Răspuns :
presupunem ca √(105n +12) ∈ Q
caz 1) n-par =>n=2k
√(105n+12)= √(....0)+12= √(......2) ∈Q (fals)
caz 2)n-impar =>n=2k+1
√(105n+12)= √(....5)+12= √(......7) ∈Q (fals)
caz 1) n-par =>n=2k
√(105n+12)= √(....0)+12= √(......2) ∈Q (fals)
caz 2)n-impar =>n=2k+1
√(105n+12)= √(....5)+12= √(......7) ∈Q (fals)
presupunem ca √(105n +12) ∈ Q
caz 1) n-par =>n=2k
√(105n+12)= √(....0)+12= √(......2) ∈Q (fals)
caz 2)n-impar =>n=2k+1
caz 1) n-par =>n=2k
√(105n+12)= √(....0)+12= √(......2) ∈Q (fals)
caz 2)n-impar =>n=2k+1