How to solve this? If in the Ring (A,+,*) the equation $x^{2} + 1 = 0$ $x^2+1=0$ has unique solution demonstrate that 1+1=0,where "1" is unit element of the ring.

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Answer 1 · 2017-03-25T15:00:34

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Let $α$ $alpha$ be a root of $x^{2} + 1 = 0$ $x^2+1 = 0$

Then ${(- α)}^{2} + 1 = α^{2} + 1 = 0$ $(-alpha)^2+1 = alpha^2 + 1 = 0$

So $- α$ $-alpha$ is a root of $x^{2} + 1 = 0$ $x^2+1 = 0$

So if $x^{2} + 1 = 0$ $x^2+1 = 0$ has only one solution, we must have:

$- α = α$ $-alpha = alpha$

Add $α$ $alpha$ to both sides to get:

$0 = α + α = 1 \cdot α + 1 \cdot α = (1 + 1) \cdot α$ $0 = alpha+alpha = 1*alpha+1*alpha = (1+1)*alpha$

Since $α \neq 0$ $alpha != 0$ we can divide both sides by $α$ $alpha$ to get:

$0 = 1 + 1$ $0 = 1+1$

How to solve this? If in the Ring (A,+,*) the equation x2+1=0x^2+1=0 has unique solution demonstrate that 1+1=0,where "1" is unit element of the ring.