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Question

[[x]n]=[xn] 


  1. False

  2. True


Solution

The correct option is B

True


In [[x]n] We want to devide [x] by m first.

Let the quotient be q and the remainder be r .

[x] = nq + r ............(1)

[[x]n] = nq+rn 

= q + rn 

 [[x]n]=[q+rn] 

We know , 0 r < n

0 rn < 1

rn  is the fractional part of   [[x]n] 

  [[x]n]=q

x = [x] + {x} ({x} - is fractional part of x) 

= nq + r + {x} (from (1) we have [x] = nq + r)

Consider    r + {x}

We know    0 < n => 0 r n - 1 

Also, 0 {x} < 1 

   = 0 r + {x} < n 

0 r+(x)n

Consider [xn]=[[x]+(x)n] 

= [nq+r+(x)n] 

= [q+r+(x)n] 

0 r+(x)n < 1 [from (1) ] 

[xn] = q

[[x]n]

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