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Take the statement "Let $a \in \mathbb{R}$ and let $z=ai.$" $(1)$

In English, this means "Let's choose any real $a$ then define $z$ such that $z=ai$ for that particular $a$ we have just chosen". All good there. We might use terms like "arbitrary constant" or "parameter" to describe $a$ too.

How do we write $(1)$ in formal logic?

I know that $\forall a \exists z (z=ai)$ is one option, but is that correct/the only one?

I know that $\forall a \forall z (z=ai)$ is wrong, and that so is $\exists z \forall a (z=ai).$

I know that "let" means to consider a specific real $a$ such that we can assume only that it's a member of $\mathbb{R},$ and $z$ if well-defined for this $a$ is defined for all $a \in \mathbb{R}.$ But then this implies $\forall a (z=ai),$ and it seems that each $a$ requires a different $z$ to make the statement true, i.e., $z$ depends on $a,$ yet $\forall a (z=ai)$ doesn't allow that.

ryang
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Nav Bhatthal
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    Usually you would use this construction in a proof or similar place, where it would write as $\forall_a ((z=ai) \implies p(a,z,..))$ where $p$ is the statement you finally have at the end. Without such general context it does not make much sense because "let" itself is not introducing a proposition. – Michal Adamaszek Aug 18 '23 at 07:57
  • With no further context, how would you interpret "Let $z=ai$ where $a \in \mathbb{R}$? – Nav Bhatthal Aug 18 '23 at 08:27
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    As a definition of $z$ given $a$ – Michal Adamaszek Aug 18 '23 at 08:56
  • But the only thing we know about $a$ is it is real, so $z=ai$ for all real $a$, right? But is that not a contradiction? We can write $\forall a$ ($ai$ is an imaginary number), but we cannot write $\forall a$ ($z=ai$ is an imaginary number) (My choice of $a$ is permitted to be arbitrary) – Nav Bhatthal Aug 18 '23 at 09:05
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    Of course if you want to be very formal then you should actually consider $z$ as a function which assigns to each real $a$ the number $ai$, and consequently call it $z(a)$, but then it would be impossible to read any mathematical text. – Michal Adamaszek Aug 18 '23 at 09:09
  • You claim "Given an arbitrary real $a$, $z = ai$", from this we cannot conclude $\forall a (z=ai)$ as that is false. In the case of induction once we have proven $P(k) \implies P(k+1)$ for arbitrary $k$ we can justify the $\forall$ quantifier, why can we not do that here? Is it correct to write $\forall a$(Define $z=ai$)? – Nav Bhatthal Aug 18 '23 at 09:12
  • A natural language sentence does not necessarily correspond to a complete proposition. It may well be that multiple sentences condense to one proposition. These discrepancies will always be there, because if there isn't, why make the effort in learning formal logic, since natural languages are already so close? – Trebor Aug 24 '23 at 04:59

3 Answers3

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"Let" statements modify the context for subsequent content by introducing symbols and assumptions that can be referenced later. They don't translate to closed formal sentences themselves; the introduced variables are free until some later step produces a closed statement using e.g. universal generalization. In your example this could be something like $\forall a\forall z(z=ai\implies zi=-a).$

Statements of the form "Let $v=\dots$" can also be regarded as introducing notational abbreviations instead of adding any formal content.

Karl
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  • I just want to write $z$ in terms of $a$, there is no further steps, i,e I wish to be able to write sentences like "In particular, for $a = 1$ we have $z = i$", here $z$ and $i$ are equal only in the sub-context for which $a$ is chosen to be $1$. – Nav Bhatthal Aug 18 '23 at 08:17
  • If I wrote "Given $a \in \mathbb{R}$ we define $z = ai$" then I cannot generalise and write $\forall a (z=ai)$ because "$z=ai$" is not some property that I am assessing whether an arbitrary real $a$ has or not, we know to prove $\forall a P(a)$ start with proving $P$ for arbitrary $a$, then generalising, but this isn't possible in my circumstance. – Nav Bhatthal Aug 18 '23 at 08:21
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    @Mathguy you're describing a definition. A definition can be expressed as a natural language sentence but not as a closed formal sentence (unless you use a function symbol $z(a)$ as Michal suggests). – Karl Aug 18 '23 at 14:45
  • How do we write the definition in natural language (English) then? – Nav Bhatthal Aug 18 '23 at 15:09
  • You mention closed, how do we write an open sentence for the definition? – Nav Bhatthal Aug 18 '23 at 15:25
  • The open sentence is just "$z=ai$". – Karl Aug 19 '23 at 00:34
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Since this is only a part of a mathematical text, I suggest to write $$a \in \mathbb R \land z=ai.$$ Since your text is not a statement, you cannot write it as a closed formula. When you prove the statement $a \in \mathbb R \land z=ai \Rightarrow \phi(a,z)$ for some formula $\phi(a,z)$, you can put quantifiers before: $\forall z \forall a (a \in \mathbb R \land z=ai \Rightarrow \phi(a,z))$; equivalently: $\forall z \, \forall a \in \mathbb R (z=ai \Rightarrow \phi(a,z))$.

I recommend you not to use formulae at places English is better understandable. From my suggestion, it is not clear that $a \in \mathbb R$ is a definition, so your sentence $(1)$ is perfectly fine. Definitions are usually written with colon, i.e., $a:=b+c$ or $$(a,b,c) \text{ is a pythagorean triple} :\Leftrightarrow a^2+b^2=c^2.$$

For your definitions, this would give $a :\in \mathbb R \land z:=ai$, but I never saw the symbol $:\in$, so it would be difficult to parse.

I think the reason why I did not see the symbol $:\in$ is that this definition is not unique. Picking an arbitrary element of a set usually leads to open formulae as I suggested.

Timotej
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Take the statement "Let $a \in \mathbb{R}$ and let $z=ai.$" $(1)$

Here, you are assigning a denotation ($z:=ai$) and a restriction ($a$ belongs to a strict subset of the discourse domain $\mathbb C$).

And in such statements, it is tacit that $a$ is arbitrary.

So, $(1)$ means

  • Let $a$ be an arbitrary real number and let $z$ denote $ai.$

In English, this means "Let's choose any real $a$ then define $z$ such that $z=ai$ for that particular $a$ we have just chosen".

Sure, but a key point of $(1)$'s intended meaning is that we are free to vary the value of $a.$

"arbitrary constant" to describe $a$

Yes: in the large, $a$ is variable ("arbitrary"), while within any iteration of what follows $(1),\,$ $a$ has a particular value ("constant").

How do we write $(1)$ in formal logic?

I know that $\forall a \exists z (z=ai)$ is one option, but is that correct/the only one?

$(1)$ is the beginning fragment of a proof, and isn't really an assertion so isn't really an FOL sentence. After Universal Generalisation has been invoked, you might then end up with, for example,

  • $\forall a{\in\mathbb R}\;\forall z{\in\mathbb C}\;(z=ai\implies iz=-a)$

or

  • $\forall a{\in\mathbb R}\;(ai^2=-a).$
ryang
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