G.3 Converting Decimals

Converting Decimals to Binary Numbers

In order to convert decimals to binaries, we reverse the process outlined in Section G.1 for converting a binary to a decimal.

41 10 =	20x2	+ 1	Dividing 41 by 2, gives the quotient 20 and remainder 1.
20 10 =	10x2	+	We again divide the current quotient 20 by 2.
10 10 =	5x2	+
5 10 =	2x2	+ 1	We repeat this procedure until ...
2 10 =	1x2	+
1 10 =	0x2	+ 1	... the quotient is 0.
41 10 =	101001 2

The divisor used in the steps above is the base of the target number system (binary, base 2). The binary value, 101001 ₂ , is represented by the remainders, with the last remainder as the left-most bit. Back substitution of the quotient gives the same result:

41 10	= (((((0x2 + 1)x2 + 0)x2 + 1)x2 + 0)x2 + 0)x2 + 1
	= 1 x2 5 + x2 4 + 1 x2 3 + x2 2 + x2 1 + 1 x2
	= 101001 2

Converting Decimals to Octal and Hexadecimal Numbers

Analogously, we can apply the above procedure for converting an octal to a binary. The conversion for the decimal number 90 can be done as follows :

90 10 =	11x8	+ 2
11 10 =	1x8	+ 3
1 10 =	0x8	+ 1
90 10 =	132 8 = 0132

The remainder values represent the digits in the equivalent octal number: 132 ₈ . This can be verified by back substitution, which gives the following result:

90 10	= ((0x8 + 1)x8 + 3)x8 + 2
	= 1 x8 2 + 3 x8 1 + 2 x8
	= 132 8 = 0132

Conversion to hexadecimal is analogous:

90 10 =	5x16	+ 10
5 10 =	0x16	+ 5
90 10 =	5a 16 = 0x5a

The remainders represent the digits of the number in the hexadecimal system: 5a . Back substitution gives the same result:

90 10	= (0x16+ 5)x16 +10
	= 5 x16 1 + a x16
	= 5a 16 = 0x5a