Binary Representation of Numbers Autar Kaw

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1 Binry Representtion of Numbers Autr Kw After reding this chpter, you should be ble to: 1. convert bse- rel number to its binry representtion,. convert binry number to n equivlent bse- number. In everydy life, we use number system with bse of. For exmple, look t the number Ech digit in hs vlue of 0 through 9 nd hs plce vlue. It cn be written s 57.76= In binry system, we hve similr system where the bse is mde of only two digits 0 nd 1. So it is bse system. A number like ( ) in bse- represents the deciml number s (( ) + ( )) ( ) = = in the deciml system. To understnd the binry system, we need to be ble to convert binry numbers to deciml numbers nd vice-vers. We hve lredy seen n exmple of how binry numbers re converted to deciml numbers. Let us see how we cn convert deciml number to binry number. For exmple tke the deciml number First, look t the integer prt: Divide 11 by. This gives quotient of 5 nd reminder of 1. Since the reminder is 1, 0 = 1.. Divide the quotient 5 by. This gives quotient of nd reminder of 1. Since the reminder is 1, 1 = Divide the quotient by. This gives quotient of 1 nd reminder of 0. Since the reminder is 0, = Divide the quotient 1 by. This gives quotient of 0 nd reminder of 1. Since the reminder is, 1. 3 = Sylor URL: Pge 1 of 1

2 Since the quotient now is 0, the process is stopped. The bove steps re summrized in Tble 1. Sylor URL: Pge of 1

3 Tble 1 Converting bse- integer to binry representtion. Quotient Reminder 11/ 5 1= 0 5/ 1= 1 / 1 0= 1/ 0 1= 3 Hence (11) = ( 3 = (11) 1 0 ) For ny integer, the lgorithm for finding the binry equivlent is given in the flow chrt on the next pge. Now let us look t the deciml prt, tht is, Multiply by. This gives The number before the deciml is 0 nd the number fter the deciml is Since the number before the deciml is 0, =0.. Multiply the number fter the deciml, tht is, by. This gives The number before the deciml is 0 nd the number fter the deciml is Since the number before the deciml is 0, = Multiply the number fter the deciml, tht is, 0.75 by. This gives 1.5. The number before the deciml is 1 nd the number fter the deciml is 0.5. Since the number before the deciml is 1, = Multiply the number fter the deciml, tht is, 0.5 by. This gives 1.0. The number before the deciml is 1 nd the number fter the deciml is 0. Since the number before the deciml is 1, = 1. Since the number fter the deciml is 0, the conversion is complete. The bove steps re summrized in Tble. Tble. Converting bse- frction to binry representtion. Sylor URL: Pge 3 of 1

4 Number Number fter deciml Number before deciml = = = = Sylor URL: Pge 4 of 1

5 Strt Input (N) Integer N to be converted to binry formt i = 0 Divide N by to get quotient Q & reminder R i = i+1 i = R No Is Q = 0? Yes n = i (N) = (... ) Sylor URL: STOP Pge 5 of 1

6 Hence (0.1875) ( = = (0.0011) ) The lgorithm for ny frction is given in flowchrt on the next pge. Hving clculted nd we hve ( 11) = (11 ) ( ) = (0.0011, ) ( ) = ( ) In the bove exmple, when we were converting the frctionl prt of the number, we were left with 0 fter the deciml number nd used tht s plce to stop. In mny cses, we re never left with 0 fter the deciml number. For exmple, finding the binry equivlent of 0.3 is summrized in Tble 3. Tble 3. Converting bse- frction to pproximte binry representtion. Number Number fter deciml Number before deciml = = = Sylor URL: Pge 6 of 1

7 = = 5 As you cn see the process will never end. In this cse, the number cn only be pproximted in binry formt, tht is, ( 0.3) = ( 5) (0.001) Q: But wht is the mthemtics behinds this process of converting deciml number to binry formt? A: Let z be the deciml number written s where z = x. y x is the integer prt nd y is the frctionl prt. We wnt to find the binry equivlent of x. So we cn write Sylor URL: Pge 7 of 1

8 Strt Input (F) Frction F to be converted to binry formt i= Multiply F by to get number before deciml, S nd fter deciml, T i= i i = R No Is T = 0? Yes n = i (F) = (... ) Sylor URL: STOP Pge 8 of 1

9 x= n n 0 n + n If we cn now find 0,..., in the bove eqution then n ( x) = ( nn ) We now wnt to find the binry equivlent of y. So we cn write y= b + b +... b If we cn now find 1 + m ( y) = ( b 1b... b m) m m b 1,..., b in the bove eqution then Let us look t this using the sme exmple s before. Exmple 1 Convert ( ) to bse. Solution To convert ( 11) to bse, wht is the highest power of tht is prt of 11. Tht power is 3, s 3 = 8 to give 11= Wht is the highest power of tht is prt of 3. Tht power is 1, s 1 = to give So 3= 11= = Wht is the highest power of tht is prt of 1. Tht power is 0, s 0 = 1 to give Hence 0 1= Sylor URL: Pge 9 of 1

10 ( 11) = = + + 1= + + = (11) To convert ( ) to the bse, we proceed s follows. Wht is the smllest negtive power of tht is less thn or equl to Tht power is 3 s 3 =0.15. So = Wht is the next smllest negtive power of tht is less thn or equl to Tht power is 4 s = So Hence Since nd we get = 3 + ( ) = ( 11) = (11 = = + = (0.0011) ) ( ) = ( ) ( ) = ( ) Cn you show this lgebriclly for ny generl number? Exmple Convert ( ) to bse. Solution For ( 13), conversion to binry formt is shown in Tble 4. Sylor URL: Pge of 1

11 Tble 4. Conversion of bse- integer to binry formt. Quotient Reminder 13/ 6 1= 0 6/ 3 0= 1 3/ 1 1= 1/ 0 1= 3 So ( 13) = (11. ) Conversion of ( 0.875) to binry formt is shown in Tble 5. Tble 5. Converting bse- frction to binry representtion. Number Number fter deciml Number before deciml = = = So ( 0.875) = (0.111 ) Sylor URL: Pge 11 of 1

12 Hence ( ) = ( ) Sylor URL: Pge 1 of 1

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