Question

For each of the following statements decide whether they are True or False and give a short argument if True, or counter example if False.

(1) ∀n ∈ Z, ∃m ∈ Z, n + m ≡ 1 mod 2.

(2) ∀n ∈ Z, ∃m ∈ Z, (2n + 1)^2 = 2m − 1.

(3) ∃n ∈ Z, ∀m > n, m^2 > 100m.

Answer 1

(1) TRUE

if n is even, then there exists odd m such that, n+m is odd, so, n+m 1 (mod 2)

If n is odd, then there exists even m such that, n+m is odd, so, n+m 1 (mod 2)

So, for all n inZ there exists m in Z, such that, n+m 1 (mod 2)

(2) TRUE

let, n be in Z, So, (2n+1)² = 4n²+4n+1 = 4n²+4n+2 - 1 = 2(2n²+2n+1) - 1

So, m = 2n²+2n+1 (which is an integer), such that, (2n+1)² = 2m - 1

(3) THIS SEEMS TO BE WRONG. THERE IS NO n INVOLVED INTO THE PROPOSITION.