Help finding mistake in this problem When I am doing math during free time, I produce a strange result when doing this:
$$(x^{\frac{1}{2}})^{2}=x+1$$ (definitely no solution) 
$$x^{\frac {1}{2}}=(x+1)^{\frac{1}{2}}$$
$$1=\frac{(x+1)^{\frac{1}{2}}}{x^{\frac{1}{2}}}$$
$$1=\frac{x^{2}+2x+1}{x^{2}}$$
$$x^{2}=x^{2}+2x+1$$
$$2x+1=0$$
$$x=-0.5$$
Can you help find my mistake in this situation?Thanks.
 A: When you went from the third line to the fourth, you took the 4th power of both sides. (I'm not sure why -- you could have just squared both). 
But $x^4 = y^4$ does not imply that $x = y$. (for instance $(-1)^4 = 1^4$, but $-1$ and $1$ are different. 
So in taking 4th powers, you may have introduced new solutions that are not real solutions of the original equation. 
Post-comment addition
To see why taking a 4th power introduces new solutions, let's look at the squaring function, $f(x) = x^2$, defined on the real numbers. 
The function $f$ is not injective, i.e., there can be two different numbers $a$ and $b$ with $f(a) = f(b)$. For instance, $a = 2$ and $b = -2$ both have the property that $f(a) = f(b) = 4$. 
If I tell you that I'm thinking of numbers $x$ and $y$ with $f(x) = f(y)$, can you conclude that $x = y$? No! 
Why not? Well, suppose that $f(x)$ and $f(y)$ are both equal to $4$. Then $x$ could be $2$ and $y$ could be $-2$, and those are different. 
Now let's go the other direction. Suppose I tell you that 
$$
x = y = 2
$$
Then we can apply $f$ to get that 
$$
x^2 = y^2 = 4.
$$
But that doesn't mean that every solution to the SECOND equation is a solution to the first one. For instance, $x = 2, y = -2$ is a solution to the second equation, but not a solution to the first. 
The general principle to derive from this example is that if you have an equation, and apply a function $f$ to both sides to get a new equation, every solution of the original equation is still a solution of the new equation (provided that they are all within the domain of $f$). But not every solution to the new equation is a solution to the original. If you want to guarantee that a well, then the function $f$ must have an inverse, i.e., a function $g$ with $g(f(x)) = x$ and $f(g(x)) = x$ for every $x$. 
Examples: 
$f(x) = x + 1$; $g(x) = x - 1$. This example shows that you can "add 1 to both sides of an equation" without changing the set of solutions. 
$f(x) = 2x$; $g(x) = x/2$. This example shows that you can "multiply both sides of an equation by 2" without changing the set of solutions. 
But since $f(x) = x^2$ does not have an inverse function, you cannot ensure that squaring both sides of an equation leaves the solution set unchanged. 
A: There is no need of all these intermediate steps to create the "paradox".
$$x=x+1\implies x^2=x^2+2x+1\implies2x+1=0.$$
Actually the squaring step is invalid, as it introduces the extra solution
$$-x=x+1.$$

When you use a squaring, like $a=b\implies a^2=b^2$, you must remember that $-a=b$ is ruled out.
A: But the domain gives $x>0$. 
Which says that our equation has no solutions.
We can see this immediately:
$$\left(x^{\frac{1}{2}}\right)^2=x+1$$ it's
$$x=x+1$$ or
$$0=1,$$ which has no solutions.
A: Note that you have the intermediate step $$1=\frac {x^{\frac 12}}{(x+1)^{\frac 12}}$$
Now that doesn't have a solution, and when you square it you get $$1=\frac {x}{x+1}$$ which again has no solution. However $$-1=\frac {x}{x+1}$$ has the solution $x=-\frac 12$, and these two equations combine into one when you square either.
In general if you have $x=y$ and square it to get $x^2=y^2$, the second equation can be rewritten $(x+y)(x-y)=0$ or $x=\pm y$. So the squared equation has a solution which does not belong to the original equation.
