The Real Numbers

Definition 1.1.1. The absolute value of a real number a is the non-negative number defined as follows

| a | = {\begin{cases} a, & if a \geq 0 \\ - a, & if a < 0. \end{cases}

From the definition of absolute value it follows immediately that

∀a ≥ 0: |x| ≤ a ⇐⇒ −a ≤ x ≤ a 1.1.1

From 1.1.1 it the Triangle inequality follows

Proposition 1.1.2 (Triangle inequality) For any real numbers a,b

|a + b| ≤ |a| + |b| 1.1.2

Proof. It sufficies to write

−|x| ≤ x ≤ |x| and −|y| ≤ y ≤ |y|

adding up both inequalities

−(|x| + |y|) ≤ x + y ≤ |x| + |y|

from eq 1.1.1 the triangle inequality 1.1.2 follows. □

Proof2.

|a + b|² = (a + b)² = a² + b² + 2ab + b² ≤ |a|² + 2|a||b| + |b|² = (|a| + |b|)²

extracting the square root we have the claim. □

Proposition 1.1.3 (Triangle inequality second form)For any real numbers a,b

|a| − |b| ≤ |a − b|

Proof. If in 1.1.2, we put x = a − b and y = b, we have

|a| ≤ |a − b| + |b| ⟶ |a| − |b| ≤ |a − b| □

The expression |a − b| represents the geometric distance between the two points a and b on a line.
Geometrically, the meaning of both triangle inequalities is that the lenght of any side of a triangle is no greater than the sum of the lenghts of the two other sides and no less than the absulote value of the difference of the lenghts of the two other sides. Since the shortest distance between two points is a straight line, any side of any triangle is always shorter thant the sum of the lenghts of the other two sides.

triangle inequality — The shortest distance from going from A to B is to move along z. The top example shows a case where z is much less than the sum x + y of the other two sides, and the bottom example shows a case where the side z is only slightly less than x + y.

Proposition 1.1.4 ℕ is ubbounded in ℝ.

Proof. If ℕ were bounded, then by the completness axiom ℕ would have a least upper bound, say sup ℕ = m; then m − 1 is not an upper bound of ℕ. Thus, there exists a positive integer n such that n > m − 1, that is, n + 1 > m, and n + 1 ∈ ℕ (n + 1 being a successor of n). Thus m is not an upper bound of ℕ, which is a contradiction. □

Proposition 1.1.5 (Archimedean Property) Let x and y be positive real numbers. Then there is a natural number n such that

nx > y.

Equivalently, given y > 0, there is an n ∈ ℕ such that ny > 1.

Proof. If the first conclusion were false, then every natural numnber would be bounded above by y/x. If the second conclusion were false, then evey natural number would be bounded above by 1/y. □

The Archimedean Property also implies that there are no infinitely small positive real numbers.

Corollary 1.1.6 (Density of ℚ in ℝ) If x and y are real numbers with x < y, then there exists a rational number q ∈ ℚ, such that x < q < y.

Proof. Since x < y, we have y − x > 0, by the Archimedean Property there is a positive integer n such that

n (y − x) > 1

Applying the the Archimedean Property again, to obtain positive integers m₁ such that m₁ > nx and m₂ > −n ⋅ x. Thus we see that nx falls between the integers m₁ and −m₂;

−m₂ ≤ nx < m₁

therefore there must be an integer m (with −m₂ ≤ m ≤ −m₁) such that

m − 1 ≤ nx < m

If we combine these inequalities, we obtain

nx < m ≤ 1 + nx < ny

Since n > 0, it follows that

x < m/n < y

this proves the claim, with q = m/n. □

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