+0

# Binary to Decimal........

0
686
6

Convert the following string of binary digits to decimal digits:11000000111001. Thanks for any help, with steps if possible please.

Dec 30, 2015

#3
+15

Better is to use the 'double and add the next digit' method.

Consider an easier example first.

1011 -> 8 + 0 + 2 + 1 = 11

Start with the 1 at the left hand end, double it and add the next digit, 2*1 + 0 = 2, double that and add the next digit, 2*2 + 1 = 5, double that and add the next (the final) digit 5*2 + 1 = 11.

Try another one, 110101 (-> 32 + 16 + 0 + 4 + 0 + 1 = 53).

On paper, it's usual to set it out in a column

1     1

1     1*2 + 1 = 3

0     3*2 + 0 = 6

1     6*2 + 1 = 13

0     13*2 + 0 = 26

1     26*2 + 1 = 53.

If you look in 'Sticky Topics - Great Questions to Learn From', there's a note regarding the conversion of decimal to binary, this is the reverse operation.

- Bertie

Dec 31, 2015

#1
+5

11,000,000,111,001, start by counting them from the RHS and must begin with 0,1,2,3.......etc. The zeros in the string have no values. Only 1's do. If we do that, you will notice that we have 14 digits starting with zero. This means the last 1 is number 13. The ranking of each 1 is the power of 2. So, we have:

2^13+2^12+2^5+2^4+2^3+2^0=12,345. Which is the equivalent decimal number.

Dec 30, 2015
#2
+5

Hola

Mira, para convertirlos puedes aprender de la siguiente manera.

Tomas cada uno de los dígitos de derecha a izquierda y comienzas a sumar los valores de aquellos que aparecen en 1.  Es decir, cada posición de tu número tendrá por llamarlo asignado un número decimal por llamarlo así. La longitud de tu número binario y la cantidad de 0 y 1 que tenga, variarán el resultados. Hay una forma sencilla de realizar esta conversión, solo basta con aprenderse las potencias de dos, separar el número binario y ponerle arriba la potencia de 2 equivalente a su posición.

Con esta imagen lo entenderás mejor. Te preguntarás que significan los números que están arriba del tu número en binario. Sencillo son las potencias de dos que van aumentando de izquierda a derecha.

2^0=1

2^1=2

2^2=4

2^3=8

....

2^13=8192

Ahora lo único que te queda por hacer es sumar aquellos que tengan el uno abajo y listo.

De izquierda a derecha  -> 1+8+16+32+4096+8192=  (Te la dejo a vos)

Saludos

Dec 30, 2015
#3
+15

Better is to use the 'double and add the next digit' method.

Consider an easier example first.

1011 -> 8 + 0 + 2 + 1 = 11

Start with the 1 at the left hand end, double it and add the next digit, 2*1 + 0 = 2, double that and add the next digit, 2*2 + 1 = 5, double that and add the next (the final) digit 5*2 + 1 = 11.

Try another one, 110101 (-> 32 + 16 + 0 + 4 + 0 + 1 = 53).

On paper, it's usual to set it out in a column

1     1

1     1*2 + 1 = 3

0     3*2 + 0 = 6

1     6*2 + 1 = 13

0     13*2 + 0 = 26

1     26*2 + 1 = 53.

If you look in 'Sticky Topics - Great Questions to Learn From', there's a note regarding the conversion of decimal to binary, this is the reverse operation.

- Bertie

Guest Dec 31, 2015
#4
+5

I just want to look at these methods, especially Bertie's method 11000000111001

#3 Guest

Better is to use the 'double and add the next digit' method.

Consider an easier example first.

1011 -> 8 + 0 + 2 + 1 = 11

Start with the 1 at the left hand end, double it and add the next digit, 2*1 + 0 = 2, double that and add the next digit, 2*2 + 1 = 5, double that and add the next (the final) digit 5*2 + 1 = 11.

Try another one, 110101 (-> 32 + 16 + 0 + 4 + 0 + 1 = 53).

On paper, it's usual to set it out in a column

1     1

1     1*2 + 1 = 3

0     3*2 + 0 = 6

1     6*2 + 1 = 13

0     13*2 + 0 = 26

1     26*2 + 1 = 53. If you look in 'Sticky Topics - Great Questions to Learn From', there's a note regarding the conversion of decimal to binary, this is the reverse operation.

-----------------------------------------------------------------------------------------

110101 = 2^5+2^4+2^2+2^0 = 32+16+4+1 = 53

Bertie's Method :  Starting from the LEFT
\(110101_2 \\ = (((((1*2+1)*2+0)*2+1)*2+0)*2+1) \\ = (((( 2^2+2+0)*2+1)*2+0)*2+1)\\ =(((2^3+2^2+1)*2+0)*2+1)\\ =((2^4+2^3+1*2+0)*2+1)\\ =2^5+2^4+2^2+1\\ =32+16+4+1\\ =53\)

I learned to do the decimal to binary this way (in reverse).  And I like that method.

I think I learned it from Heureka but it could have been from you.

BUT

This is really weird Bertie, I can see that it works but I think most school students would be better off to make sure that  they understand Our German guests methods.  That is how they would normally be taught at school and it is much more intuitive.

Now I might as well try to do the original question this way:

11000000111001    14 digits

I am going to use the table tab that is provided for us in the ribbon with 14 rows

 1 1 1 1*2+1=3 0 2*3+0=6 0 6*2+0=12 0 12*2+0=24 0 24*2+0=48 0 48*2+0=96 0 96*2+0=192 1 192*2+1=385 1 385*2+1=771 1 771*2+1=1543 0 1542*2+0=3086 0 3084*2+0=6172 1 6168*2+1=12345

Ok I got the same as our first guest that is great!

NOW i am going to do it MY way - A different way again!

I am going to change it to hexidecimal FIRST

By grouping the digits into 4s starting from the RIGHT

11   0000   0011   1001 = 3 0 3 9 base 16

\(3*16^3 + 3*16 + 9 = 12288+48+9 = 12345   \)

If you want me to explain this better then just ask Thanks for showing us your method Bertie ,

I never would have thought to do it like that. :))

Jan 1, 2016
#5
+5

Bertie's "double and add the next digit" method is exactly the same thing we do in base 10.....consider  the number 1267

Starting from the left.....

Multiply the 1 by 10  = 10........add the next digt = 12 .........multiply this by 10   = 120.....add the next digit  = 126........multiply this by 10  = 1260 .......add the last digit = 1267......!!!

This method might be rightly called ...."multiply by the base, add the next digit"

Here's a way to understand why this "works"

(1 x 10) + 2

(1x10  + 2)10 + 6

(1x10^2 + 2*10 + 6)10  + 7

(1x 10^3 + 2x10^2 + 6x10) + 7  =   1267   Jan 1, 2016
edited by CPhill  Jan 1, 2016
#6
+5

Yes, we are all in agreement :)

Bertie's method works :)

It works in base ten and all other bases as well. :)

The other two methods also work well and if it was going to be taught at school I would use #2 guests method because it is easiest to understand.

In order to understand Chris's or my own reasoning demonstration of Bertie's method you need to understand #2 method.

The #2 guest's method brings the problem down to a basic conceptual knowledge of place value.

Jan 2, 2016
edited by Melody  Jan 2, 2016
edited by Melody  Jan 2, 2016