Convert 132 from decimal to binary
(base 2) notation:
Power Test
Raise our base of 2 to a power
Start at 0 and increasing by 1 until it is >= 132
20 = 1
21 = 2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
27 = 128
28 = 256 <--- Stop: This is greater than 132
Since 256 is greater than 132, we use 1 power less as our starting point which equals 7
Build binary notation
Work backwards from a power of 7
We start with a total sum of 0:
27 = 128
The highest coefficient less than 1 we can multiply this by to stay under 132 is 1
Multiplying this coefficient by our original value, we get: 1 * 128 = 128
Add our new value to our running total, we get:
0 + 128 = 128
This is <= 132, so we assign our outside coefficient of 1 for this digit.
Our new sum becomes 128
Our binary notation is now equal to 1
26 = 64
The highest coefficient less than 1 we can multiply this by to stay under 132 is 1
Multiplying this coefficient by our original value, we get: 1 * 64 = 64
Add our new value to our running total, we get:
128 + 64 = 192
This is > 132, so we assign a 0 for this digit.
Our total sum remains the same at 128
Our binary notation is now equal to 10
25 = 32
The highest coefficient less than 1 we can multiply this by to stay under 132 is 1
Multiplying this coefficient by our original value, we get: 1 * 32 = 32
Add our new value to our running total, we get:
128 + 32 = 160
This is > 132, so we assign a 0 for this digit.
Our total sum remains the same at 128
Our binary notation is now equal to 100
24 = 16
The highest coefficient less than 1 we can multiply this by to stay under 132 is 1
Multiplying this coefficient by our original value, we get: 1 * 16 = 16
Add our new value to our running total, we get:
128 + 16 = 144
This is > 132, so we assign a 0 for this digit.
Our total sum remains the same at 128
Our binary notation is now equal to 1000
23 = 8
The highest coefficient less than 1 we can multiply this by to stay under 132 is 1
Multiplying this coefficient by our original value, we get: 1 * 8 = 8
Add our new value to our running total, we get:
128 + 8 = 136
This is > 132, so we assign a 0 for this digit.
Our total sum remains the same at 128
Our binary notation is now equal to 10000
22 = 4
The highest coefficient less than 1 we can multiply this by to stay under 132 is 1
Multiplying this coefficient by our original value, we get: 1 * 4 = 4
Add our new value to our running total, we get:
128 + 4 = 132
This = 132, so we assign our outside coefficient of 1 for this digit.
Our new sum becomes 132
Our binary notation is now equal to 100001
21 = 2
The highest coefficient less than 1 we can multiply this by to stay under 132 is 1
Multiplying this coefficient by our original value, we get: 1 * 2 = 2
Add our new value to our running total, we get:
132 + 2 = 134
This is > 132, so we assign a 0 for this digit.
Our total sum remains the same at 132
Our binary notation is now equal to 1000010
20 = 1
The highest coefficient less than 1 we can multiply this by to stay under 132 is 1
Multiplying this coefficient by our original value, we get: 1 * 1 = 1
Add our new value to our running total, we get:
132 + 1 = 133
This is > 132, so we assign a 0 for this digit.
Our total sum remains the same at 132
Our binary notation is now equal to 10000100
