GrahamH wrote:
Can you give an example of a "non-contigent truth" and show it to be true?
I gave the example before of a non-contingent truth (although I didnt call it such) earlier, when I said emperical demonstration was impossible for 'an odd plus an even always gives an odd'
We had some fun with that; recall I suggested it cant be shown emperically but can be proved for real numbers as follows
simple proof for real numbers
Let n be any real whole number.
Then 2n is always even because 2n/2 = n which is a whole number. The definition of ‘even’ is met.
And 2n+1 is always odd because (2n+1)/2 = 2n/2 + ½ = n + 1/2 . The definition of ‘odd’ is met.
Therefore any odd plus any even is given by
(2n) + (2n + 1)
But since (2n + 2n + 1)/2 = (4n +1)/2 = 2n + ½ the definition of odd is met.
Therefore proving that for real numbers any odd plus any even will always be odd.
Via PM a member (Newolder) tried to suggest it was possible to dispute me using complex numbers, we disagreed over content of his PM, I thought his 'disproof' actually proved my point, he did not. So I asked Newolder's permission and presented the disproof/proof for complex numbers to a mathematician friend of mine for him to check the validity, and I asked for him to suggest a simple proof of a non-contingent truth.
Here is his (entire) response; My explainations {added}
Mehdi, mathematician wrote:
Hi Steve:
First of all, given that I force myself to swallow the concept of odd&even to other than integers, it would then be defined as:
Z=a+ib is even(odd) if b=0 and a is even(odd).
Which puts back the problem in the integer world and there is no point discussing it as a complex numbers case.
As a non-contingent truth to be proved without empirical observation, I would suggest: the irrationality of √2
Proof:
If we suppose that it is a rational number, then:
√2 =a/b a fraction {which has been} fully simplified …(*) {i.e. there are no common factors to a and b, they cant both be even, or the fraction can be further simplified}
Than 2= a2/b2
Which gives: a2=2b2 ie: a2 {is} even implying a is even
On the other hand, if a is even that means that a can be written as:
a=2k (k being an integer)
substituting it back into the equation: 2= a2/b2
we get: 2= 4k2/b2
which leads to: b2 = 2 k2
so b2 is even and so is b
a and b being both even numbers is a statement that contradict our assumption in (*)
Therefore we cannot express √2 as a rational number (fraction), hence its irrationality.
Note: the irrationality of a number being also equivalent to the finite or recurrence of its decimals, one can try to find the recurrence of √2 and pi’s decimals for ever without ever observing any pattern and the proof above is the only way to attest it.
Hope this answers your question.
Cheers,
Mehdi
As can be seen with a google search for proof of the irrationality of √2, this is more or less the classical proof. Here is a
link to the university of Utah offering a similar proof. Here is the definition of numbers
from the same universtity site
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