#1- irrational
#2- whole
#3- rational
#4- natural and whole
a. Assume a is even, so a = 2k for some integer k. Now let a and b be integers such that a divides b and a + b is odd.
Since a divides b, b = an for integer n, and in turn b = 2nk, which means <u>b is even</u> and hence a + b is also even. But this contradicts our initial assumption, so a must be odd.
b. Let n be even, so that n = 2k for some integer k. Then
n² = (2k)² = 4k²
so that n² ≡ 0 (mod 4).
Now let n be odd, so n = 2k + 1 for integer k. Then
n² = (2k + 1)² = 4k² + 4k + 1
so that n² ≡ 1 (mod 4).
Therefore n² is never congruent to 2 (mod 4).
1.
Answer B.
2.
Answer B.
3.
Answers A. and C.
Answer : b: How many push-ups can a sixth grade girl do? and d:What are the weights of the sixth grade boys?