On 14 Apr 2017, at 19:31, David Nyman wrote:
On 14 April 2017 at 17:59, Bruno Marchal <[email protected]> wrote:
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!
Possibly, but I think I've recovered from any lingering PMSD ;)
Good!
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!
Fortunately you had explained well that it was just a coding
system.
Yes. But notice that I have tried to avoid, this time, the term
"code", because it is more a translation, like in German or in a
natural language. It is partly conventional, but the structural
relations between the numbers will be isomorphic in different
languages, and is not conventional.
In fact in the past I have managed to disabuse myself of unnecessary
confusions of this sort by realising that something was simply a
notation or naming convention. Are irrational numbers, after all,
insane?
They are not insane, but the fact that 2xx = yy has no solutions
(except x = 0 = y) is not conventional, and does not depend on a
coding. That is important for having later real pain and pleasure for
the machine, and not conventional behavior only.
And why is x^0 always 1? (that way you can always add the exponents).
Yes, it is a good theory.
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!).
I see (and so do you, I trust). Good health!
Thanks. If only we knew the jaws has much well as the eyes ... The
human jaws articulation is less known than even the brain ....
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.
Yes, I suspected I was succumbing to the use/mention
distinction, which you warned me about. But alas I looked
too quickly and managed to miss where the symbol for 0 had been
mentioned, so I relied on the expectation that you would kindly
correct me. Thanks!
Fair enough :)
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 we were equally sloppy in this case! Mea culpa though.
The teachers can be as sloppy as the learners, but usually only the
learners get bad notes (grin).
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).
And fruit flies like a banana!
I continue to think on an intermediate level for the modal logics. I
have three forthcoming articles deadline, and that might help me too.
Your interest is helpful, many thanks.
Bruno
David
Bruno
David
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