Maths Where Pythagorean Triples Don't Exist
Are there maths where the Pythagorean triples don’t exist?
I ask because of the following:
Let A,B,C, and K be elements of the Natural Numbers
1.∃K(K≤1)∧∃A,B,C(A^K+B^K=C^K )
2.∃A,B,C(A^K+B^K=C^K )∧∃K(K≤1)
3.∃A,B,C(A^K+B^K=C^K )→∃K(K≤1)
4.∀K(K>1)→∀A,B,C(A^K+B^K≠C^K )
Premise 1 is true because at K equal to 1 I can find natural numbers A,B, and C such that A+B=C. Premise 2 is true because of the commutative property. Premise 3 is true because of Premise 2 and Premise 1. Which would mean Premise 4 is true since it is the contrapositive of Premise 3. Yet this seems wrong though since the Pythagorean Triples exist. So it is either the case that the reasoning is fallacious, or at least one premise is wrong, or there are maths where the Pythagorean triples don’t exist.
9 comments
Your formalization is wrong. In fact, it's nonsensical. The predicate variable K should be bound by the existential quantifier associated with the proposition A^K+B^K=C^K. You cannot just separate K from the proposition with a conjunction operator and then shuffle them around, arguing from commutativity. In your "Premise 2", K is not bound at all in the first conjunction operand. You want K to mean the same thing in both operands, and with your formulas, that's not the case.
There are other problems here, but that's the most glaring one.
There are other problems here, but that's the most glaring one.
Yes. You’re right. I should have written it like this:
Let A,B,C, and K be elements of the Natural Numbers.
1. ∃K(K≤1∧∃A,B,C(A^K+B^K=C^K ))
2. ∃K(∃A,B,C(A^K+B^K=C^K )∧K≤1)
3. ∃K(∃A,B,C(A^K+B^K=C^K )→K≤1)
4. ∃K(K>1→∀A,B,C(A^K+B^K≠C^K ))
Premise 4 is trivially true for the same reason premise 3 is: choose K=1, then K>1 is false, and false implies anything.
Suppose I would pick K=2 though. What would happen then?
#3 and #4 say there exists a K for which the interior statement is true.
K=2 doesn't make the interior true for either, so choosing it doesn't tell us whether #3 and #4 are true or false.
However, if we choose K=1 then we prove #3 and #4 are true.
K=2 doesn't make the interior true for either, so choosing it doesn't tell us whether #3 and #4 are true or false.
However, if we choose K=1 then we prove #3 and #4 are true.
Pythagorean triples exist for natural numbers, but other exotic number sets lack them.
For example, integers modulo 3: the square of both 1 and 2 is 1, so the sum of two nonzero squares is always 1 + 1 = 2, which is not a square.
Looking for infinite number sets without Pythagorean triples should be fun.
For example, integers modulo 3: the square of both 1 and 2 is 1, so the sum of two nonzero squares is always 1 + 1 = 2, which is not a square.
Looking for infinite number sets without Pythagorean triples should be fun.
On the positive side, in integers modulo 3, we have
n³ = n
1³ + 2³ = 0³Pythagorean triples must be three nonzero numbers, and the exponent must be 2.
If you are interested in other equations, don't call them the same way; and more importantly, while generalizing to other exponents and sums of more than two numbers is reasonable, allowing 0 makes the problem trivial because all powers of 0 are 0 and the sum of any number of zeros is 0.
If you are interested in other equations, don't call them the same way; and more importantly, while generalizing to other exponents and sums of more than two numbers is reasonable, allowing 0 makes the problem trivial because all powers of 0 are 0 and the sum of any number of zeros is 0.
> If you are interested in other equations, don't call them the same way;
I didn’t do that. I implicitly alluded to Fermat’s last theorem, expecting readers to pick that up.
> because allowing 0 makes the problem trivial because all powers of 0 are 0
The example I gave isn’t trivial. The trivial ones are
I didn’t do that. I implicitly alluded to Fermat’s last theorem, expecting readers to pick that up.
> because allowing 0 makes the problem trivial because all powers of 0 are 0
The example I gave isn’t trivial. The trivial ones are
x³ + 0³ = x³