Theorem:    For any finite x there is a unique r Î R such that x » r.

Proof:        First, show existence:

                 For x Î F take S = {y Î R: y £* x}

                 This has an upper bound in R.

                 If r is the least of these then x » r.

                 Suppose ~(x » r); then $qÎ R x <* q £* r

                 So q is an upper bound for S and r can't be the least one.

                 Second, show uniqueness:

                 If (x » r) and (x » s) then (r » s).

                 For standard r, s, this means r = s.

Definition

From the previous theorem we find that each x Î F has a decomposition into a standard and a non-standard part.

            x = s + i for s Î R, i Î I

Call s the standard part of x, st(x) 

Some results for st follow quite easily from previous demonstrations:

1.    st: F Þ R

2.    x Î I Þ st(x) = 0

3.    st(x +* y) = st(x) + st(y)

4.    st(x .* y) = st(x) . st(y)