There's a deeper concept behind this fake proof

Question

Whenever I start to get cocky about my math knowledge, some basic property of the complex plane puts me back in my place. I just came across this fake proof that 2 = 0:

In the shallowest sense, I think I know what the problem is here. The author deceptively exploits how squaring is a 2-to-1 mapping on $C \setminus 0$, and flips the sign of a root somewhere. If we use $-i$ as the square root of $-1$ instead of $i$, we get a tautology rather than a contradiction.

But what I'd like is a fuller description of the phenomenon underlying the trick. My guess is that the proof smuggles in some special property of $R$ that is so basic that the naive reader assumes it holds in $C$, even though it doesn't. Could anyone state what this property is? Is my guess on the right track?

Answer 1

The proof tacitly assumes that there is a function $sqrt:\mathbb{C}\rightarrow\mathbb{C}$ (which it calls "$\sqrt{\cdot}$") with the following two properties:

A: $sqrt$ gives square roots: for all $z$ we have $sqrt(z)^2=z$.

B: $sqrt$ distributes over multiplication: for all $z_0,z_1$ we have $sqrt(z_0)\cdot sqrt(z_1)=sqrt(z_0\cdot z_1)$.

I'll call such a function (if it exists) a good square rooter.

If there were such a function, then the proof would work - so in fact what's being shown is that no such function exists. This can be a stumbling point because of course over $\mathbb{R}_{\ge 0}$ there is such a function, namely the function sending $x$ to its unique nonnegative square root.

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OK, so what is it about $\mathbb{C}$ as opposed to $\mathbb{R}_{\ge0}$ that makes the former have no good square rooter?

Well, it turns out that the issue is exactly that elements of $\mathbb{C}$ have multiple square roots in $\mathbb{C}$ in general, whereas each element of $\mathbb{R}_{\ge0}$ has exactly one square root in $\mathbb{R}_{\ge0}$. As soon as we're forced to "make a choice," we lose any hope of having a good square rooter.

To be precise:

Suppose $A$ is a commutative semiring in which every element has at least one square root. Then the following are equivalent:

1. Every element in $A$ has exactly one square root.
2. There is a good square rooter $sqrt_A:A\rightarrow A$.

Proof: The direction $2\rightarrow 1$ is basically just the argument in the OP! Suppose we have a good square rooter $sqrt_A$, and pick $a,b,c\in A$ with $a^2=b^2=c$. We have $$sqrt_A(c)=sqrt_A(a)\cdot sqrt_A(a)=sqrt_A(b)\cdot sqrt_A(b)$$ by condition B of good-square-rooter-ness, but we also have $$sqrt_A(a)\cdot sqrt_A(a)=a\mbox{ and }sqrt_A(b)\cdot sqrt_A(b)=b$$ by condition A. Put together we get $a=b$ as desired.

In the other direction, suppose $(1)$ holds. Then we can define a function $s: A\rightarrow A$ by $s(a)=$ the unique $b$ with $b^2=a$. This trivially satisfies condition A of good-square-rooter-ness, so we just have to show that $$s(a)\cdot s(b)=s(a\cdot b)$$ for every $a,b\in A$.

And this is nice and easy! By definition of $s$, we have $$[s(a)s(b)]^2=[s(a)^2][s(b)^2]ab=[s(ab)]^2.$$ So $s(a)s(b)$ and $s(ab)$ are elements of $A$ which square to the same thing (namely $ab$), which means .... that they're equal by our assumption that we're in case $(1)$.

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"But wait!", you might reasonably say, "what about $\mathbb{R}_{\ge0}$? Positive real numbers do have multiple square roots even though we have a good square rooter in $\mathbb{R}_{\ge0}$. What gives?"

The point is that we get extra square roots for positive reals only when we step outside of $\mathbb{R}_{\ge 0}$. Within $\mathbb{R}_{\ge0}$ itself, every element has exactly one square root. The proposition is very carefully phrased to be about what's going on inside the commutative semiring $X$, not about how $X$ sits inside some yet larger commutative semiring.

So we always have to pay attention to where the solutions to various equations exist!

Answer 2

The invalid assumption is that if $a^2 = m$ then $\sqrt m = a$. That is not true. (Example: $(-5)^2 = 25$ but $\sqrt {25} \ne -5$.

With that invalid assumption we make an invalide rule of arithmetic:$\sqrt{ab} = \sqrt a \sqrt b$. That is not true and it is based on an invalid assumption.

The reason we think it is true is because $(\sqrt{a}\sqrt b)^2 = (\sqrt a)^2 \cdot (\sqrt b)^2=ab$. That is true. But $(\sqrt a\sqrt b)^2 = ab$ does not mean that $\sqrt{ab} = \sqrt a \sqrt b$.

But it is valid if $a,b$ are positive (well, actually non-negative).

This is because in real numbers the say $\sqrt a$ and $\sqrt b$ exist at all, we have to have $a \ge 0$ and $b\ge 0$. In complex numbers we don't have to have that conclusion.

And if $\sqrt a\ge 0$ and $\sqrt b\ge 0$ then $\sqrt a \sqrt b \ge 0$. We have to reach that conclusion in real numbers if $\sqrt a$ and $\sqrt b$ even exist. But in complex numbers we dont have to have that conclusion.

Now, in real numbers we have $\sqrt a = m$ if i) $m^2 = a$ but ALSO if ii) $m\ge 0$. So we must in real numbers reach the conclusion that $\sqrt {ab} = \sqrt a \sqrt b$ if $\sqrt a$ and $\sqrt b$ exist at all.

But in complex numbers we have $\sqrt a = m$ if i) $m^2 = a$ and also the angle argument of $m$ is in the upper half of the complex plane (well, maybe, different texts have different definitions.) We do not have the requirement that $\sqrt a \sqrt b$ satisfies that conditions. So we do not have that $\sqrt {ab} =\sqrt a\sqrt b$> Mayb $\sqrt {ab} = -\sqrt a \sqrt b$.

It does come down to there are always 2 different numbers where $m^2 = a$ and it is arbitrary which one we call "the" square root. In reals where $\sqrt{negative}$ do not exist that is not an issue when claiming $\sqrt{ab}=\sqrt a\sqrt b$ and $a,b\ge 0$ is implied. In complex where $\sqrt{negative}$ is allowed we can't assume it distributes like that.

Answer 3

The proof asserts that $$1+\sqrt{(-1).(-1)}=1+\sqrt{-1}\sqrt{-1}$$ for the reason that $$\sqrt{a. b}=\sqrt a\times\sqrt b.$$

This is wrong. The equation $\sqrt{a. b}=\sqrt a\times\sqrt b$ does not hold for all complex (or even all real) numbers $a$ and $b$. In particular, it doesn’t hold when $a=b=-1$, which is where it is incorrectly used.