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Excerpt from Lewis Carroll's book:

"Let me see:four times five is twelve,and four times six is thirteen,and four times seven is-oh dear! I shall never get to twenty at that rate!"

My questions:

What mathematical machinery might the writer have had in mind when writing down the sentences above? Could one reengineer Carroll's thoughts behind? Is there any explanatory hint/link/allusion in Carroll's work or somewhere else? I know that Carroll was a mathematician. So, there must be something out there...

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    $\begingroup$ It's not correct, and there is no modulus in which it is correct. "Adding four is adding one" is only true mod 3 and trivially mod 1. $\endgroup$ – Patrick Stevens Oct 1 '15 at 20:38
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    $\begingroup$ Well, if we interpret times as plus and work in base $7$. But that is probably not intended. $\endgroup$ – André Nicolas Oct 1 '15 at 20:44
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    $\begingroup$ What exactly are you asking? Do you mean "Is there an interpretation of mathematics in which Alice's statement makes sense and is true?" Because obviously four times five is not twelve. $\endgroup$ – mweiss Oct 1 '15 at 20:52
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    $\begingroup$ "I am emailing Lewis Carrol to ask him what he meant." Okay, now you're just trolling. $\endgroup$ – mweiss Oct 1 '15 at 21:02
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    $\begingroup$ @Jhon: your e-mail won't get to Prof. Dodgson if you persist in misspelling his nom de plume $\ddot{\smile}$ $\endgroup$ – Rob Arthan Oct 1 '15 at 21:28
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This blog has a suggestion

The idea is that she is doing calculations in base 10 but the answers are coming out in different bases ...

$4 \times 5 = 12 $ ( in base 18 )
$4 \times 6 = 13 $ ( in base 21 )
$4 \times 7 = 14 $ ( in base 24 )

So she is expressing $4n$ in base $3+3n$

indeed she can't get to 20 that way - if she could, $n$ would be a solution to $4n=6+6n$ which has solution $n=-3$

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In theory there are other possibilities...

Imagine that

$$x\circ y=5x-8^{6-y}7^{y-5}$$

and it reads as $x$ times $y$.

So,

$$4\circ y=20-8^{6-y}7^{y-5}.$$

In concreto $$4\circ 5=20-8\cdot7^{0}=12 \text{ and } 4\circ 6=20-8^{0}7^{1}=13$$

(and $4\circ 7=20-8^{-1}7^{2}=20-\frac{49}8...)$


PS: It can be shown easily that $x\circ y$ $(y=5,6,7,...)$ will never reach $4x$.

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    $\begingroup$ I would like to know a question that all philosophers agree is well posed, philosophically. $\endgroup$ – guest Oct 1 '15 at 22:01
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    $\begingroup$ Here is one: "Is there a question that all philosophers agree is well posed, philosophically?" $\endgroup$ – zoli Oct 1 '15 at 22:04
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We have $4\cdot 5=20,\ldots ,4\cdot 10=40$. Assuming that one "has to stop" there (because one does stop there in elementary school), it would correspond to $12,13,14,15,16$ and $17$ in the "wrong" counting. So we will never reach $20$. Of course, there are many other possibilities.

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