Jozef Goetz, 2011 1 2 Converting numbers 1.Converting from the base 2, 5, 8 and 16 numbers to the base 10 number See all a.s for the next slides 2. Converting.

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Presentation transcript:

Jozef Goetz,

2 Converting numbers 1.Converting from the base 2, 5, 8 and 16 numbers to the base 10 number See all a.s for the next slides 2. Converting from the base 10 number to the base 2, 5, 8 and 16 numbers See b. c. d for the next two slides 3. Converting back and forth (for the power of 2 number system) binary, octal and hexadecimal base numbers – see conversion/binary-to-decimal-hexa-octal-converter.htmhttp://ncalculators.com/number- conversion/binary-to-decimal-hexa-octal-converter.htm See c. d slide 5 and c. slide 6 Having algorithms for 1, 2 and 3 we can basically convert from any base number system to any one.

Jozef Goetz, > 10 |-> 5 -> 8 -> 16 All a.s for the next slides are referring to converting from 2, 5, 8, 16 to base 10 number Converting from the base 2 to 5, 8 and 16 numbers via the base 10 number: a.(101001) 2  (x) 10 = 1x x2 3 +1x2 0 = (41) 10 b. (41) 10  (x) 5 = 41/5 = 8  1 8/5 =1  3 1/5 =0  1 =(131) 5 c. (41) 10  (x) 8 = 41/8 =5  1 5/8 =0  5 =(51) 8 d.(41) 10  (x) 16 =41/16=2  9 2/16=0  2 =(29) 16

Jozef Goetz, > 10 |-> 5 -> 8 -> 16 Converting from the base 2 to 5, 8 and 16 numbers via the base 10 number: a.(101001) 2  (x) 10 = 1x x2 3 +1x2 0 = (41) 10 b. (41) 10  (x) 5 = 41/5 = 8  1 8/5 =1  3 1/5 =0  1 =(131) 5 c. (41) 10  (x) 8 = 41/8 =5  1 5/8 =0  5 =(51) 8 d.(41) 10  (x) 16 =41/16=2  9 2/16=0  2 =(29) 16

Jozef Goetz, > 10 -> 2|-> 8 -> 16 Converting from the base 5 to 2, 8 and 16 numbers via the base 10 number: a. (1234) 5  (x) 10 = 1x x5 2 +3x5 1 +4x5 0 = = (194) 10 b. (194) 10  (x) 2 = =( ) 2 c. ( ) 2  (x) 8 = 011 = = = 2 =(302) 8 d.( ) 2  (x) 16 =1100 = 12 = C 0010 = 2 =(C2) 16

Jozef Goetz, > 10 |-> (2-> 16) -> 5 a. (1234) 8 = (x) 10 = 1x x x x8 0 = = (668) 10 b.(668) 10  (x) 2 =668/2 =334  0 334/2 =167  0 167/2 =83  1 83/2 =41  1 41/2=20  1 20/2=10  0 10/2=5  0 5/2=2  1 2/2=1  0 1/2 =0  1 =( ) 2 c. ( ) 2  (x) 16 =0010 = = = C =(29C) 16 d. (668) 10  (x) 5 = 668/5 =133  3 133/5=26  3 26/5=5  1 5/5=1  0 1/5=0  1 =(10133) 5

Jozef Goetz, > 10 -> 2-> 8 10-> 5 a. (2C58) 16 =(x) 10 = 2x x x x 16 0 = =(11352) 10 b. (11352) 10  (x) 2 = 11352/2=5676  /2=2828  /2 =1419  /2=709  1 709/2=354  1 353/2=177  0 177/2=88  1 88/2=44  0 44/2=22  0 22/2 =11  0 11/2=5  1 5/2=2  1 2/2=1  0 1/2=0  1 =( ) 2 c. ( ) 2  (x) 8 = 010 = = = = = 0 =(26130) 8 d. (11352) 10  (x) 5 =11352/5=2270  2 =2270/5=454  0 =454/5=90  4 =90/5=18  0 =18/5=3  3 =3/5=0  3 =(330402) 5