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Zorluk: OrtaEven-Odd Properties and Sign Rules

Let mm and nn be integers such that m<0m < 0, n>0n > 0, (−1)m=−1(-1)^m = -1, and m+nm + n is an even integer. Which of the following expressions must be a positive even integer?

  1. A
    m⋅nm \cdot n
  2. B
    (−m)n(-m)^n
  3. n−mn - mCevap
  4. D
    (m+n)2(m + n)^2
  5. E
    m2−nm^2 - n

Cevap

The expression n−mn - m must be a positive even integer.
The expression n−mn - m subtracts a negative odd integer from a positive odd integer, which equals adding two positive odd integers. The sum of two positive odd integers is always a positive even integer.

Adım Adım Çözüm

1
Determine the parity and sign of mm.
Since m<0m < 0 and (−1)m=−1(-1)^m = -1, mm must be a negative odd integer.
An odd exponent on −1-1 yields −1-1.
2
Determine the parity and sign of nn.
Since n>0n > 0 and m+nm + n is even, nn must be a positive odd integer.
The sum of two integers is even if and only if both integers have the same parity. Since mm is odd, nn must also be odd.
3
Evaluate the sign and parity of n−mn - m.
n−m=n+(−m)n - m = n + (-m). Since n≥1n \ge 1 and −m≥1-m \ge 1, n−m≥2n - m \ge 2 (strictly positive). Also, odd minus odd is always even.
Combining the sign rules (n>0n > 0 and −m>0-m > 0) with the even-odd subtraction rule confirms n−mn - m is always a positive even integer.

Anahtar Kavram

Even-Odd Properties and Sign Rules
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Even-Odd Properties and Sign Rules

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