Triangle Inequality in Number Theory

Triangle Inequality in Number Theory

Triangle Inequality in Number Theory

For any real numbers a and b, the Triangle Inequality states:

|a + b| ≤ |a| + |b|


Proof

We will consider different cases based on the signs of a and b:

Case 1: a ≥ 0 and b ≥ 0

|a + b| = a + b = |a| + |b|

Case 2: a ≥ 0 and b < 0 (or vice versa)

If a + b ≥ 0, then:

|a + b| = a + b ≤ a - b = |a| + |b|

If a + b < 0, then:

|a + b| = -(a + b) = -a - b = |a| + |b|

Case 3: a < 0 and b < 0

|a + b| = -(a + b) = -a - b = |a| + |b|

Algebraic Proof

Using the definition of absolute values:

|a + b|2 ≤ (|a| + |b|)2

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

(|a| + |b|)2 = |a|2 + 2|a||b| + |b|2

Since |a|2 = a2 and |b|2 = b2:

a2 + 2ab + b2 ≤ a2 + 2|a||b| + b2

Given |a||b| ≥ ab, we conclude that:

a2 + 2ab + b2 ≤ a2 + 2|a||b| + b2

Taking square roots on both sides:

|a + b| ≤ |a| + |b|

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