If Six Was Nine
"If Six Was Nine" is the name of a Jimi Hendrix song from Axis: Bold as Love (1967). How could six have been nine? Hendrix's title plays on the symmetry of the Arabic numerals, "$6$" and "$9$": each is obtained by rotation of the other through 180 degrees. But the possibility of converting a representation $r$ to a representation $r^{\circ}$ doesn't automatically correspond to some important relation between what they refer to. That's a use/mention confusion.
Following Frege and Russell, (finite) cardinal numbers are the cardinalities of (finite) sets, and cardinalities are obtained by abstraction over the equivalence relation (on sets) of equinumerousness: i.e., there is a bijection $f : A \rightarrow B$. Writing $A \sim B$ to mean this, the guiding axiom is Hume's Principle: $card(A) = card(B) \leftrightarrow A \sim B$. So, for example, $0$ is defined as $card(\emptyset)$.
Suppose that $A = \{a_1, a_2, a_3, a_4, a_5, a_6\}$, with $a_i \neq a_j$ for $i \neq j$, and $B = \{b_1, b_2, b_3, b_4, b_5, b_6, b_7, b_8, b_9\}$, with $b_i \neq b_j$ for $i \neq j$. So $card(A) = 6$ and $card(B) = 9$. But there is no injection $f : B \rightarrow A$. So, $card(A) \neq card(B)$ and therefore six isn't nine.
But could six have been nine, even though it actually isn't? I don't think so, because pure mathematical objects are modally invariant. Unlike "concreta", they don't change their properties from world to world. Concreta have "counterparts". Though Quine-in-the-actual-world $w^{\ast}$ was a logician, for some other world $w$, Quine-in$w$ was not a logician. Quine-in-$w^{\ast}$ and Quine-in-$w$ are mutual counterparts. But abstract mathematical entities like six and nine are just what they are, and couldn't have been different. Concrete worlds are like planets embedded in a fixed background mathematical universe: mathematics is the spacetime of modality.
There are ways, however, of making the linguistic representation "$6 = 9$" true, if we change the interpretation of the symbols. Suppose we have the ring $\mathbb{Z}_3$ of integers modulo three. The ring $\mathbb{Z}_n$ of integers modulo $n$ involves treating integers that differ by adding a multiple of $n$ as equivalent. We write: $p \equiv k \text{ (mod } n)$ as short for $\exists a(p = k + a \times n)$. So, for example, $1\equiv 4 \text{ (mod } 3)$. Then, if we define terms of the language so that, roughly the term "$+n$" is "$0 + 1 + 1 ... + 1$'', with $n$ occurrences of "$+1$" (and similarly for $-n$), then what corresponds to the terms "$6$" and "$9$" both refer to $0$ in $\mathbb{Z}_3$. In that sense, "$6 = 9$" is true in the structure $\mathbb{Z}_3$.
Still, the truth of "$6 = 9$" in $\mathbb{Z}_3$ isn't what is meant by wondering whether 6 could have been 9 (or 6 might be 9, even though we don't know). That question concerns whether the finite cardinal numbers 6 and 9 could have been identical, and the answer to that is no.
Unfortunately, there isn't a Youtube video of Jimi Hendrix's "If Six Was Nine", but there is an Eddie van Halen version,
Following Frege and Russell, (finite) cardinal numbers are the cardinalities of (finite) sets, and cardinalities are obtained by abstraction over the equivalence relation (on sets) of equinumerousness: i.e., there is a bijection $f : A \rightarrow B$. Writing $A \sim B$ to mean this, the guiding axiom is Hume's Principle: $card(A) = card(B) \leftrightarrow A \sim B$. So, for example, $0$ is defined as $card(\emptyset)$.
Suppose that $A = \{a_1, a_2, a_3, a_4, a_5, a_6\}$, with $a_i \neq a_j$ for $i \neq j$, and $B = \{b_1, b_2, b_3, b_4, b_5, b_6, b_7, b_8, b_9\}$, with $b_i \neq b_j$ for $i \neq j$. So $card(A) = 6$ and $card(B) = 9$. But there is no injection $f : B \rightarrow A$. So, $card(A) \neq card(B)$ and therefore six isn't nine.
But could six have been nine, even though it actually isn't? I don't think so, because pure mathematical objects are modally invariant. Unlike "concreta", they don't change their properties from world to world. Concreta have "counterparts". Though Quine-in-the-actual-world $w^{\ast}$ was a logician, for some other world $w$, Quine-in$w$ was not a logician. Quine-in-$w^{\ast}$ and Quine-in-$w$ are mutual counterparts. But abstract mathematical entities like six and nine are just what they are, and couldn't have been different. Concrete worlds are like planets embedded in a fixed background mathematical universe: mathematics is the spacetime of modality.
There are ways, however, of making the linguistic representation "$6 = 9$" true, if we change the interpretation of the symbols. Suppose we have the ring $\mathbb{Z}_3$ of integers modulo three. The ring $\mathbb{Z}_n$ of integers modulo $n$ involves treating integers that differ by adding a multiple of $n$ as equivalent. We write: $p \equiv k \text{ (mod } n)$ as short for $\exists a(p = k + a \times n)$. So, for example, $1\equiv 4 \text{ (mod } 3)$. Then, if we define terms of the language so that, roughly the term "$+n$" is "$0 + 1 + 1 ... + 1$'', with $n$ occurrences of "$+1$" (and similarly for $-n$), then what corresponds to the terms "$6$" and "$9$" both refer to $0$ in $\mathbb{Z}_3$. In that sense, "$6 = 9$" is true in the structure $\mathbb{Z}_3$.
Still, the truth of "$6 = 9$" in $\mathbb{Z}_3$ isn't what is meant by wondering whether 6 could have been 9 (or 6 might be 9, even though we don't know). That question concerns whether the finite cardinal numbers 6 and 9 could have been identical, and the answer to that is no.
Unfortunately, there isn't a Youtube video of Jimi Hendrix's "If Six Was Nine", but there is an Eddie van Halen version,
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