On 2/28/2014 4:07 AM, John Gabriel wrote: > On Friday, 28 February 2014 07:17:41 UTC+2, Martin Shobe wrote: > >>>> I agree that 6 - 4 = 2 wouldn't show that 2 + 3 = 6. > >> Yes 2 + 3 = 6 is false. That's why the fact that it fits your axioms is >> such a problem. > > You do not understand the first axiom.
The axiom under question here is the third, not the first.
> The difference (or subtraction) of two numbers (2 and 3) is that number (6) which describes how much the larger exceeds the smaller.
> So why can't you have 2 + 3 = 6?
> Because, 6 - 4 = 2, but we KNOW that the numbers are 2 and 3, not 2 and 4.
> Can you have 6 - 3 = 3?
> No. Because you KNOW that the numbers are 2 and 3, not 3 and 3.
But you didn't say that. You said that I could use either number as the subtrahend and have the difference be either number. For the 2 + 3 = 6 case, 6 - 3 = 3 fits that. If that isn't what you mean, you should reword your axioms so that it does say what you mean.
> Moreover, you cannot say 1 - 3 or 0 - 3 because the smaller is subtracted ('-') from the larger.
> The first axiom establishes the primitive operator '-' from which all the remaining operations are derived.
At least you got your axiom reference right here.
>> Your axiom was, "The difference (or subtraction) of two numbers is that >> number which describes how much the larger exceeds the smaller." Three >> is most definitely two larger than one, so 1-3 = 2. Similarly, three is >> three larger than 0, so 0 - 3 = 3.
> You don't do this even in the flawed Peano axioms. What makes you think you can do it in mine? The axiom states clearly the smaller must be subtracted from the larger: "Difference" is defined as "how much the larger exceeds the smaller"
Yes, Peano's axioms don't have this flaw. But what makes me think I can do it in yours, is that your axioms say I can. It's a rather trivial derivation from what you said to 0 - 3 = 3. If you didn't mean what you said, you should change it to say what you mean.