Triangle Inequality/Real Numbers

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Theorem

Let $x, y \in \R$ be real numbers.

Let $\size x$ denote the absolute value of $x$.


Then:

$\size {x + y} \le \size x + \size y$


General Result

Let $x_1, x_2, \dotsc, x_n \in \R$ be real numbers.

Let $\size x$ denote the absolute value of $x$.


Then:

$\ds \size {\sum_{i \mathop = 1}^n x_i} \le \sum_{i \mathop = 1}^n \size {x_i}$


Proof 1

\(\ds \size {x + y}^2\) \(=\) \(\ds \paren {x + y}^2\) Square of Real Number is Non-Negative
\(\ds \) \(=\) \(\ds x^2 + 2 x y + y^2\) Square of Sum
\(\ds \) \(=\) \(\ds \size x^2 + 2 x y + \size y^2\) Square of Real Number is Non-Negative
\(\ds \) \(\le\) \(\ds \size x^2 + 2 \size {x y} + \size y^2\) Negative of Absolute Value
\(\ds \) \(=\) \(\ds \size x^2 + 2 \size x \cdot \size y + \size y^2\) Absolute Value Function is Completely Multiplicative
\(\ds \) \(=\) \(\ds \paren {\size x + \size y}^2\) Square of Sum


Then by Order is Preserved on Positive Reals by Squaring:

$\size {x + y} \le \size x + \size y$

$\blacksquare$


Proof 2

This can be seen to be a special case of Minkowski's Inequality for Sums, with $n = 1$.

$\blacksquare$


Proof 3

We have that Real Numbers form Ordered Integral Domain.

Therefore Sum of Absolute Values on Ordered Integral Domain applies directly.

$\blacksquare$


Proof 4

We do a case analysis.


$(1): \quad x \ge 0, y \ge 0$

\(\ds x\) \(\ge\) \(\ds 0\)
\(\ds y\) \(\ge\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \size {x + y}\) \(=\) \(\ds x + y\)
\(\ds \) \(=\) \(\ds \size x + \size y\)

$\Box$


$(2): \quad x \le 0, y \le 0$

\(\ds x\) \(\le\) \(\ds 0\)
\(\ds y\) \(\le\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \size {x + y}\) \(=\) \(\ds -x - y\)
\(\ds \) \(=\) \(\ds \size x + \size y\)

$\Box$


$(3): \quad x \ge 0, y \le 0$

We have that $\size x = x$ and $\size y = -y$.

In this case we show:

$\size {x + y} \le \max \set {\size x, \size y}$


Let $\size x \le \size y$.

Then:

\(\ds x\) \(\le\) \(\ds -y\)
\(\ds \leadsto \ \ \) \(\ds y\) \(\le\) \(\ds y + x\) as $x \ge 0$
\(\ds \) \(\le\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \size {x + y}\) \(=\) \(\ds -\paren {x + y}\)
\(\ds \) \(\le\) \(\ds -y\)
\(\ds \) \(=\) \(\ds \size y\)


Let $\size x \ge \size y$.

Then:

\(\ds x\) \(\ge\) \(\ds -y\)
\(\ds \leadsto \ \ \) \(\ds x\) \(\ge\) \(\ds x + y\) as $y \le 0$
\(\ds \) \(\ge\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \size {x + y} =\) \(=\) \(\ds x + y\)
\(\ds \) \(\le\) \(\ds x\)
\(\ds \) \(=\) \(\ds \size x\)


We have $\max \set {a, b} \le a + b$ for positive real numbers $a$ and $b$.

The result follows by taking $a = \size x$ and $b = \size y$.

$\Box$


$(4): \quad x \le 0, y \ge 0$

Follows by symmetry from the case $(3)$.

$\blacksquare$


Proof 5

From Negative of Absolute Value, it is sufficient to prove that:

$\size x + \size y \ge x + y$

and:

$\size x + \size y \ge -\paren {x + y}$


By definition of absolute value:

$x \le \size x$

and:

$y \le \size y$

Then:

$x + y \le \size x + \size y$


We also have that:

\(\ds -x\) \(\le\) \(\ds \size {-x}\)
\(\ds \) \(=\) \(\ds \size x\) Absolute Value of Negative
\(\ds -y\) \(\le\) \(\ds \size {-y}\)
\(\ds \) \(=\) \(\ds \size y\) Absolute Value of Negative
\(\ds \leadsto \ \ \) \(\ds -\paren {x + y}\) \(=\) \(\ds \paren {-x} + \paren {-y}\)
\(\ds \) \(\le\) \(\ds \size x + \size y\)

Hence the result.

$\blacksquare$


Examples

Example: $\size {-1 + 3}$

$2 = \size {-1 + 3} \le \size {-1} + \size 3 = 1 + 3 = 4$


Sources

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