G
Guest
Guest
Can someone please explain how to convert base two numerals into base eight numerals?
1001
10101010
110110
101111
Al are base two
1001
10101010
110110
101111
Al are base two
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- Feb 4, 2004
- Messages
- 16,582
There are probably various methods, but one would be to convert from binary (base-2) to decimal (base-10) and then to octal (base-8).
For instance:
. . . . .1001<sub>2</sub>
. . . . .1001<sub>2</sub> = 1×2<sup>3</sup> + 0×2<sup>2</sup> + 0×2<sup>1</sup> + 1×2<sup>0</sup>
. . . . .1001<sub>2</sub> = 1×8 + 0×4 + 0×2 + 1×1
. . . . .1001<sub>2</sub> = 8 + 1 = 9<sub>10</sub>
. . . . .9<sub>10</sub> = 1×8 + 1×1
. . . . .9<sub>10</sub> = 1×8<sup>1</sup> + 1×8<sup>0</sup>
. . . . .9<sub>10</sub> = 11<sub>8</sub>
Do the others the same way.
Eliz.
For instance:
. . . . .1001<sub>2</sub>
. . . . .1001<sub>2</sub> = 1×2<sup>3</sup> + 0×2<sup>2</sup> + 0×2<sup>1</sup> + 1×2<sup>0</sup>
. . . . .1001<sub>2</sub> = 1×8 + 0×4 + 0×2 + 1×1
. . . . .1001<sub>2</sub> = 8 + 1 = 9<sub>10</sub>
. . . . .9<sub>10</sub> = 1×8 + 1×1
. . . . .9<sub>10</sub> = 1×8<sup>1</sup> + 1×8<sup>0</sup>
. . . . .9<sub>10</sub> = 11<sub>8</sub>
Do the others the same way.
Eliz.
Hello, bware!
Eliz is right: there is another way.
. . There's a trick to it, known to programmers.
Group the digits in sets-of-three, starting from the right.
Convert each three-digit binary number to octal (or decimal).
(d) 101,111<sub>2</sub>
. . We have: . (101) (111)
. . . . . . . . . . . . \ / . . . \ /
. . . . . . . . . . . . .5 . . . .7
. . Therefore: .101,111<sub>2</sub> .= .57<sub>8</sub>
(b) .10,101,010<sub>2</sub>
. . We have: . ( 10)(101)(010)
. . . . . . . . . . . . .| . . \ / . . .\ /
. . . . . . . . . . . . .2 . . 5 . . . .2
. . Therefore: .10,101,010<sub>2</sub> .= .252<sub>8</sub>
Eliz is right: there is another way.
. . There's a trick to it, known to programmers.
Group the digits in sets-of-three, starting from the right.
Convert each three-digit binary number to octal (or decimal).
(d) 101,111<sub>2</sub>
. . We have: . (101) (111)
. . . . . . . . . . . . \ / . . . \ /
. . . . . . . . . . . . .5 . . . .7
. . Therefore: .101,111<sub>2</sub> .= .57<sub>8</sub>
(b) .10,101,010<sub>2</sub>
. . We have: . ( 10)(101)(010)
. . . . . . . . . . . . .| . . \ / . . .\ /
. . . . . . . . . . . . .2 . . 5 . . . .2
. . Therefore: .10,101,010<sub>2</sub> .= .252<sub>8</sub>