Prove that between any two elements of an ordered field, there areinfinitely man

Question

Prove that between any two elements of an ordered field, there areinfinitely many elements of that field by proving the following: a. Prove that if x and y are elements of an orderedfield F with x<y then there exists z in F so thatx<z<y b. Let x and y be elements of an ordered field F withx<y. Prove that for every n in the natural numbers, thereare elements z1, z2,z3...zn, all different, so that for all i =1,2,...n, x < zi < y a. Prove that if x and y are elements of an orderedfield F with x<y then there exists z in F so thatx<z<y b. Let x and y be elements of an ordered field F withx<y. Prove that for every n in the natural numbers, thereare elements z1, z2,z3...zn, all different, so that for all i =1,2,...n, x < zi < y

Explanation / Answer

First, we show 0 < 1. If not, then -1 is positive, hence -1* -1 = 1 is positive, and -1 and 1 cannot both be positive at thesame time. Hence, 0 < 1. In fact, all squares are positive. If awas any non zero element, and a > 0, then a*a > a*0 = 0. If awas negative, then -a is positive, hence a*a = (-a)*(-a) >(-a)*0 = 0 which shows a*a > 0 again.
Now define 2 as the element 1 + 1 and denote itsmultiplicative inverse by 1/2. We claim: 0 < 1/2 < 1.
Since 0 < 1, we have 0 + 1 < 1 + 1 which implies 1 <2 (***), hence 2 is positive as well. Now 1/2 is positive because(1/2)*(1/2) is a square, hence positive, hence 2*(1/2)*(1/2) =(1/2) is positive. Thus, 0 < 1/2. To see that 1/2 < 1,multiply (***) by 1/2.
Now for the general case, if x < y, then 0 = x - x < y -x, so y - x is positive and not 0 because y is different fromx. Now multiply 0 < 1/2 < 1 by (y-x). You get
0 < (1/2)*(y-x) < y-x. Now add x to everything giving x< (1/2)*(y-x) + x < y. To finish off show that (1/2)*(y-x) +x is not x and not y. You do that part. Just argue that if not,then you could undo the adding x and multiplying by (y-x) to get acontradiction.
b. x < (1/2)*(y-x) + x < y by part a. Now apply part ato (1/2)*(y-x) + x < y to get an element inbetween (1/2)*(y-x) +x and y. Repeat as long as your heart desires.

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