Is |a - b| < |a| + |b| ? (1) ab< 0 (2) a^b < 0

Question :Is |a - b| < |a| + |b| ?

On the right side of equation we will always see the sum of absolute values of a and b
But on the left hand side the |a - b| to be smaller, it's important to have the same sign because in case of opposite signs of a and b, left side of equation will be same as the right side of the equation

i.e. for the condition of the question to be true it's must for a and b to have same sign therefore,

Question REPHRASED : Do a and b have the same sign ?

Statement 1: ab< 0

i.e. a and b have opposite sign only then the product of a and b will be negative hence

SUFFICIENT

Statement 2: a^b < 0
\((-1)^1 < 0\)
Also, \((-1)^{-1} < 0\)
i.e. and b may have the same sign as well as opposite signs hence

NOT SUFFICIENT

Answer: Option A

\(\left| {a - b} \right|\,\,\mathop < \limits^? \,\,\,\left| a \right| + \left| b \right|\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,ab\,\,\mathop > \limits^? \,\,\,0\)

(*) This equivalence will be PROVED at the end of this post. Ignore this proof if you don´t like math!

\(\left( 1 \right)\,\,ab < 0\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\, \Rightarrow \,\,\,\,{\text{SUFF}}.\)

\(\left( 2 \right)\,\,\,{a^b} < 0\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {a,b} \right) = \left( { - 1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,\,\,{{\left( { - 1} \right)}^1} = - 1\,\,\,} \right] \hfill \cr \\
\,{\rm{Take}}\,\,\left( {a,b} \right) = \left( { - 1, - 1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,\,\,{{\left( { - 1} \right)}^{ - 1}} = {1 \over {{{\left( { - 1} \right)}^1}}} = - 1\,\,\,} \right]\,\, \hfill \cr} \right.\)

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

POST-MORTEM:

\(\left( * \right)\,\,\,\left\{ \matrix{\\
\,\left( i \right)\,\,\,\,\left| {a - b} \right|\,\, < \,\,\,\left| a \right| + \left| b \right|\,\,\,\,\,\,\, \Rightarrow \,\,\,\,ab > 0 \hfill \cr \\
\,\left( {ii} \right)\,\,\,\,ab > 0\,\,\,\, \Rightarrow \,\,\,\,\left| {a - b} \right|\,\, < \,\,\,\left| a \right| + \left| b \right|\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left| {a - b} \right|\,\, \ge \,\,\,\left| a \right| + \left| b \right|\,\,\,\,\, \Rightarrow \,\,\,\,\,ab \le 0 \hfill \cr} \right.\,\)

\(\left( i \right)\,\,\,\,\left| {a - b} \right|\,\, < \,\,\,\left| a \right| + \left| b \right|\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{squaring}}} \,\,\,\,{\left( {a - b} \right)^2} < \,\,\,{a^2} + 2\left| {ab} \right| + {b^2}\,\,\,\, \Rightarrow \,\,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,\, - ab < \left| {ab} \right|\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,ab > 0\)

\(\left( {ii} \right)\,\,\,\,\left| {a - b} \right|\,\, \geqslant \,\,\,\left| a \right| + \left| b \right|\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{squaring}}} \,\,\,\,{\left( {a - b} \right)^2} \geqslant \,\,\,{a^2} + 2\left| {ab} \right| + {b^2}\,\,\,\, \Rightarrow \,\,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,\, - ab \geqslant \left| {ab} \right|\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,ab \leqslant 0\,\,\,\)

I got the explanation but I messed up my solution when I removed the modulus with + and - signs for each of them! I mean when we remove modulus then there are 2 possibilities right? |a| => a and -a. Why we did not do this?

Hi MrCleantek

Just a suggestion that I share with many of my students... Be logical first and use maths second... So I tend to avoid opening Modulus till it's not needed.

And the need to open modulus will be seen in less than 10% questions of inequality

Agreed! I will now remember to not open modulus by default. Thank you..

Since both sides of the question stem are NONNEGATIVE, we can square the inequality:
\((|a-b|)^2 < (|a| + |b|)^2\)
\(a^2 + b^2 - 2ab < a^2 + b^2 + 2|ab|\)
\(-ab < |ab|\)
The resulting inequality is true only if ab > 0.
Question stem, rephrased:
Is ab > 0?

Statement 1:
The answer to the rephrased question stem is NO.
SUFFICIENT.

Statement 2:
If a=-1 and b=1, the answer to the rephrased question stem is NO.
If a=-1 and b=-1, the answer to the rephrased question stem is YES.
INSUFFICIENT.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.