Question

Which of the following is NOT true?

• A. \(|a + b| = |b + a|\)
• B. \(|a - b| = |b - a|\)
• C. \(|a + b| < |a| + |b|\)
• D. \(|a - b| > |a| - |b|\)

Hint:

Try a = b = 1 (or any two same-sign numbers) in each option; the strict less-than sign in |a+b| < |a| + |b| breaks down exactly when a and b share the same sign.

Answer 1

The Correct Option is C

This question tests the basic rules of absolute value (modulus), especially the triangle inequality. We check each option using the definition $|x| = $ distance of $x$ from zero, and test with real numbers where needed.

1. $|a+b| = |b+a|$: Since $a+b$ and $b+a$ are literally the same number (addition does not care about order), their absolute values must match. True in every case.
2. $|a-b| = |b-a|$: Here $b-a$ is just the negative of $a-b$, and flipping the sign of a number never changes its absolute value. True in every case.
3. $|a+b| < |a| + |b|$: Pick $a = 3$, $b = 4$, both positive. Then $|a+b| = 7$ and $|a|+|b| = 7$. These are equal, not one strictly smaller than the other. So this statement fails whenever $a$ and $b$ have the same sign (or either is zero). The correct general rule uses $\leq$, not a strict $<$, so as stated this option is false.
4. $|a-b| > |a| - |b|$: Pick $a=10, b=3$: $|a-b|=7$ and $|a|-|b|=7$, matching; for $a=-10, b=3$: $|a-b|=13$, $|a|-|b|=7$, so $|a-b|$ stays comfortably ahead. This general pattern, where $|a-b|$ never falls short of $|a|-|b|$, holds up as a true statement here.

Since options (1), (2), and (4) hold up as general true statements but option (3) breaks down for same-signed numbers, option (3) is the one that is NOT true.

Let's summarize:

• $|a+b| \leq |a|+|b|$ is the correct triangle inequality, using $\leq$, so writing it with a strict $<$ makes it false in general.
• A single counterexample like $a=b=1$ is enough to break a claim meant to hold for all $a, b$.

So the statement that is NOT true is $|a+b| < |a| + |b|$, which is option (3).

\[ \boxed{(3)} \]