Even-Odd Properties and Sign Rules

22 questions

Question 21Question

Let rr and ss be integers such that (1)r+s=1(-1)^{r+s} = -1 and r2s+rr^2 s + r is an odd integer. Which of the following statements must be true? Select all such statements.

Select all that apply

Show answer & explanation

Answer: rsr - s is an odd integer.; r2+s2r^2 + s^2 is an odd integer.; r2s+sr^2 s + s is an even integer.

Answer

The statements that must be true are 'rsr - s is an odd integer', 'r2+s2r^2 + s^2 is an odd integer', and 'r2s+sr^2 s + s is an even integer'.
From (1)r+s=1(-1)^{r+s} = -1, the sum r+sr+s must be odd, meaning rr and ss have opposite parity. Factoring r2s+rr^2 s + r yields r(rs+1)=oddr(rs + 1) = \text{odd}, which requires both rr and rs+1rs + 1 to be odd. Hence, rr is odd, which forces ss to be even. Testing the options shows that subtracting an even number from an odd number gives an odd number, adding the squares of an odd and an even number gives an odd number, and multiplying any integer by the even number ss gives an even number.

Step-by-Step Solution

1
Determine the parity of r+sr + s from (1)r+s=1(-1)^{r+s} = -1
r+sr + s is an odd integer
For (1)k=1(-1)^k = -1, the exponent kk must be an odd integer. Therefore, r+sr + s is odd, which implies that one variable is even and the other is odd.
2
Analyze the given expression r2s+rr^2 s + r
rr is odd and ss is even
Factor r2s+rr^2 s + r as r(rs+1)r(rs + 1). For the product of two integers to be odd, both factors must be odd. Thus, rr must be odd. Since r+sr + s is odd and rr is odd, ss must be even. (Verification: if ss is even and rr is odd, rs+1rs + 1 is even + 1 = odd, so r(rs+1)r(rs + 1) is odd ×\times odd = odd).
3
Evaluate each given statement using r=oddr = \text{odd} and s=evens = \text{even}
Statements 'rsr - s is an odd integer', 'r2+s2r^2 + s^2 is an odd integer', and 'r2s+sr^2 s + s is an even integer' are true.
1) oddeven=odd\text{odd} - \text{even} = \text{odd} (True).
2) odd+2(even)=odd+even=odd\text{odd} + 2(\text{even}) = \text{odd} + \text{even} = \text{odd} (False for even).
3) (odd)2+(even)2=odd+even=odd(\text{odd})^2 + (\text{even})^2 = \text{odd} + \text{even} = \text{odd} (True).
4) (even)(odd+1)=even×even=even(\text{even})(\text{odd} + 1) = \text{even} \times \text{even} = \text{even} (False for odd).
5) r2s+s=s(r2+1)=even×even=evenr^2 s + s = s(r^2 + 1) = \text{even} \times \text{even} = \text{even} (True).

Key Concept

Parity rules for integer addition, multiplication, and exponents
Question 22Question

Let xx and yy be non-zero integers such that x5y2<0x^5 y^2 < 0 and x+yx + y is an odd integer. Which of the following statements MUST be true? Select all that apply.

Select all that apply

Show answer & explanation

Answer: xx is a negative integer; xyxy is an even integer; x2+y2x^2 + y^2 is an odd integer

Answer

The statements 'xx is a negative integer', 'xyxy is an even integer', and 'x2+y2x^2 + y^2 is an odd integer' MUST be true.
The statement specifying that 'xx is a negative integer' is true because y2y^2 is strictly positive for any non-zero integer yy, forcing x5<0x^5 < 0 and thus x<0x < 0. The statement 'xyxy is an even integer' is true because x+yx + y being odd requires one variable to be even and the other to be odd, making their product even. The statement 'x2+y2x^2 + y^2 is an odd integer' is true because the square of an even number is even and the square of an odd number is odd, and their sum is always odd.

Step-by-Step Solution

1
Determine the sign of xx from the given inequality x5y2<0x^5 y^2 < 0.
xx must be negative (x<0x < 0).
Since yy is a non-zero integer, y2>0y^2 > 0. Dividing the inequality by y2y^2 gives x5<0x^5 < 0, which means xx must be negative.
2
Analyze the parity (even/odd nature) of xx and yy using x+yx + y is odd.
One of x,yx, y is even and the other is odd.
The sum of two integers is odd if and only if one integer is even and the other is odd.
3
Evaluate the statement 'xyxy is an even integer'.
The product xyxy is always even.
The product of an even integer and an odd integer is always even.
4
Evaluate the statement 'x2+y2x^2 + y^2 is an odd integer'.
The sum of squares x2+y2x^2 + y^2 is always odd.
Squaring an even integer yields an even integer, and squaring an odd integer yields an odd integer. Adding an even number and an odd number yields an odd number.

Key Concept

Even-Odd Parity Rules and Exponent Sign Properties
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