlfkushevich Feb 20 at 10:26
  • $\begingroup$ I meant multiplying to right side of the equation to get a new equation :) $\endgroup$ – Treehee Feb 20 at 10:27
  • $\begingroup$ In some contexts it is allowed to say $\langle undefined\rangle\cdot0=0$, but the context should be well delimited with very careful definitions. $\endgroup$ – egreg Feb 20 at 10:32
  • $\begingroup$ could you explain what you mean specifically with "some contexts"? Could you give an example of such a context? $\endgroup$ – Treehee Feb 20 at 10:33
  • $\begingroup$ 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). $\endgroup$ – Ahlfkushevich Feb 20 at 10:33
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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.

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$y=0$ 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.

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  • $\begingroup$ 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?) $\endgroup$ – Treehee Feb 20 at 10:31
  • $\begingroup$ 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. $\endgroup$ – PackSciences Feb 20 at 10:39
  • $\begingroup$ I'm rather confused by that comment, could you rephrase? $\endgroup$ – Treehee Feb 20 at 10:41
  • $\begingroup$ If $g(x) \cdot a$, it doesn't change its domain, even if $a$ is zero. $\endgroup$ – PackSciences Feb 20 at 10:42
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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.

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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 $$rr=r^2=x.$$

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  • $\begingroup$ 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. $\endgroup$ – Treehee Feb 20 at 10:44
  • $\begingroup$ @Treehee Since you don't want to work with $i,$ then an equation involving $i$ would be undefined, too. $\endgroup$ – Allawonder Feb 20 at 10:46
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Indeed,

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

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