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We may analyze the statement with the expression "the square root of $x$" according to Russell's analysis of definite description; see Definite description.

The "general form" is :

$\psi(\iota x(\phi(x))$ is equivalent to $\exists x \forall y ( \phi(y) \leftrightarrow y = x \land \psi(y))$.

In our example, we have that $\phi(x)$ is "the square of $x$ is equal to $-1$" and $\psi(y)$ is "$y$ is equal to $1$", so that : $\iota x(x^2 = -1) = 1$.

Russellian analysis , when applied to the above statement, is : "The square root of $-1$ is equal to $1$" says that some $x$ is such that it is the square root of $-1$, and that any $y$ is the square root of $-1$ only if $y = x$, and that $x = 1$:

$\exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y=x) \land x = 1]$

This is false in the real field, since it is not the case that some $x$ is such that $x^2 = -1$.

The negation of this sentence, i.e. "The square root of $-1$ is not equal to $1$", is ambiguous in natural language. It could mean one of two things, depending on where we place the negation 'not'.

On one reading, it could mean that there is no one who is the square root of $-1$ and equal to $1$:

$\lnot \exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y = x) \land x = 1]$.

On this disambiguation, the sentence is true (since there is indeed no $x$ that is square root of $-1$).

On a second reading, the negation could be construed as attaching directly to '$ = 1$', so that the sentence means that there is presently a square root of $-1$, but that this number fails to be $= 1$ :

$ \exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y = x] \land x \neq 1]$.

On this disambiguation, the sentence is false (since there is no $x$ that is a square root of $-1$

Thus, whether "the square root of $-1$ is equal to $1$" is true or false depends on how it is interpreted at the level of logical form: if the negation is construed as taking wide scope (as in $\lnot \exists x [... \land x = 1]$, it is true, whereas if the negation is construed as taking narrow scope (with the existential quantifier taking wide scope, as in $\exists x [... \land x \neq 1]$), it is false. In neither case does it lack a truth value.

So we do not have a failure of the Law of Excluded Middle: "the square root of $-1$ is equal to $1$" is false, because there is no present square root of $-1$. The negation of this statement is the one in which 'not' takes wide scope. This statement is true because there does not exist anything which is the square root of $-1$.

We may analyze the statement with the expression "the square root of $x$" according to Russell's analysis of definite description; see Definite description.

The general form of "the unique $x$ such that $\phi(x)$" is symbolized as $\iota(\phi(x))$ and:

$\psi(\iota x(\phi(x))$ is stipulated to be equivalent to $\exists x \forall y ( \phi(y) \leftrightarrow y = x \land \psi(y))$.

In our example, we have that $\phi(x)$ is "the square of $x$ is equal to $-1$" and $\psi(y)$ is "$y$ is equal to $1$", so that "The square root of $-1$ is equal to $1$" will be : $\iota x(x^2 = -1) = 1$.

Russellian analysis , when applied to the above statement, is : "The square root of $-1$ is equal to $1$" says that some $x$ is such that it is the square root of $-1$, and that any $y$ is the square root of $-1$ only if $y = x$, and that $x = 1$:

$\exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y=x) \land x = 1]$

This is false in the real field, since it is not the case that some $x$ is such that $x^2 = -1$.

The negation of this sentence, i.e. "The square root of $-1$ is not equal to $1$", is ambiguous in natural language. It could mean one of two things, depending on where we place the negation 'not'.

On one reading, it could mean that there is no one who is the square root of $-1$ and equal to $1$:

$\lnot \exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y = x) \land x = 1]$.

On this disambiguation, the sentence is true (since there is indeed no $x$ that is square root of $-1$).

On a second reading, the negation could be construed as attaching directly to '$ = 1$', so that the sentence means that there is presently a square root of $-1$, but that this number fails to be $= 1$ :

$ \exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y = x] \land x \neq 1]$.

On this disambiguation, the sentence is false (since there is no $x$ that is a square root of $-1$).

Thus, whether "the square root of $-1$ is equal to $1$" is true or false depends on how it is interpreted at the level of logical form: if the negation is construed as taking wide scope (as in $\lnot \exists x [... \land x = 1]$, it is true, whereas if the negation is construed as taking narrow scope (with the existential quantifier taking wide scope, as in $\exists x [... \land x \neq 1]$), it is false. In neither case does it lack a truth value.

So we do not have a failure of the Law of Excluded Middle: "the square root of $-1$ is equal to $1$" is false, because there is no present square root of $-1$. The negation of this statement is the one in which 'not' takes wide scope. This statement is true because there does not exist anything which is the square root of $-1$.

We may analyze the statement with the expression "the square root of $x$" according to Russell's analysis of definite description; see Definite description:

Russellian analysis , when applied to the above statement, is : "The square root of $-1$ is equal to $1$" says that some $x$ is such that it is the square root of $-1$, and that any $y$ is the square root of $-1$ only if $y = x$, and that $x = 1$:

$\exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y=x) \land x = 1]$

This is false in the real field, since it is not the case that some $x$ is such that $x^2 = -1$.

The negation of this sentence, i.e. "The square root of $-1$ is not equal to $1$", is ambiguous in natural language. It could mean one of two things, depending on where we place the negation 'not'.

On one reading, it could mean that there is no one who is the square root of $-1$ and equal to $1$:

$\lnot \exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y = x) \land x = 1]$.

On this disambiguation, the sentence is true (since there is indeed no $x$ that is square root of $-1$).

