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$ 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.

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4  
If you want a proof, you must first decide (or find out) what the definition of $\sqrt 1$ you want to prove it from is. –  Henning Makholm Sep 21 '12 at 13:22
10  
Back in my school days, I learned that $\sqrt{y}$ is defined as the unique non-negative number such that $(\sqrt{y})^2 = y$, and if that is your definision, then case 1 has only one solution: $x=1$. Case 2 still has two solutions, though, $x=\pm 1$. –  Arthur Sep 21 '12 at 13:27
    
In its present formulation, this is NARQ. –  Did Sep 21 '12 at 13:35
    
$ f(x) = \sqrt{1} $ then f(x) = 1 only .. –  Kushashwa Ravi Shrimali Sep 21 '12 at 13:37
2  
@ParthKohli: That doesn't mean that $-1$ would be an answer, since the implication from right to left does not hold. –  Thomas E. Sep 21 '12 at 13:47

2 Answers 2

up vote 1 down vote accepted

$x=\sqrt1$ 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

$x^2-1=0$

$\implies (x+1)\cdot(x-1)=0$

$\implies x=\pm1$

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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 $x=-1$.

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