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