QuadAxiom=1[1].6.00

QuadAxiom=1[1].6.00 - 1 Quadratic Axioms Harvey M. Friedman...

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1 Quadratic Axioms Harvey M. Friedman 1/3/00 We axiomatize EFA in strictly mathematical terms, involving only the ring operations, without extending the language by either exponentiation, finite sets of integers, or polynomials. The two axioms beyond commutative rings with unit, are essentially the least element and least common multiple principles for certain quadratics in, respectively, two and four variables. Strictly speaking, one cannot point to a specific place in the literature where these statements are made, even though they are very elemental and implicit in the literature. So in this sense they are impeachable. In contrast, our axiomatizations that use finite sets of integers are almost entirely immune to this objection. We conjecture that an axiomatization can be given entirely within the ring axioms which consists of a finite set of sentences all of which are well known facts stated explicitly in the literature. Let PQV (positive quadratic values) be the system 1. Axioms for commutative ring with unit. 2. For every a,b,c,d,e,f, the values of axy + bx + cy + d, |x| <= e, |y| <= f, have a least positive upper bound. 3. For every a,b,c,d,e,f,g,h,i,j,k,m, the values of axy + bzw + cx + dy + ez + fw + g in [1,h], with |x| <= i, |y| <= j, |z| <= k, |w| <= m, have a least positive common multiple. In 2, we use the most common definition of least upper bound. In partciular, if there are positive values, then the least positive upper bound is the max. If there are no positive values, then the least positive upper bound is 1. In 3, we use the most common definition of common multiple. In particular, if there are no such values then the least positive common multiple is 1. LEMMA 1. The interval (0,1) is empty.
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2 Proof: Suppose 0 < x < 1. Let x be the least nonzero value of x in [0,1], for 0 < x < 1. I.e., x is least such that 0 < x. But then 0 < x^2 < x, which is a contradiction. QED LEMMA 2. Let a,b,c,d,e,f,g,h,i be given. Then axy + bx + cy + d assumes a least value >= e on the rectangle [f,g]dot[h,i], provided there is such a value. Proof: By translation, we can assume that the rectangle is of the form [-j,j+alpha]dot[-k,k+beta], where alpha,beta in {0,1}. But then we only have to handle the cases where the rectangle is [-j,j]dot[-k,k], {j+1}dot[-k,k], [-j,j]dot{k+1}. By translation, the last two cases can be redued to the first case. So we need only prove that axy + bx + cy + d assumes a least value >= e with |x| <= p and |y| <= q, assuming that it assumes some such value. But by axiom 2, axy + bx + cy + d - e + 1 assumes a least positive value with |x| <= p and |y| <= q, assuming that it assumes some such value. But positive values of axy + bx + cy + d - e + 1 correspond exactly to values of axy + bx + cy + d that are >= e. They are the same up to translation. QED LEMMA 3. For every a,b,c,d,e,f,g,h,i,j,k,m,n,p,q,r, the values of axy + bzw + cx + dy + ez + fw + g in [1,h], with i <= x <= j, k <= y <= m, n <= z <= p, q <= w <= r, have a least positive common multiple, assuming i-j, m-k, p-n, r-q
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QuadAxiom=1[1].6.00 - 1 Quadratic Axioms Harvey M. Friedman...

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