On a second reading, the negation could be construed as attaching directly to '$ = 1$', so that the sentence means that there is presently a square root of $-1$, but that this number fails to be $= 1$ :

$ \exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y = x] \land x \neq 1]$.

On this disambiguation, the sentence is false (since there is no $x$ that is a square root of $-1$

Thus, whether "the square root of $-1$ is equal to $1$" is true or false depends on how it is interpreted at the level of logical form: if the negation is construed as taking wide scope (as in $\lnot \exists x [... \land x = 1]$, it is true, whereas if the negation is construed as taking narrow scope (with the existential quantifier taking wide scope, as in $\exists x [... \land x \neq 1]$), it is false. In neither case does it lack a truth value.

So we do not have a failure of the Law of Excluded Middle: "the square root of $-1$ is equal to $1$" is false, because there is no present square root of $-1$. The negation of this statement is the one in which 'not' takes wide scope. This statement is true because there does not exist anything which is the square root of $-1$.

We may analyze the statement with the expression "the square root of $x$" according to Russell's analysis of definite description; see Definite description.

The "general form" is :

$\psi(\iota x(\phi(x))$ is equivalent to $\exists x \forall y ( \phi(y) \leftrightarrow y = x \land \psi(y))$.

In our example, we have that $\phi(x)$ is "the square of $x$ is equal to $-1$" and $\psi(y)$ is "$y$ is equal to $1$", so that : $\iota x(x^2 = -1) = 1$.

Russellian analysis , when applied to the above statement, is : "The square root of $-1$ is equal to $1$" says that some $x$ is such that it is the square root of $-1$, and that any $y$ is the square root of $-1$ only if $y = x$, and that $x = 1$:

$\exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y=x) \land x = 1]$

This is false in the real field, since it is not the case that some $x$ is such that $x^2 = -1$.

The negation of this sentence, i.e. "The square root of $-1$ is not equal to $1$", is ambiguous in natural language. It could mean one of two things, depending on where we place the negation 'not'.

On one reading, it could mean that there is no one who is the square root of $-1$ and equal to $1$:

$\lnot \exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y = x) \land x = 1]$.

On this disambiguation, the sentence is true (since there is indeed no $x$ that is square root of $-1$).

On a second reading, the negation could be construed as attaching directly to '$ = 1$', so that the sentence means that there is presently a square root of $-1$, but that this number fails to be $= 1$ :

$ \exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y = x] \land x \neq 1]$.

On this disambiguation, the sentence is false (since there is no $x$ that is a square root of $-1$

Thus, whether "the square root of $-1$ is equal to $1$" is true or false depends on how it is interpreted at the level of logical form: if the negation is construed as taking wide scope (as in $\lnot \exists x [... \land x = 1]$, it is true, whereas if the negation is construed as taking narrow scope (with the existential quantifier taking wide scope, as in $\exists x [... \land x \neq 1]$), it is false. In neither case does it lack a truth value.

So we do not have a failure of the Law of Excluded Middle: "the square root of $-1$ is equal to $1$" is false, because there is no present square root of $-1$. The negation of this statement is the one in which 'not' takes wide scope. This statement is true because there does not exist anything which is the square root of $-1$.

Assuming that the statement :

$\forall x (x \in \{-1, -2 \} \rightarrow SQRT(x) = 1)$

is false, its negation :

$\exists x (x \in \{-1, -2 \} \land SQRT(x) \neq 1)$

must be true, provided that the formula has a meaning.

As said by amWhy, the function $SQRT(x)$ is not defined in the real field, so the formula is meaningless.

And if we interpret it in the complex domain, now is the "order" that is not defined.

We may analyze the statement with the expression "the square root of $x$" according to Russell's analysis of definite description; see Definite description:

Russellian analysis , when applied to the above statement, is : "The square root of $-1$ is equal to $1$" says that some $x$ is such that it is the square root of $-1$, and that any $y$ is the square root of $-1$ only if $y = x$, and that $x = 1$:

$\exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y=x) \land x = 1]$

This is false in the real field, since it is not the case that some $x$ is such that $x^2 = -1$.

The negation of this sentence, i.e. "The square root of $-1$ is not equal to $1$", is ambiguous in natural language. It could mean one of two things, depending on where we place the negation 'not'.

On one reading, it could mean that there is no one who is the square root of $-1$ and equal to $1$:

$\lnot \exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y = x) \land x = 1]$.

On this disambiguation, the sentence is true (since there is indeed no $x$ that is square root of $-1$).

On a second reading, the negation could be construed as attaching directly to '$ = 1$', so that the sentence means that there is presently a square root of $-1$, but that this number fails to be $= 1$ :

$ \exists x [x^2 = -1 \land \forall y (y^2 = -1 \rightarrow y = x] \land x \neq 1]$.

On this disambiguation, the sentence is false (since there is no $x$ that is a square root of $-1$

Thus, whether "the square root of $-1$ is equal to $1$" is true or false depends on how it is interpreted at the level of logical form: if the negation is construed as taking wide scope (as in $\lnot \exists x [... \land x = 1]$, it is true, whereas if the negation is construed as taking narrow scope (with the existential quantifier taking wide scope, as in $\exists x [... \land x \neq 1]$), it is false. In neither case does it lack a truth value.

So we do not have a failure of the Law of Excluded Middle: "the square root of $-1$ is equal to $1$" is false, because there is no present square root of $-1$. The negation of this statement is the one in which 'not' takes wide scope. This statement is true because there does not exist anything which is the square root of $-1$.