Transfer 1011110.110010101 from binary in octal number system

This transfer is possible in two ways: direct transfer and using the decimal system.

first, let\'s make a direct transfer.

Fill in the number with missing zeros on the left

let\'s make a direct translation from binary to post-binary like this:

001011110.110010101₂ = 001 011 110. 110 010 101 = 001₍₌₁₎ 011₍₌₃₎ 110₍₌₆₎. 110₍₌₆₎ 010₍₌₂₎ 101₍₌₅₎ = 136.625₈

the Final answer: 1011110.110010101₂ = 136.625₈

now let\'s make the transfer using the decimal system.

let\'s translate to decimal like this:

0∙2⁸+0∙2⁷+1∙2⁶+0∙2⁵+1∙2⁴+1∙2³+1∙2²+1∙2¹+0∙2⁰+1∙2^-1+1∙2^-2+0∙2^-3+0∙2^-4+1∙2^-5+0∙2^-6+1∙2^-7+0∙2^-8+1∙2^-9 = 0∙256+0∙128+1∙64+0∙32+1∙16+1∙8+1∙4+1∙2+0∙1+1∙0.5+1∙0.25+0∙0.125+0∙0.0625+1∙0.03125+0∙0.015625+1∙0.0078125+0∙0.00390625+1∙0.001953125 = 0+0+64+0+16+8+4+2+0+0.5+0.25+0+0+0.03125+0+0.0078125+0+0.001953125 = 94.791015625₁₀

got It: 001011110.110010101₂ =94.791015625₁₀

Translate the number 94.791015625₁₀ в octal like this:

the Integer part of the number is divided by the base of the new number system:

94	8
-88	11	8
6	-8	1
	3

the Fractional part of the number is multiplied by the base of the new number system:


	0.	791015625*8
	6	.32813*8
	2	.625*8
	5	.0*8

the result of the conversion was:

94.791015625₁₀ = 136.625₈

the Final answer: 1011110.110010101₂ = 136.625₈

0
(	)	^	←
C	±	÷	X
7	8	9	-
4	5	6	+
1	2	3	=
0		.	=