$ x = \sqrt{1} $ then x = ?
and
$ x^2 = 1 $ then x = ?
please help
I am puzzled. I know that in first case we will get x = 1 and in second case we will get x = $ \pm 1 $
But, I need the proof for the first case.
$ x = \sqrt{1} $ then x = ?
and
$ x^2 = 1 $ then x = ?
please help
I am puzzled. I know that in first case we will get x = 1 and in second case we will get x = $ \pm 1 $
But, I need the proof for the first case.
$x=\sqrt1$ is a single order equation so, it can have only one solution of $x$
and that is $1$
But, in case of $x^2 = 1$, it is a second order equation and by theory, it will have two solution (may be same.
Now, $x^2=1$ can be represented as
$x^2-1=0$
$\implies (x+1)\cdot(x-1)=0$
$\implies x=\pm1$
That $\sqrt{1}$ means the positive square root of $1$ is a matter of convention, and probably some will maintain that it's only convention.
$1^2=1$ and $(-1)^2=1$ and there are no other numbers whose square is $1$, so there are exactly two square roots of $1$. The fact that there are no others is why you can say that if $x^2=1$ then either $x=1$ or $x=-1$.