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Now substitute it into the original equation



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Almost better to ask: what, if anything, is correct on any of the lines of this "argument"? – RecklessReckoner Apr 28 '13 at 5:16
@PeterTamaroff I do not think OP's interpretation of what $\sqrt{x}$ means can be ascertained from this question. OPs proof of $\sqrt{x}=-1\implies x=1$ is a valid logical deduction, the trick here is that an equation with no solution will always be false and hence anything may be obtained from it, including things like $x=1$, or $x=$ any value really. – anon Apr 28 '13 at 7:44
A similar case: take $x=x+1$. This equation has no solution. Squaring both sides yields $x^2=x^2+2x+1$, which yields $0=2x+1$, which yields $x=-1/2$ though obviously $-1/2$ is not a solution. If you cube the equation and then solve you would get $x=(-3\pm i\sqrt{3})/6$, totally different values. Again the issue is that $a=b\implies a^2=b^2$ but $a^2=b^2\not\Rightarrow a=b$. Obtaining $x=\square$ from a given equation through valid deduction involving squaring is not sufficient for $x=\square$ to entail the truth of the original equation, which is what it means for $x=\square$ to be a solution. – anon Apr 28 '13 at 7:48

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 $x$.

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I think this point outshines all of the others answered thus far. When you perform an operation on both sides of the equation, you obtain a logical consequence of whatever you started with. Thus, the OP has correctly concluded that $\sqrt{x}=-1\implies x=1$. However, it is an elementary mistake to believe a logical consequence is logically equivalent to the original equation - here, it isn't, and squaring generally weakens things logically speaking. Going from $f(x)=a$ to $x=b$ is not sufficient for $x=b$ to be a solution; we must also have $f(b)=a$, which we do not have here. +1. – anon Apr 28 '13 at 7:41

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

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That is a highly misleading and self contradictory statement to be made. First thing: you are asserting that it's a function. Now, by definition, a function has just ONE value for an element in the domain. Now saying that $\sqrt x$ is not a single-valued function just goes against the definition. Oxymoron-ish! – Parth Thakkar May 5 '13 at 9:03
In the field of complex numbers, the square root function has a branch point at the origin and is not a function but a multifunction. Believing that this $\sqrt{x}$ on $\mathbb{C}$ is indeed a function is the root of the "paradoxes" found in OP and other similar calculations. – vadim123 May 9 '13 at 4:08

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.

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exactly. in the first step the LHS is positive (in standard conventions) and the RHS is negative. – Ant Apr 28 '13 at 11:09

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?

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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"

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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

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Sorry, The reason is NOT that $ \sqrt x $ is defined only for positive numbers but the fact that showing $P\implies Q$ does not allow one to use $Q\implies P$. See @anon's comments for more on this. – Did May 5 '13 at 9:10
I agree that what anon said is not wrong, not at all. But I don't understand why you say my reasoning is wrong. – Parth Thakkar May 5 '13 at 12:34
Imagine that $\sqrt{\ }$ is defined also on the negative real numbers, say by the formula $\sqrt{x}=\mathrm i\sqrt{|x|}$ for every $x\lt0$. Then $[\sqrt{x}=-1\implies x=1]$ would still be true since the assertion that $[\sqrt{x}=-1]$ would be false for every real number $x$. Hence the first sentence of your answer is (true but) not the reason why the proof in the OP that $1=-1$ fails. – Did May 5 '13 at 14:35

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