Error: $x^2 + 1 = 0$ has solution set $\{-1;1\}$
So, is it correct that the solution set of $x^2 + 1 = 0$ is [-1;1] ?
Is the error in equation development or in the solution set?
Help me, I don't know how to proceed in this question.
$\endgroup$ 14 Answers
$\begingroup$All the implications are true but you are drawing the wrong conclusion. You have proved that $x^{2}+1=0$ implies $x=1$ or $x =-1$ but the converse of this implication is not true. The fact is there is no real number $x$ with $x^{2}+1=0$.
$\endgroup$ 9 $\begingroup$When they multiplied by $(1-x^2)$, they introduced two additional roots for the equation $\pm 1$.
Note that
$$x^4=1$$
has four roots in the complex domain which are $\pm 1$ and also $\pm i$.
$\endgroup$ 1 $\begingroup$The final implication should be "$\implies x\in\{1,-1,i,-i\}$." That is, the solutions to $x^2+1=0$ (if any) are among the elements of this set, not that all the elements of the set are automatically solutions of the equation.
$\endgroup$ $\begingroup$No. It is not as plugging in $1$ and $-1$ will give you $(-1)^2 + 1 = 2$ and $1^2 +1 =2$.
The problem is that multiplying by $x^2 -1$ gives extraneous solutions and $1, -1$ are the solutions to $x^2 -1 =0$ which was brought in from nowhere.
This would b similar to doing this:
Suppose $(x -3)(x-2) = x^2 -5x + 6 = 0$ (So the solutions are $x = 3$ or $x = 2$. As that is zero we multiply it by $x+3057$ So
$(x^2 - 5x+6 ) = 0$ so
$(x^2 - 5x + 6)(x+3057) = 0\cdot (x+3057)=0$ so
$x^3 - 3052x^2 -1579x +18342 = 0$
If we tried to solve $x^3 - 3052x^2 -1579x +18342 = 0$ we would get $x = 3$ or $x = 2$ or $x =-3057$.
The third solution came in when we multiplied by $x+3057$. That is because $x = -3057$ is a solution to $x+3057=0$. So by multiplying $0$ by $x+3057$ we add a new solution to the problem.
So in this "false proof":
So $x^2 + 1 = 0$ has no real solutions. (It has complex solutions, $x = i$ or $x = -i$ but no real solutions).
But $x^2 -1=0$ has two real solutions; $x = 1$ and $x = -1$.
By mulitplying both sides of the equation $x^2 + 1= 0$ by $x^2 -1$ we are taking all the original solutions (there are no real solutions but there were complex $x=i$ and $x=-i$) and adding the solutions $x = 1$ and $x =-1$. So we end up with solutions $x =1$ and $x = -1$ but they were both artificially added when there no real solutions in the first place.
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