Currently, I am self-studying Terence Tao's Analysis. I proves several of exercises but encountered difficulties with the others. Could you please check my proofs and suggest the solution for the remaining ones ? This is the proposition that we need to prove.
Proposition 2.2.12 (Basic properties of order for natural numbers).
Proposition 2.2.12 (Basic properties of order for natural numbers).
Let a, b, c be natural numbers. Then
- (a) (Order is reflexive) a ≥ a.
- (b) (Order is transitive) If a ≥ b and b ≥ c, then a ≥ c.
- (c) (Order is anti-symmetric) If a ≥ b and b ≥ a, then a = b.
- (d) (Addition preserves order ) a ≥ b if and only if a + c ≥ b + c.
- (e) a < b if and only if a++ ≤ b.
- (f ) a < b if and only if b = a + d for some positive number d.
Tao gives the following definition.
Definition 2.2.11 (Ordering of the natural numbers). Let n and m be
natural numbers. We say that n is greater than or equal to m, and
write n ≥ m or m ≤ n, iff we have n = m + a for some natural number a.
We say that n is strictly greater than m, and write n>m or m
I have the proofs for the first three statements. Could you please check it.
-
a) a $\ge$ a
Proof: if a$\ge$ a, then a = a + n for some natural n. This is true for n = 0. -
b) a $\ge$ b and b $\ge$ c, then a $\ge$ c
Proof: From the antecedent, a = b + n and b = c + m, for some natural n,m Then a =
c + (n + m). Hence, a $\ge$ c -
c) If a $\ge$ b and b $\ge$ a, then a = b
Proof: Antecedent implies
that a = b + n and b = a + m for some natural n, m. Hence a = a + m
n which is only true for n, m = 0.
I have doubts regarding the last proof since I have not used Peano axioms.
Best Answer
The last two proofs look good. The first is a little sloppy.
You can't start with “if $a \geq a$” if you want to show that $a \geq a$ is true. So instead say: “Since $a = a+0$ and $0$ is a natural number, $a \geq a$.“