Read the following algorithm: Algorithm for converting a fraction x into binary with p binary places....

 

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Read the following algorithm: Algorithm for converting a fraction x into binary with p binary places. Inputs: fraction x EQ, 0<r<1 and number of places p E N. Outputs: bits b1, b2,..., by such that (0.b1b2... bp)2 is the Method: best approximation to x using p binary places. Initialise: j1, b1, b2,..., bp 0 Loop: if jp+1 stop. x+2x If≥1 [b,1, x+x-1] ==== 5. Can you spot how to j+j+1 Repeat loop (a) Verify that the algorithm converts to 0.110102 when express exactly in repeating binary? (b) Express 0.8510 as accurately as possible with 8 binary places. Read the following algorithm: Algorithm for converting a fraction x into binary with p binary places. Inputs: fraction x EQ, 0<r<1 and number of places p E N. Outputs: bits b1, b2,..., by such that (0.b1b2... bp)2 is the Method: best approximation to x using p binary places. Initialise: j1, b1, b2,..., bp 0 Loop: if jp+1 stop. x+2x If≥1 [b,1, x+x-1] ==== 5. Can you spot how to j+j+1 Repeat loop (a) Verify that the algorithm converts to 0.110102 when express exactly in repeating binary? (b) Express 0.8510 as accurately as possible with 8 binary places. Read the following algorithm: Algorithm for converting a fraction x into binary with p binary places. Inputs: fraction x EQ, 0<r<1 and number of places p E N. Outputs: bits b1, b2,..., by such that (0.b1b2... bp)2 is the Method: best approximation to x using p binary places. Initialise: j1, b1, b2,..., bp 0 Loop: if jp+1 stop. x+2x If≥1 [b,1, x+x-1] ==== 5. Can you spot how to j+j+1 Repeat loop (a) Verify that the algorithm converts to 0.110102 when express exactly in repeating binary? (b) Express 0.8510 as accurately as possible with 8 binary places.

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The algorithm you provided is essentially a binary search algorithm It starts by initializing all bi... View the full answer