# True, false and meaningless statements in math.

Consider statement for every x belonging to set consisting of {-1 , -2 } : square root of x is equal to one. Of course it is false, given that square root is not defined over negative nubers which takes real value. So, the opposite statement should be true: exists x belonging to the set {-1 , -2 } : such that square root of x is not equal to one. So it should be less or more than one. This statement is also not true. So, can we conclude that statement is meaningless?

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are you talking about square root or square? – dato datuashvili Feb 8 '14 at 16:38
Dato, I am talking about square root. (P.S. qartvelebs gaumarjos) – nkoreli Feb 8 '14 at 16:49
@kinokijuf Unicode says those are georgian letters. – canaaerus Feb 8 '14 at 20:29
The language seems to be Georgian. Nkoreli is typing Georgian in the Latin alphabet. – user54609 Feb 8 '14 at 20:29
This issue can be stated in a simpler way: 1/0 = 0 is not true, 1/0 != 0 is not true either. – usr Feb 8 '14 at 22:41

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 $-1$.

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We can still say, because of the non-definition of the square root, that $\sqrt{-1} \neq 1$...This is indeed true. – amWhy Feb 8 '14 at 17:07
Mauro, thanks for the answer. But, I cannot get how one statement be with meaning, and the negation be meaningless?? Do you have other examples of such statements? – nkoreli Feb 8 '14 at 17:08
The issue with this is that there is not likely to be any "SQRT" function symbol; there is not one in the normal language of fields, for example. – Carl Mummert Feb 8 '14 at 17:09
@nkoreli - You are right; following Carl perfect answer, I should have written : "Assuming that the statement ... is false ...". – Mauro ALLEGRANZA Feb 8 '14 at 17:15
@randomatlabuser - You are right: I've improved my answer, in order to avoid the previous sloppiness. Instead of "meaningless", we have to appeal to semantics for first-order logic. In defining truth of a sentence in a model, we require that every name occurring in the sentence has a denotation. Moreover, function signs are taken always to represent operations everywhere defined in the domain: there is no place in standard semantics for "non-denoting" terms in the sentences involved [this is different for free logic]. – Mauro ALLEGRANZA Feb 10 '14 at 17:10

The issue here has to do with "undefined terms". Suppose that we are working with the field of real numbers (without non-real complex numbers). The signature for this field has symbols for addition and multiplication, the order relation $<$, and the equality sign.

This signature does not have a symbol for square roots. So the phrase "the square root of $-1$ is not equal to $1$" does not have any direct translation into the formal language at hand. If every number did have a unique square root, we could introduce a new function symbol $\sqrt{}$, and interpret it so that $\sqrt{x}$ is always a square root of $x$.

In contrast, every real number does have a square, so we could add a new unary function $S$ such that $S(x)$ gives $x\cdot x$. We cannot do that for square roots. In the field of real numbers, we cannot refer to $\sqrt{-1}$, as we have no symbol $\sqrt{}$.

We could translate the phrase "the square root of $-1$ is not equal to $1$" as "for all $x$, if $x^2 = -1$ then $x \not =1$". That new quoted statement can be directly translated to the language of fields, and it is true in the field of real numbers. On the other hand, the statement "the square root of $-1$ is not equal to $1$" can also be translated as "there is an $x$ with $x^2 = 1$ and $x \not = 1$". That statement is false in the field of real numbers. So the translation that we choose will affect the truth value.

In the general, an English sentence that talks about "the" object with a particular property makes an existential assumption that there is a unique object with that property. When that existential assumption is correct, it usually straightforward to translate the English sentence into a formal sentence, and the truth of the formal sentence will not depend on which reasonable translation we use. But, when the existential assumption is not satisfied, the exact translation that we use can make a difference.

Another approach to handling issues like real square roots (which don't always exist) is to introduce partial function symbols - function symbols that can be undefined on certain inputs. This is the subject of "free logic". But normal first-order logic is not a free logic: it assumes that a function symbol applied to an object always returns an object.

Yet another approach is to add a new function symbol $\sqrt{}$ and define it in some arbitrary way when it would not normally be defined. The, all we can say is that if $x$ has a square root then $\sqrt{x}$ is a square root of $x$. In that case, the statement "the square root of $-1$ is not equal to $1$" will depend on exactly how we arbitrarily assign values for numbers that don't have square roots.

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Carl Mummert, thank you very much for this comment. It clarified the issue. – nkoreli Feb 8 '14 at 17:18

"...So it should be less or more than one". That (added) conclusion is what's untrue.

What is true is that the square root of $-1$, just like the square root of $-2$ is not equal to one (and in this case, that is because their square roots do not even exist).

Of course there are meaningless statements, in the sense that the statements themselves do not admit a truth value assignment: E.g.:

• "Stop!",

• "Is my proof correct?"

• "Prove that $\sqrt 2$ is irrational." etc.

We can say nothing about the truth value of commands, questions, or imperative. Also problematic in this regard arestatements like "I wonder if...", "I hope that...", and the like (addressing states of mind).

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Thanks for the answer, but I am still confused, since I thought that statements (x != 3) and (x>3 or x<3) are equivalent. It is from definition of relation != . – nkoreli Feb 8 '14 at 16:42
Yes, that is true providing $x$ exists and is real. While $x \in \{-1, -2\}$ exists, $\sqrt{-1}, \sqrt{-2}$ does not exist in the reals, so it cannot be compared to the number $1$. What we can say for certain is that $\sqrt{-1} \neq 1$ and $\sqrt{-2} \neq 1$. Even if we allow the complex roots, complex numbers are not comparable save for $z_1 = z_2$ or $z_1\neq z_2$ (the complex numbers are not an ordered field, like the reals are, "less than" and "larger than" are meaningless in the complex numbers.) – amWhy Feb 8 '14 at 16:51
In the normal setting of the field of real numbers, we can't say "$\sqrt{-1} = 1$" or "$\sqrt{-1} \not = 1$", of course, as we have no square root function. Actually we can't even say this in the complex numbers until we extend the language with a square root function in some way. – Carl Mummert Feb 8 '14 at 17:17
Just because a statement does not admit a truth value (or is not, in fact, a statement, like the examples you cited) does not mean it's meaningless. – tomasz Feb 9 '14 at 0:03