05:48:21 27/06/2022

# ERROR: \$X^2 + 1 = 0\$ HAS SOLUTION SET  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

\$\$x^4=1\$\$

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.

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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