On 13 Apr 2017, at 19:26, David Nyman wrote:
On 12 April 2017 at 20:59, Bruno Marchal <[email protected]> wrote:
2 ^(2^9) * (3^2) * (5^17) * (7^2) * (11^21) *(13^2) * (17*17) *
3 ^ <exercise! :) >
Oh, I see what you mean:
3 ^(2^17) * (3^0) * (5^21) * (7^0) * (11^17)
Now, you see it was easy, and, like me, you go faaaar ... too much
quickly.
Even without quoting the dictionary we can suspect some mistake. And I
hope I will not put my finger where it might hurt in case you have
been mentally abused by a sadistic mathematical teacher!
That one would ask you what is the value of (3^0) ? and what is the
value of (7^0)?
And you would panic, thinking like ... hmm... er... 3^2 = 3 * 3, 3^1 =
3, 3^0 = ...... !!! ..... gosh that one is tricky!
Well, (a^b) * (a ^ c) = a ^(b + c), that is obvious for a, b, c
natural numbers, and if you want keep this true for all integers, you
will have that 3^0 = 3^(5 - 5) = 3^(5 +(-5)) = (3^5) *(3^(-5)) =
(again to keep that law on the integers) = (3^5)/(3^5) = 1.
same reasoning for (7^0) = 1.
But 1 is not a prime number, and if we allow it in our coding, it
would become ambiguous. In fact 1 has been thrown out of the prime to
get a simple enunciation of the fundamental theorem of arithmetic:
there is only one decomposition into prime factors. if 1 was a prime
number, you can add it, as a factor where you want!
That is why, normally, 0 has been represented by non null number, as
any number ^ 0 = 1.
I say normally, I will now search the dictionary logic-arithmetic
(seen as languages), and I pray I did not make a typo!
I said:
So let us denote the logical and arithmetical symbols v, &, ->, ~,
E, A, t, f, "(", ")" =, 0, s, +, *
by the first odd numbers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23,
25, 27, 29,
So, I did not (at least here). 0 has been represented by 23. (If my
glasses does not fail me). Let us verify * is 29, + is 27, s is 25, 0
is 23.
(A chance my recent cataract surgery has succeed! I have an implant!).
So you should have written:
3 ^(2^17) * (3^23) * (5^21) * (7^23) * (11^17)
(I guess my talk about was not necessary, .. you did just forget to
represent 0, with a notion known by the machine.
PS Are "(" and ")" meant to be the same?
Ah! Now, that is *my* error, which you should not have copied, given
that you detect it. ts, ts, ts ....
So the correct answer is
3 ^(2^17) * (3^23) * (5^21) * (7^23) * (11^19)
And so the proof
Ax(x=x)
(0 = 0)
is translated in arithmetic-language in the number
2 ^[(2^9) * (3^2) * (5^17) * (7^2) * (11^21) *(13^2) * (17*17)]
* 3 ^ [(2^17) * (3^23) * (5^21) * (7^23) * (11^19)]
The only important things is to see that this is just one number,
written s(s(s(s(s(s(s(s ... s(0)))))...), and quite huge (!). Once the
machine believes in addition and multiplication, it will denote the
proof, and the machine will be able to answer question like "does a
variable occur in that proof?", etc.
Very good David, welcome to logic! (and sorry for being long, and
probably too short in my next answer to the other posts as ... time
flies like an arrow).
Bruno
David
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