Beautiful Irrational Number (3)

Number Theory Level 3

Given that $S=2+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ ... } } } } } } }$

Find the value of $\sqrt{S}-\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ ... } } } } } } }$ .

The answer is 1.

1 solution

Rindell Mabunga
Sep 4, 2014

From here , we can say that $S = \phi + 1$ where $\phi$ is the golden ratio which satisfies the equation ${\phi}^2 = \phi + 1$

Back to the problem, $S = \phi + 1 = {\phi}^2$ therefore $\sqrt {S} = \phi$ . The second expression is equal to $\sqrt {S} - (\phi - 1) = \phi - \phi + 1 = \boxed{1}$

How can you tell that the second expression is equal to $phi-1$ ??

Anik Mandal - 6 years, 9 months ago

Beautiful Irrational Number (3)

The answer is 1.

1 solution

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