Triangle Inequality/Real Numbers/General Result

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Theorem

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

Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

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


$\map P 1$ is true by definition of the usual ordering on real numbers:

$\size {x_1} \le \size {x_1}$


Basis for the Induction

$\map P 2$ is the case:

$\size {x_1 + x_2} \le \size {x_1} + \size {x_2}$

which has been proved in Triangle Inequality for Real Numbers.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

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


Then we need to show:

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


Induction Step

This is our induction step:


\(\ds \size {\sum_{i \mathop = 1}^{k + 1} x_i}\) \(=\) \(\ds \size {\sum_{i \mathop = 1}^k x_i + x_{k + 1} }\) Definition of Indexed Summation
\(\ds \) \(\le\) \(\ds \size {\sum_{i \mathop = 1}^k x_i} + \size {x_{k + 1} }\) Basis for the Induction
\(\ds \) \(\le\) \(\ds \sum_{i \mathop = 1}^k \size {x_i} + \size {x_{k + 1} }\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^{k + 1} \size {x_i}\) Definition of Indexed Summation


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

$\blacksquare$


Also see


Sources

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