We need to show that 4 divides 1-n2 whenever n is an odd positive integer. If n is an odd positive integer then by definition n = 2k+1 for some non negative integer, k. Now 1 - n2 = 1 - (2k+1)2 = -4k2-4k = 4 (-k2-4k). k is a
The square of an integer is 30 more than the integer. Find the integer. Is there some sort of equation to use to solve this? And how can the square of an interger actually be 30 more than the integer? I am confused!
Disprove: For every integer a, if 8 divides a, then 6 divides a. im still stuck. i wrote: Disproof. Suppose 8|a. by definition of divisibility, we know 8|a means there is an integer b with a=8b. There is not value of y.
The number 1 is both the square of an integer and the cube of an integer. What is the next larger interger which is both a square and a cube of a positve integer. plz help, the only number i can think of was 0 but its smaller than
Suppose n is an integer. Select all statements below that are true: (choose 3) A) n^2 + n is always an even integer*** B) n^2 + n is always an even integer when n is even*** C) n^2 + n is always an even integer when n is odd*** D)
Prove by contradiction that for any even integer a and any odd integer b, 4 does not divide (a^2 + 2b^2). Proposition: That 4k (k is any integer) = a^2 +2b^2, and a is even, and b is odd. But 4k is even (product of any integer and
1. Assume that n is a positive integer. Use the proof by contradiction method to prove: If 7n + 4 is an even integer then n is an even integer. 2. Prove: n is an even integer iff 7n + 4 is an even integer. (Note this is an if and