Fractions and Fractions

Algebra Level 1

$\LARGE \color{#3D99F6}{1 + \dfrac{1}{1 + \dfrac{1}{9}}}$

Simplify the number above.

\dfrac{10}{19}

\dfrac{9}{10}

\dfrac{10}{9}

\dfrac{19}{10}

4 solutions

Mehul Arora
Apr 28, 2015

$1+ \dfrac{1}{1+ \dfrac{1}{9}}= 1 + \dfrac{10}{9}$

Because $1+\dfrac{1}{9}=\dfrac{10}{9}$

$= 1+ \dfrac{9}{10}$

Because $\dfrac{1}{\dfrac{a}{b}}=\dfrac{b}{a}$

This is Equal to $\boxed {\dfrac{19}{10}}$

Moderator note:

Good. What would be its value if the fraction nested on itself indefinitely, that is what is the value of the fraction below?

$\large 1 + \frac {1}{1 + \frac {1}{9 + \frac {1}{1 + \frac {1}{1 + \frac {1}{9 + \frac {1}{\ldots} }} }}}$

Note the numbers $1,1,9,1,1,9, \ldots$ .

Challenge Master, @Calvin Lin sir, Is the answer $\dfrac{1}{10}(9+\sqrt{101})?$

Mehul Arora - 6 years, 1 month ago

it has to be the positive one ;)

Otto Bretscher - 6 years, 1 month ago

Yes. It does Indeed. Thanks! Is the answer Correct?

Mehul Arora - 6 years, 1 month ago

@Mehul Arora – Yes, I believe it is correct. I don't know whether we have to convince Mr. Calvin that the limit actually exists... we just found the fixed point.

Otto Bretscher - 6 years, 1 month ago

@Otto Bretscher – Well, Okay. That Went over my head :P I haven't learnt Any Calculus Right now . Anyway. Thanks Sir!

Mehul Arora - 6 years, 1 month ago

@Mehul Arora – The question is: If you consider more and more terms of your continued fraction, are you getting closer to your value $x_{equ}=\frac{1}{10}(9+\sqrt{101})$ ? It boils down to this: If you iterate the function $f(x)=1+\frac{1}{1+\frac{1}{9+\frac{1}{x}}}=\frac{19x+2}{10x+1},$ starting with the "seed" $x=1$ , are you approaching $x_{equ}$ ? We found the fixed point by letting $f(x)=x$ , but that's not enough to guarantee convergence (unless you know some theorems about the convergence of continued fractions).

Otto Bretscher - 6 years, 1 month ago

let the equation u gave be equal to x ,So we can write 1+\frac { 1 }{ 1+\frac { 1 }{ 9+\frac { 1 }{ x } } } =x

Harshit Shukla - 6 years, 1 month ago

observe that a(n+1)=1+1+1/(1+a(n)) assuming that the limit exists then the limit is a. Solving for a we get a = 2^0.5 since a>0.

Siddharth Iyer - 6 years, 1 month ago

Greg Wallace
Dec 19, 2015

Start at the component most inside of the equation and work your way out:

$1 + \frac{1}{9} = \frac{9}{9} + \frac{1}{9} = \frac{10}{9}$

$\frac{1}{\frac{10}{9}} = \frac{9}{10}$

$1 + \frac{9}{10} = \frac{10}{10} + \frac{9}{10} = \boxed{\frac{19}{10}}$

Ubaidullah Khan
Apr 29, 2015

Keith Daggett
Apr 29, 2015

9/9x1/(1+1/9)=9/10>>>

>>1=10/10>>> >>10/10+9/10=19/10

Fractions and Fractions

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