Converting number bases

Computer number system: base 2 (binary)

Base (radix) k = 2, digits di = (0, 1)

Given: 13ten

Find: binary representation

b5

b4

b3

b2

b1

b0

weight

25

24

23

22

21

20

digit

0

0

1

1

0

1

8 + 4 + 1 = 13ten

result

remainder

try it!

start

13

13/2

6

1

b0

6/2

3

0

b1

3/2

1

1

b2

1/2

0

1

b3

0/2

0

0

b4

0/2

0

0

b5

Why does this work?

13ten

=             b5 * 32 + b4 * 16 + b3 * 8 + b2 * 4 + b1 * 2 + b0

(2 * 6 + 1)ten

=      2 * (b5 * 16 + b4 *  8 + b3 * 4 + b2 * 2 + b1) + b0

Therefore, dividing by 2 gives a result of the expression in parentheses and a remainder of b0

6ten

=        b5 * 16 + b4 * 8 + b3 * 4 + b2 * 2 + b1

= 2 * (b5 * 8 + b4 * 4 + b3 * 2 + b2) + b1

Dividing by 2 again gives a remainder of b1 and so forth.

3ten

=      b5 * 8 + b4 * 4 + b3 * 2 + b2

= 2 * (b5 * 4 + b4 * 2 + b3) + b2

1ten

=      b5 * 4 + b4 * 2 + b3

= 2 * (b5 * 2 + b4) + b3

0ten

= b5 * 2 + b4

What do all higher bits have to be?