Never-ending fraction

Algebra Level 3

If a a and b b are integers such that

− 2 + 2 − 2 + 2 − 2 + 2 − 2 + ⋯ = a − b , -2+\cfrac{2}{-2+\cfrac{2}{-2+\cfrac{2}{-2+\cdots}}}=a-\sqrt{b},

find a + b a+b .


The answer is 2.

Victor Loh by Victor Loh , 19, Singapore

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Adrian Stefan
Jul 6, 2014

I t i s o b v i o u s t h a t − 2 + 2 a − b = a − b . It\quad is\quad obvious\quad that\quad -2+\frac { 2 }{ a-\sqrt { b } } =a-\sqrt { b } .\quad

S i n c e a − b c a n n o t b e 0 , w e g e t t h a t ( a − b ) 2 + 2 ( a − b ) − 2 = 0. Since\quad a-\sqrt { b } cannot\quad be\quad 0,\quad we\quad get\quad that\quad ({ a-\sqrt { b } ) }^{ 2 }+2(a-\sqrt { b } )-2=0. \quad

S o l v i n g t h i s w e g e t t h a t a − b = − 1 − 3 . Solving\quad this\quad we\quad get\quad that\quad a-\sqrt { b } =-1-\sqrt { 3 } .

T h u s w e g e t t h a t a + b = 3 − 1 = 2. Thus\quad we\quad get\quad that\quad a+b=3-1=2.

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Renz jimwel Mina
Jul 23, 2014

-2 + 2/(a-√b)=a-√b Let x= a-√b We get x^2+2x-2=0 Since a-√b is a root of this equation, a+√b must be another root And their sum is -2 and product is -2. Thus, 2a=-2 a=-1 a^2-b=-2 b=3 so a+b=3+(-1)=2

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Why don't you use LaTeX ?

Sai Ram - 5 years, 10 months ago
Julian Poon
Jul 9, 2014

Can somebody help verify this solution?

Since it is an unending fraction,

− 2 + 2 − 2 + 2 − 2 + 2 . . . = x = − 2 + 2 x -2+\frac { 2 }{ -2+\frac { 2 }{ -2+\frac { 2 }{ ... } } } =x=-2+\frac { 2 }{ x }

Solving:

x = − 2 + 2 x x=-2+\frac { 2 }{ x }

Gives: x = − 1 − 3 x=-1-\sqrt { 3 } or x = 3 − 1 x=\sqrt { 3 } -1

So it could be 2 2 answers. But either ways, the final answer is always 2 \boxed { 2 } for both cases.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I did it the same way as you can very well see, but there is only one answer, -1 -sqrt(3) because -1 + sqrt(3) would mean -sqrt(b)=sqrt(3), which would mean that sqrt(b)= - sqrt(3), which means b would be complex, but we are given that both a and b are integers.

Adrian Stefan - 6 years, 11 months ago

That's similar to the manipulation I used. I verified it numerically using an old set of recurrence formulae due to Wallis and Euler: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
 
a=2
b=-2

A_old, A_older = b, 1
B_old, B_older = 1, 0
for n in range ( 8 ) :
    A_new = b * A_old + a * A_older
    B_new = b * B_old + a * B_older
    print 1.0 * A_new / B_new
    A_old, A_older = A_new, A_old
    B_old, B_older = B_new, B_old

This converges quickly to the negative value you obtained.

Bill Bell - 6 years, 7 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...