Prove that there exists an irrational r for every natural number k(>=2) such that [rm]≡-1mod(k) for every natural number m. Here [x] is the greatest integer less than or equal to x.
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1 Answers
1For every r > 1 , I assume that ,
r + 1r = a
Again , [ r ] = r + 1r - 1
So , [ r m ] = r m + 1r m - 1
= a m - 1
So , [ r m ] + 1 = a m - 1 + 1 = a m ≡ 0 ( mod k ) ...........Given
Now , a 0 = 2 ; a 1 = a ; a n + 2 = a a n + 1 - a n
Obviously , we can choose such an " a m " which satisfies the above relations .
But for that to happen , we must have ,
a ≡ 0 ( mod k )
So , lets put a = n k .
Again , we can choose exactly such an " a " for which D = a 2 - 4 is not a perfect square also .
Hence , we get our required " r " in terms of k .
r = a + √D2 = 12 { n k + ( n 2 k 2 - 4 ) 1 / 2 }