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Difficulty: HardEven-Odd Properties and Sign Rules

Let pp, qq, and rr be integers such that p<0<q<rp < 0 < q < r. If pp is an odd integer, qq is an even integer, and rr is an odd integer, which of the following expressions must be negative? Select all that apply.

  1. pq(q−r)p^q(q - r)Answer
  2. B
    p(p−q)rp(p - q)^r
  3. (p−r)(−p)r(p - r)(-p)^rAnswer
  4. pr(r−p)qp^r(r - p)^qAnswer
  5. E
    (−p)q(q−p)(-p)^q(q - p)

Answer

The expressions that must be negative are pq(q−r)p^q(q - r), (p−r)(−p)r(p - r)(-p)^r, and pr(r−p)qp^r(r - p)^q.
Expressions pq(q−r)p^q(q - r), (p−r)(−p)r(p - r)(-p)^r, and pr(r−p)qp^r(r - p)^q evaluate to the product of a positive factor and a negative factor in every case, making their values strictly negative.

Step-by-Step Solution

1
Analyze the given signs and parities of variables
p<0p < 0 (negative, odd), q>0q > 0 (positive, even), r>0r > 0 (positive, odd), with q<rq < r.
Establishing the domain and sign/parity properties of each variable is essential before evaluating exponential and subtractive terms.
2
Evaluate the sign of pq(q−r)p^q(q - r)
pq>0p^q > 0 because an even exponent yields a positive result for non-zero bases. Since q<rq < r, (q−r)<0(q - r) < 0. Thus, positive×negative=negative\text{positive} \times \text{negative} = \text{negative}.
Demonstrates that this expression is guaranteed to be negative.
3
Evaluate the sign of p(p−q)rp(p - q)^r
p<0p < 0. p−q<0p - q < 0, and raising a negative number to an odd exponent rr gives a negative result. Thus, p(p−q)r=negative×negative=positivep(p - q)^r = \text{negative} \times \text{negative} = \text{positive}.
Shows that this expression is positive, so it cannot be negative.
4
Evaluate the sign of (p−r)(−p)r(p - r)(-p)^r
p−r<0p - r < 0 because subtracting a positive number from a negative number is negative. −p>0-p > 0, so (−p)r>0(-p)^r > 0. Thus, negative×positive=negative\text{negative} \times \text{positive} = \text{negative}.
Demonstrates that this expression is guaranteed to be negative.
5
Evaluate the sign of pr(r−p)qp^r(r - p)^q
pr<0p^r < 0 because a negative number raised to an odd exponent is negative. r−p>0r - p > 0, so (r−p)q>0(r - p)^q > 0. Thus, negative×positive=negative\text{negative} \times \text{positive} = \text{negative}.
Demonstrates that this expression is guaranteed to be negative.
6
Evaluate the sign of (−p)q(q−p)(-p)^q(q - p)
−p>0  ⟹  (−p)q>0-p > 0 \implies (-p)^q > 0. q−p>0q - p > 0. Thus, positive×positive=positive\text{positive} \times \text{positive} = \text{positive}.
Shows that this expression is always positive.

Key Concept

Sign rules for bases raised to even vs. odd powers, and order of operations with signed quantities.
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Topics

Even-Odd Properties and Sign Rules

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