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The mathematical statement $\exists h{\in} \mathbb{R} \quad \dfrac{1}{h}=5\tag*{}$ seems a mathematical statement that is true. However, its negation $\forall h{\in} \mathbb{R} \quad \dfrac{1}{h}\neq 5\tag*{}$ is not a mathematical statement, because it claims that $\displaystyle\frac10\ne5,$ which is nonsense since $0$ does not have a multiplicative inverse.

Does the above contain mathematical nonsense, or are the two statements actually not negations of each other?

And how do I formalise “there exists a real number whose multiplicative inverse is $5$” ?

ryang
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boyler
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    For this expression to be a mathematically meaningful sentence, we should be able to check it for every real number. Some forms of intuitionism might interest you. – Shaun Jan 11 '21 at 14:09
  • By saying “meaningful”, I meant nothing actually. I just meant that “For this expression to be a mathematical sentence, we should...”. So contrarily, I feel nothing about math. A sentence is either meaningful (i.e. we are allowed to write it) in math or not. For example, $1=0$ is a wrong mathematical sentence but $\dfrac{1}{0}=5$ is not even a mathematical sentence, i.e. mathematically meaningless. – boyler Jan 11 '21 at 14:21
  • $(\forall h\in \mathbb{R}) ; \left(5h\neq1\right)$ – saulspatz Jan 11 '21 at 14:27
  • Although $\frac 10$ may be non-sensical, $\frac 10 \ne 5$ so the statement is still true. – Vishu Jan 11 '21 at 14:31
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    @Tavish No, an expression that is not well-formed has no truth value—from a formal standpoint, it's a nonsensical string of symbols with no meaning whatsoever, neither true nor false. – Daniel Hast Jan 11 '21 at 14:37
  • @DanielHast Really? It still feels unconvincing though, as if we suppose the statement isn’t true, then that would mean $\frac 10=5$ which is a contradiction. – Vishu Jan 11 '21 at 14:39
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    It's not a statement in the formal logical sense at all—it's a nonsensical string of symbols with no meaning, from which we can derive no conclusions whatsoever. (That is, unless we resolve the issue by, say, interpreting division in the extended real numbers rather than the standard real numbers so we have a formal interpretation of $1/0$, but then there's no syntactic issue at all.) – Daniel Hast Jan 11 '21 at 14:41
  • @Tavish We didn’t define the object $\dfrac{1}{0}$. If you $\textbf{define}$ it to be equal to 7, then you can say that it is not equal to 5. But otherwise $\dfrac{1}{0}=5$ is mathematically the same as “rgjrhj=eghjy”. I mean we don’t know what is rgjrhy as we dont know what is $\dfrac{1}{0}$. Because we didn’t define them. – boyler Jan 11 '21 at 15:03
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    your statements are not well typed, that is the main issue! – Nico Apr 25 '23 at 18:44
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    um, it is actually true that $\frac{1}{0} \neq 5$. do not overthink this – RyRy the Fly Guy Jun 23 '23 at 06:01
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    Not a logic expert by any means, but isn't it the case that the negation of $\exists h \in \mathbb{R} : \frac1h = 5$ is $\neg \exists h \in \mathbb{R} : \frac1h = 5$, which one could say is equivalent to $\forall h \in \mathbb{R} : \neg \frac1h = 5$? So if by the predicate $\frac1h \not= 5$ we really mean $\neg \frac1h = 5$, is there a problem? – Brian Tung Sep 24 '24 at 05:23
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    The first statement $\exists h{\in} \mathbb{R} \quad \dfrac{1}{h}=5\tag{}$ is just as problematic as the second: functions in logic should be total, so if you take your domain that the quantifier (whether existential or universal) quantifies over to be all of $\mathbb{R}$, you obviously have a problem. So: *both statements are bad. – Bram28 Oct 14 '24 at 14:52
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    @BrianTung: There is no problem. Every single expert in mathematical logic today sees no problem with allowing 1/0 to have a value, in order to stay within FOL, so "1/0 = 5" has a truth-value but we don't know which, contrary to both Vishu and Daniel Hast. It is only those who don't really grasp FOL that see a problem. Of course, if we want to capture normal mathematical practice, then we need some additional restrictions, but then we must give up essentially all of the standard equivalencies of FOL! See this for more details. – user21820 Aug 30 '25 at 06:16
  • @Bram28: Hello! There is no problem for a logician; see my explanation linked above. After all, do you want "∀x∈ℝ ∃y∈ℝ ( x+y = 0 )" to be true? Do you want it to be equivalent to "∀x ( x∈ℝ ⇒ ∃y ( x+y = 0 ∧ y∈ℝ ) )"? How about "∀x ( ∀y ( x+y ≠ 0 ∨ y∉ℝ ) ⇒ x∉ℝ )"? =) – user21820 Aug 30 '25 at 06:20

