Higher Mathematics 2nd Paper · Classes 11-12 · Chapter 1 · 1.2

Axioms of Real Numbers.

Nobody proves that 2 + 3 = 3 + 2: it is taken as given. A handful of such givens, the axioms, hold up every calculation with real numbers. This note covers those axioms, which sets obey which of them, and a few short proofs built only from them.

1. Two kinds of axiom

The axioms of the real numbers come in two kinds. One kind rests on algebraic properties and is called the field axioms; these depend on addition (+) and multiplication (.). The other kind are the order axioms, which say which number is bigger than which.

2. The field axioms

Closure: if a, b ∈ ℝ then a + b ∈ ℝ and a.b ∈ ℝ. Adding or multiplying never takes you out of the set. Commutative: a + b = b + a and ab = ba; swap the order and nothing changes.

Associative: (a + b) + c = a + (b + c) and (ab)c = a(bc); move the brackets and nothing changes. Identity: a + 0 = 0 + a = a, so 0 is the additive identity; a.1 = 1.a = a, so 1 is the multiplicative identity.

Inverse: every a has −a ∈ ℝ with a + (−a) = (−a) + a = 0, and every non-zero a has a⁻¹ = 1/a ∈ ℝ with a.a⁻¹ = a⁻¹.a = 1. Distributive: a(b + c) = ab + ac (left) and (b + c)a = ba + ca (right); for example 2(3 + 4) = 2.3 + 2.4 = 14.

Uniqueness: if a = b and c = d then a + c = b + d and ac = bd. Add or multiply equals by equals and the results are equal. The proofs below lean on this again and again.

Boards often show one example and ask which law it is. Send the eight equations below to their boxes.

3. How far ℕ, ℤ and Q obey them

ℤ is closed under addition and multiplication, obeys the associative, commutative, distributive and uniqueness laws, and has both identities. It has additive inverses too, −a ∈ ℤ. But apart from −1 and 1, no member of ℤ has a multiplicative inverse inside ℤ: the inverse of 2 is 1/2 ∉ ℤ.

All rational numbers obey the axioms of the real numbers. Positive rationals (Q₊) and positive reals have multiplicative inverses; ℕ does not. In the book's words “ℕ and ℤ have no multiplicative inverse”, which is true once ±1 is set aside. And ℕ is closed under addition and multiplication but not subtraction: 2 − 3 = −1 ∉ ℕ. ℤ is closed under all three: addition, subtraction and multiplication. One counter-example is enough; go and find one below.

4. The order axioms

O₁: for any a, b ∈ ℝ exactly one of a > b, a = b, a < b is true. O₂: if a > b and b > c then a > c; if a < b and b < c then a < c.

O₃: if a > b then a + c > b + c; adding the same number to both sides keeps the direction. O₄: if a > b and c > d then a + c > b + d.

O₅: if a > b and c > 0 then ac > bc; but if c < 0 then ac < bc, and the direction flips. O₆: if a > 0 and b > 0 then ab > 0. O₇: for every a ∈ ℝ, a² ≥ 0. The last one is short but does a lot of work.

5. From a square ≥ 0 to an extreme value

Write the quadratic as (square) + constant. O₇ says the square part is never negative, so with a positive x² coefficient the smallest value comes when the square is 0, leaving only the constant. For example 4x² − 2x + 1 = 4(x − 1/4)² + 3/4, so the minimum is 3/4; 2x² + 3x + 1 = 2(x + 3/4)² − 1/8, minimum −1/8. With a negative coefficient it turns round: 1 + 3x − 2x² = 17/8 − 2(x − 3/4)², maximum 17/8. In the same way 4x² + 4ax + b² = (2x + a)² + b² − a², with minimum b² − a².

6. Proofs from the axioms

The identity is unique. Suppose 0 and 0′ are both additive identities. Put x = 0 in x + 0′ = x to get 0 + 0′ = 0, and x = 0′ in x + 0 = x to get 0′ + 0 = 0′. By commutativity 0 + 0′ = 0′ + 0, so 0′ = 0. Multiplication works the same way: if e and e′ are both multiplicative identities then e′e = e′ and ee′ = e; commutativity gives e′e = ee′, so e′ = e.

The inverse is unique. If y and z are both additive inverses of x, then z = z + 0 = z + (x + y) = (z + x) + y = 0 + y = y. For multiplication: if y and z are both inverses of a non-zero x, then z = z.1 = z(xy) = (zx)y = 1.y = y. Each equals sign rests on one law; put the additive proof in order yourself below.

a.0 = 0. Multiply both sides of 1 + 0 = 1 by a: a.1 + a.0 = a.1, that is a + a.0 = a. Add −a to both sides: (−a + a) + a.0 = −a + a, so 0 + a.0 = 0, hence a.0 = 0.

−(a + b) = −a − b. Simplify (a + b) + (−a − b) with the commutative, associative and inverse laws and you get 0. So −a − b is the additive inverse of a + b, that is −(a + b) = −a − b.

Cancellation. If a + b = a + c, add (−a) to both sides (uniqueness): (−a) + (a + b) = (−a) + (a + c). Associativity gives ((−a) + a) + b = ((−a) + a) + c, the inverse law 0 + b = 0 + c, the identity law b = c. Multiplication goes the same way: if ac = bc and c ≠ 0, multiply both sides by c⁻¹, (ac)c⁻¹ = (bc)c⁻¹; associativity gives a(cc⁻¹) = b(cc⁻¹), the inverse law a.1 = b.1, the identity law a = b.

If ab = 0 then a = 0 or b = 0. If a ≠ 0, multiply by a⁻¹: a⁻¹(ab) = a⁻¹.0 = 0 (because a.0 = 0). Associativity gives (a⁻¹a)b = 0, that is 1.b = 0, so b = 0. Last, the sign rules: ab + (−a)b = (a + (−a))b = 0.b = 0, so (−a)b = −(ab); likewise (−a)(−b) = −((−a)b) = −(−(ab)) = ab.

7. Check yourself

Three questions in the exam's style. The first is about the identity.

In the second, look for inverses inside ℤ.

In the last, see what moved: the order or the brackets.

The whole note on eight cards.

Source: Higher Mathematics 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Real Numbers and Inequalities”, section 1.2.