1.3 Natural Numbers 23Then:
(1) p(l) is true, since ~m E NF 3 m < l.
(2) Suppose p(k) is true. That is,m < k::::} k - m E NF.
To prove p(k + 1), suppose m < k + l. Then m - 1 < k, so by p(k),
k - (m - 1) E NF.That is, (k + 1) - m E NF. But that means
m < (k + 1)::::} (k + 1) - m E NF.
That is, p(k + 1) is true. We have thus proved that p(k) ::::} p(k + 1).
Therefore, by the principle of mathematical induction, \:/n E NF, p(n) is
true.
(b) Let n E NF. Suppose, for contradiction, that 3m E NF 3 n < m <
n + l. Subtracting n from all sides of these inequalities, we then have
O<m-n<l.
By (a), m-n E NF. But, by Theorem 1.3.5 (b), there is no natural number less
than l. Thus, we have a contradiction. Therefore, ~ m E NF 3 n < m < n + l.Theorem 1.3.8 In any ordered field,(a) NF is closed under addition.(b) NF is closed under multiplication.(d) NF is not closed under subtraction or division.*Proof. (a) Let n E NF be fixed. \:/m E NF we let p(m) denote the
proposition p(m): n + m E NF. Then:
(1) p(l) is the statement n+l E NF, which is true since NF is an inductive
set.
(2) Suppose p(k) is true. Then n + k E NF. Since NF is an inductive set,
(n + k) + 1 E NF; i.e.,
n + (k + 1) E NF.
That is, p(k + 1) is true. We have thus proved that p(k) ::::} p(k + 1).
Therefore, by the principle of mathematical induction, \:/m E NF, p(m) is
true. That is ,
\:/m E NF, n + m E NF.Therefore, NF is closed under addition.