Explain the error in the following "proof" that 1=-1 : -1=i·…
Explain the error in the following "proof" that $1=-1$: $\ -1=i\cdot i=\sqrt{-1}\cdot\sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt1=1$.
Explanation of Method
The identity $\sqrt a\cdot\sqrt b=\sqrt{ab}$ is a theorem about the nonnegative real square root, proven precisely because both sides are nonnegative and their squares agree. It is not a general algebraic law, and there is no way to extend it to complex square roots that preserves it for all inputs simultaneously — the "proof" silently assumes it does.
Step 1 · Where the real-number identity comes from
For $a,b\ge0$, $\sqrt a$ and $\sqrt b$ denote the unique nonnegative numbers whose squares are $a$ and $b$. Then $(\sqrt a\sqrt b)^2=ab$ and $\sqrt a\sqrt b\ge0$, so by uniqueness of the nonnegative square root, $\sqrt a\sqrt b=\sqrt{ab}$. This argument uses "nonnegative" twice — it has no analogue once negative numbers are allowed.
Step 2 · There is no consistent choice of $\sqrt{-1}$ making $\sqrt a\sqrt b=\sqrt{ab}$ always true
Suppose we try to define $\sqrt z$ for all complex $z$ so that $\sqrt a\sqrt b=\sqrt{ab}$ holds universally. Apply it to $a=b=-1$:
$$ \sqrt{-1}\cdot\sqrt{-1}\overset{?}{=}\sqrt{(-1)(-1)}=\sqrt1=1 $$
But by definition $\sqrt{-1}\cdot\sqrt{-1}=(\sqrt{-1})^2=-1$ (that is the entire point of calling it a square root of $-1$). So the assumed identity forces $-1=1$ — a contradiction. This shows no consistent extension of "$\sqrt{\ }$" to negative (or complex) numbers can satisfy $\sqrt a\sqrt b=\sqrt{ab}$ unconditionally.
Step 3 · Diagnose the specific step that fails
In the chain $\sqrt{-1}\cdot\sqrt{-1}=\sqrt{(-1)(-1)}$, the left side equals $i\cdot i=-1$ by the defining property of $i$. The right side, if evaluated correctly as $\sqrt{(-1)(-1)}=\sqrt1=1$ using the ordinary nonnegative real square root, is $1$. These are genuinely different numbers ($-1\neq1$); the error is asserting $\sqrt{-1}\cdot\sqrt{-1}=\sqrt{(-1)(-1)}$ at all — that step, not any later one, is invalid.
Step 4 · The correct general rule
Using the principal branch $\sqrt z=|z|^{1/2}e^{i\arg z/2}$ with $\arg z\in(-\pi,\pi]$, one instead gets the correct (weaker) law:
$$ \sqrt a\,\sqrt b=\sqrt{ab}\cdot e^{i\epsilon}, \qquad \epsilon\in\{0,\pm2\pi\}\text{ depending on whether }\arg a+\arg b\text{ leaves }(-\pi,\pi] $$
For $a=b=-1$: $\arg(-1)=\pi$, so $\arg a+\arg b=2\pi$, which is outside the principal range $(-\pi,\pi]$ and must be reduced by $2\pi$ to $0$ before halving — exactly the branch jump that the naive manipulation ignored.
Result
The fallacy is a branch-cut error: multiplying two square roots can silently jump the argument outside the principal range, which is exactly what happens when both factors are $-1$.