Question

If \(a\) and \(b\) are negative, and \(c\) is positive, which of the following statement(s) is/are true?
I) \(a - b < a - c\)
II) if \(a < b\), then \(\dfrac{a}{c} < \dfrac{b}{c}\)
III) \(\dfrac{a}{b} > \dfrac{a}{c}\)

• A. I only
• B. II only
• C. III only
• D. II and III only

Hint:

Test the three statements using a small set of negative values for a and b and a positive value for c, then check the general rule behind each.

Answer 1

The Correct Option is D

Step 1: Check statement I.
Since \(c\) is positive, \(a - c\) is smaller than \(a\) (we subtract a positive number, so the value goes down). Since \(b\) is negative, \(a - b\) means we subtract a negative number, so \(a - b\) is bigger than \(a\). So \(a - c < a < a - b\), which gives \(a - c < a - b\), the opposite of what statement I claims. Statement I is false.

Step 2: Check statement II.
Dividing both sides of an inequality by the same positive number never flips the sign. Since \(c\) is positive and we are told \(a < b\), dividing both sides by \(c\) keeps the direction unchanged, so \(\dfrac{a}{c} < \dfrac{b}{c}\) always holds. Statement II is true.

Step 3: Check statement III.
\(a\) is negative and \(b\) is negative, so \(\dfrac{a}{b}\) is a negative divided by a negative, which is positive. \(a\) is negative and \(c\) is positive, so \(\dfrac{a}{c}\) is negative. A positive number is always greater than a negative number, so \(\dfrac{a}{b} > \dfrac{a}{c}\) always holds. Statement III is true.

Final Answer:
Only statements II and III hold true for every choice of negative \(a\), negative \(b\) and positive \(c\). \[ \boxed{\text{II and III only}} \]