How to prove that m n is an odd integer?

To this end, let n = 2 k + 1 and m = 2 ℓ + 1, where k, ℓ ∈ Z. Thus, we have the following: where η = ( 2 ℓ k + ℓ + k) ∈ Z. Hence, m n is an odd integer, and we have proven ¬ Q → ¬ P, thus proving P → Q, as desired. ◼ You can prove the contrapositive: if m = 2 a + 1 and n = 2 b + 1 both are odd, then n m has to be odd too.

Is 1/m + 1/n = K a positive integer?

If 1/m + 1/n = k is an integer, since m > 1 and n > 1, k must be positive integer. In order k to be a positive integer, we must have:

What is the value of m and N in this equation?

(x + m) (x + n) = x 2 + 5x + mn and x ≠ 0. The numbers m and n are integers. As we have no other information about the values of m and n from the question stem, let us analyse the individual statements.

How many distinct integers does 1/M+1/n have?

Conclusion: There are two distinct integers m and n, namely m=1 and n=-1, such that 1/m+1/n is an integer. Know someone who can answer?

How to prove that a(n) holds for all positive integers n?

Let A(n) be an assertion concerning the integer n. If we want to show that A(n) holds for all positive integer n, we can proceed as follows: Induction basis: Show that the assertion A(1) holds. Induction step: For all positive integers n, show that A(n) implies A(n+1). 3 Standard Example

Is m n = 2(p) + 1 even or odd?

By the definition of odd, m n = 2 ( p) + 1 is odd, which contradicts m n is even. Hence the supposition is false, therefore the theorem is true. No, this is not a valid proof.

How do you prove that b(n+1) holds?

Expanding the right hand side yields n3/3 + 3n2/2 + 13n/6 + 1 One easily verifies that this is equal to (n+1)(n+2)(2(n+1)+1)/6 Thus, B(n+1) holds. Therefore, the proof follows by induction on n. 8 Tip How can you verify whether your algebra is correct?