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But if anyone desires a stricter demonstration, they will be able to proceed in this manner.
Given (for the sake of distinction) z=x absolute.
Let (g) be taken as any quantity greater than (a). I say that the next (g) (produced by our method) is always less than the preceding, but greater than (a), and accordingly converges to the true root.
It will be g-z=a by hypothesis, and ggg-3ggz+3gzz-zzz=aaa=b.
From there, -3ggz+3gzz-zzz=b-ggg. Therefore -z + 3gzz-zzz/3gg = (b-ggg/3gg = -x)
Or the convergent theorem. Therefore -z + 3gzz-zzz/3gg = -x. Let (g) be added to both sides.
There will be g-z + 3gzz-zzz/3gg = g-x = new g; and accordingly the new (g) is less than the preceding, but also a + 3gzz-zzz/3gg = new (g); therefore the new (g) is greater than (a), that is, the whole by its part. Q.E.D.
Let (g) be taken as any quantity less than (a). I say that the next (g) (produced by our method) is always greater than the preceding, but less than (a), and accordingly converges to the true root.
By hypothesis g+z=a. There will be bg-ggg+b-3gg * z-3gzz-zzz=ba-aaa=c.
Therefore b-3gg * z-3gzz-zzz=c+ggg-bg. Therefore +z - 3gzz+zzz/b-3gg = (c+ggg-bg/b-3gg = +x)
Or the convergent theorem, then +z = +x + 3gzz+zzz/b-3gg. Let (g) be added to both sides.
It will result in g+z=x = g+x + 3gzz+zzz/b-3gg. But the new (g) = g+x, which is greater than the preceding by the quantity (x), but less than (a) by the quantity 3gzz+zzz/b-3gg, the part by its whole. Q.E.D.
From these demonstrations, various corollaries can be elicited, such as:
I. In equations where the sum power is denied, if (g) is less than (a), it generally ascends to the root.
II. In extracting pure roots, if (g) is less than (a), it always (in the first operation) ascends beyond the root; then, by descending, it converges.
Many other things can be deduced that are also most conducive to use; not a few things can also be found in other matters from this method, which, however, I leave to others, believing that I have shown the way, if I am not mistaken, sufficiently wide.