Absolute Error of Sum is not Greater than Sum of Absolute Errors

Theorem
Let $X_1, X_2, \ldots, X_n$ be approximations to a collection of (true) values $x_1, x_2, \ldots, x_n$ respectively.

Let $\Delta X_i$ be the absolute error of $X_i$ in $x_i$ for $i \in \set {1, 2, \ldots, n}$.

Then:
 * $\ds \map \Delta {\sum_{i \mathop = 1}^n X_i} \le \sum_{i \mathop = 1}^n \Delta X_i$

That is, the absolute error of a sum of approximations is not greater than the sum of the absolute errors of those approximations.

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
 * $\ds \map \Delta {\sum_{i \mathop = 1}^n X_i} \le \sum_{i \mathop = 1}^n \Delta X_i$

for appropriately defined $X_1, X_2, \ldots, X_n$ and $x_1, x_2, \ldots, x_n$.

Basis for the Induction
$\map P 2$ is the case:
 * $\map \Delta {X_1 + X_2} \le \Delta X_1 + \Delta X_2$

We have:

Thus $\map P 2$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\ds \map \Delta {\sum_{i \mathop = 1}^k X_i} \le \sum_{i \mathop = 1}^k \Delta X_i$

from which it is to be shown that:
 * $\ds \map \Delta {\sum_{i \mathop = 1}^{k + 1} X_i} \le \sum_{i \mathop = 1}^{k + 1} \Delta X_i$

Induction Step
This is the induction step:

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

Therefore:
 * $\forall n \in \N: n \ge 2: \ds \map \Delta {\sum_{i \mathop = 1}^n X_i} \le \sum_{i \mathop = 1}^n \Delta X_i$