Robustness Math Example 4

Follow the full solution, then compare it with the other examples linked below.

Example 4

medium

Prove that the statement '

n^2 > n

for all

n > 1

' is robust to replacing

>

with

\ge

: does '

n^2 \ge n

for all

n \ge 1

' still hold?

Solution

1
Original: $n^2 > n \Leftrightarrow n(n-1) > 0$ , which requires $n > 1$ (since $n > 0$ is assumed). True for $n > 1$ .
2
Weakened version: $n^2 \ge n \Leftrightarrow n(n-1) \ge 0$ . For $n \ge 1$ : both $n \ge 0$ and $n-1 \ge 0$ , so the product $\ge 0$ . True.
3
At $n = 1$ : $1^2 = 1 = n$ , so $n^2 \ge n$ holds with equality. The weakened statement is robust — it holds on the larger domain $n \ge 1$ .

Answer

n^2 \ge n \text{ for all } n \ge 1 \text{ (robust under weakening } > \text{ to } \ge\text{)}

Robustness of a mathematical claim means it remains valid under small modifications (such as weakening strict inequalities). Testing boundary cases like

n=1

is essential for verifying robustness.

About Robustness

The property of a result, algorithm, or model remaining valid or approximately correct even when its assumptions are slightly violated.

Learn more about Robustness →

More Robustness Examples

Example 1 easy

You estimate [formula] instead of [formula] in the formula [formula] with [formula] cm. Compute the

Example 2 medium

Show that the median is more robust than the mean as a measure of centre when outliers are present.

Example 3 easy

If an input [formula] has a measurement error of [formula], find the range of [formula] and assess w

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Sensitivity