Răspuns
Explicație pas cu pas:
√3+rad. (4-√3)=√3+rad. (4-√12)=√3+√[(4+2)/2]-√[(4-2)/2]=
=√3+√3-1=2√3-1; dar √3∈(R-Q)
Aratam ca √3∉Q
Pp. ca √3∈Q => exista a/b∈Q unde a, b∈Z si (a, b)=1.
=> (√3)²=(a/b)² <=> 3=a²/b² => a²=3b² => 3|a² =>3|a => a=3k (1) k∈Z
3/1=a²/b² => <=> (3k)²/b²=3/1 <=> b²=3k² => 3|b² => 3|b => b=3q (2) q∈Z
Avem 3|a si 3|b adica (a, b)≠1 ceea ce contrazice presupunerea ca (a, b)=1
=> √3∈(R-Q)
=> 2·√3 ∈(R-Q) => 2·√3-1 ∈(R-Q)
