wykaż,że pierwiastek 3+2√2to wszystko pod pierwiastkiem-pierwiastek3-2√2 to też pod pierwiastkiem jest liczbą naturalną

[latex] sqrt{3+2 sqrt{2} } - sqrt{3-2 sqrt{2} } = sqrt{1+2 sqrt{2}+2 } - sqrt{1-2 sqrt{2} +2}= \= sqrt{(1+ sqrt{2})^2 }- sqrt{(1- sqrt{2})^2 } =1+ sqrt{2} -|underbrace{1- sqrt{2}}_{<0}|= \=1+ sqrt{2}-[-(1- sqrt{2})]=1+ sqrt{2}+1- sqrt{2} =2 in mathbb{N} [/latex]

[latex]sqrt{3+2sqrt{2}}-sqrt{3-2sqrt{2}} = sqrt{1+2sqrt{2}+2}-sqrt{1-2sqrt{2}+2}}=\\=sqrt{(1+sqrt{2})^{2}}-sqrt{(1-sqrt{2})^{2}} = |1+sqrt{2}|-|1-sqrt2}|=\\=|1+sqrt{2}|-|sqrt{2}-1|= 1+sqrt{2}-(sqrt{2}-1) = \\=1+sqrt{2}-sqrt{2}+1} = 2 in N[/latex]