How is this proof flawed?

Question

$\sqrt{x}=-1$

$\sqrt{x}^2=(-1)^2$

$x=1$

Now substitute it into the original equation

$\sqrt{1}=-1$

$1=-1$

Answer 1

Because $\sqrt{x}$ is not a single-valued function on $\mathbb{C}$.

Answer 2

Your last step didn't square the right-hand side of the equation.

Also, the $\sqrt{x}$ function has an implicit sign of + when none is present. So the first equation is as nonsensical as starting with "4 = 5"

Answer 3

Generally, $\sqrt{x}$ is defined as the positive number $y$ such that $y^2=x$. You can choose to define it in such a way that you pick the negative root instead, but then you have to stick with that convention. So your first step is technically true if you choose that convention, but then you can't switch back to the positive root in your last step.

Answer 4

My favorite numbers are $1$ and $-1$.

$$\begin{array}{cccc}&\text{Alex's favorite number} &= &1\\ \text{ thus it follows that }&-1&=&1\end{array}$$

Where is the flaw in the above argument?

Answer 5

When solving an equation, what we mostly do is that at each step we transform an equation to another which has the same set of solutions. For example we transform $$x+4 = 0$$ into $$x = -4$$ by "subtracting 4 from both sides", which is an operation that preserves the set of solutions. This way the solutions we obtain at the end are the solutions of the original equation.

If you do an operation on a equation that doesn't preserve this property, the result obviously doesn't solve the original equation. Such an operation is squaring both sides of an equation. Consider for example $$x = -x$$ Obviously, this equation has a single real solution $x\in\{0\}$. But if we square both sides, we get $$x^2 = x^2$$ which is true for any $x\in\mathbb{R}$.

This is a common mistake, and there are many similar. For example, multiplying or dividing both sides by a number is an operation that preserves solutions only if the number is nonzero. By multiplying both sides by 0 you can "prove" anything, such as \begin{align} x &= 4 \\ x (x-x) &= 4(x-x) \\ 0 &= 0 \\ \end{align} which would make us "conclude" that $x=4$ is true for any real $x$.

Answer 6

The reason is that $ \sqrt x $ is defined only for positive numbers. Now, when you square it, you get $ |x| $ and not $x$, since $x$ by itself can be anything: $+ve$ or $-ve$. ( In the case of $ \sqrt x $, the very fact that the expression is valid implies $x$ is $+ve$ : you see the difference? ).

I've committed this mistake a hell lot of times, and my teacher has scolded us (I obviously wasn't the only person to repeatedly commit the same mistake) a lot of times for the same. So I know it! :D