2 Answers2

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Assuming we interpret division to mean the usual notion of division in the real numbers, you're correct that the statement $(\exists h \in \mathbb{R})(1/h = 5)$ doesn't quite make sense, for the reason you stated. A more formal way of putting it is that $1/h = 5$ is not, in fact, a predicate with domain $\mathbb{R}$, so we can't quantify it over $\mathbb{R}$.

There are a few ways to correct this:

  1. Rewrite the statement as $(\exists h \in \mathbb{R} \setminus \{0\})(1/h = 5)$ instead, so the domain matches the domain of division of real numbers.
  2. Rewrite the statement as $(\exists h \in \mathbb{R})(1 = 5h)$, noting that $1/h = 5$ is equivalent to $1 = 5h$ for all $h \neq 0$, but the latter also makes sense (though it is false) for $h = 0$. I'd say this is the most straightforward way to formalize "there exists a real number whose multiplicative inverse is $5$".
  3. Interpret the division symbol to mean division in the real projective line $\hat{\mathbb{R}}$, in which $1/0$ is a meaningful expression and is equal to $\infty$, a "point at infinity" that we add to the standard real numbers. With the division symbol interpreted in this way, $1/h = 5$ now makes sense for all $h \in \mathbb{R}$ (in fact, for all $h \in \hat{\mathbb{R}}$).

In practice, issues like this are often somewhat informally glossed over, because the intended meaning of the statement is clear and there's no ambiguity, as all these ways of resolving the formal syntactical issue lead to the same conclusion. But it's good to understand how a slightly informal notation like that can be correctly rewritten/reinterpreted to resolve any formal syntactical problems.

Daniel Hast
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how do I formalise “there exists a real number whose multiplicative inverse is $5$” ?

Directly, without invoking reciprocals: $$\exists h{\in} \mathbb{R} \quad 5h=1.$$

$\exists h{\in} \mathbb{R} \quad \dfrac{1}{h}=5\tag1$ $\forall h{\in} \mathbb{R} \quad \dfrac{1}{h}\neq 5\tag2$ Do the two statements contain mathematical nonsense,

or are they actually not negations of each other?

Unlike sentences like $3\text{ implies and }1+1=2\tag*{}$ that violates syntactic rules, sentences (1) and (2) are well-formed, so certainly are logical negations of each other, having opposite truth values under every interpretation in which the expressions involved are well defined.

However, statement $(2),$ as well as

  • $\frac10=5$
  • $\forall x{\in}\left\{0,\frac15\right\}\;\frac1x\ne5$
  • The reciprocal function with domain $\mathbb R{\setminus}\{0\}$ is continuous at $0$
  • $A=\{x\in\mathbb R\mid x \text{ is “large”}\},$

are arguably ill-posed and semantically invalid (non-meaningful rather than merely false), as they refer to objects or concepts that are either undefined or ill-defined. Similarly, talking about a table's tail would be an illegal attribution.

Nonetheless, in mathematical practice, it is fine to treat these statements as false—and this is indeed often the case—because the intended meaning is usually clear since the context generally excludes the "problematic" cases from consideration.

ryang
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