How do you prove that #(1/2)sqrt(2+sqrt(2+sqrt 2))=sqrt((1+sqrt((1+1/sqrt 2)/2)) /2)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Shwetank Mauria Nov 9, 2016 Please see below. Explanation: #(1/2)sqrt(2+sqrt(2+sqrt2))# = #sqrt((2+sqrt(2+sqrt2))/4)# = #sqrt((1+sqrt((2+sqrt2))/2)/2)# = #sqrt((1+sqrt((2+sqrt2)/4))/2)# = #sqrt((1+sqrt((1+sqrt(2/4))/2))/2)# = #sqrt((1+sqrt((1+sqrt(1/2))/2))/2)# Answer link Related questions How do you simplify #\frac{2}{\sqrt{3}}#? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division of Radicals? How do you simplify #7/(""^3sqrt(5)#? How do you multiply #(sqrt(a) +sqrt(b))(sqrt(a)-sqrt(b))#? How do you rationalize the denominator for #\frac{2x}{\sqrt{5}x}#? Do you always have to rationalize the denominator? How do you simplify #sqrt(5)sqrt(15)#? How do you simplify #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? See all questions in Multiplication and Division of Radicals Impact of this question 2258 views around the world You can reuse this answer Creative Commons License