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 $x^2+1=0$.


When they multiplied by $(1-x^2)$, they introduced two additional roots for the equation $pm 1$.

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Note that


has four roots in the complex domain name which are $pm 1$ và also $pm i$.


The final implication should be "$implies xin1,-1,i,-i$." 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 mix are automatically solutions of the equation.


No. It is not as plugging in $1$ & $-1$ will give you $(-1)^2 + 1 = 2$ & $1^2 +1 =2$.

The problem is that multiplying by $x^2 -1$ gives extraneous solutions và $1, -1$ are the solutions lớn $x^2 -1 =0$ which was brought in from nowhere.

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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

$(x^2 - 5x + 6)(x+3057) = 0cdot (x+3057)=0$ so

$x^3 - 3052x^2 -1579x +18342 = 0$

If we tried khổng lồ 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 địa chỉ a new solution khổng lồ 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 $x=-i$) và adding the solutions $x = 1$ and $x =-1$. So we end up with solutions $x =1$ và $x = -1$ but they were both artificially added when there no real solutions in the first place.

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