Decimal to binary and hexadecimal conversion

[weller7503]

I have just started a new class, I have never done this before I will give an example of the problems. Can someone please help me understand how to do this?

Thanks



Decimal

2678 need to convert this to binary and hexadecimal

 

[Unknown008]

Binary is easy, you'll see

What is the largest base 2 number you can get in 2678?

It's 11, and \(\displaystyle 2^{11} = 2048\)

So, in binary, you'll need at least 12 digits (11+1), and this would look like:

100000000000

This is 2048. What is left now? 2678 - 2048 = 630

Repeat again, and you'll see that \(\displaystyle 2^9 = 512\)

Since it's 9, you put a '1' in the (9+1)th digit, starting from the right:

101000000000

This is 2048+512.

Rinse and repeat, you'll get 118 left, meaning \(\displaystyle 2^6\);

101000010000

Again and you'll get 54, meaning \(\displaystyle 2^5\);

101001100000

Again and you'll get 22, meaning \(\displaystyle 2^4\);

101001110000

Again and you'll get 6, meaning \(\displaystyle 2^2\);

101001110100

Again and you'll get 2, meaning \(\displaystyle 2^1\);

101001110110

And now that's it since you are left with nothing this time! (2-2 = 0)

If you want to get the decimal number again, just do: \(\displaystyle 2^{11} + 2^9+2^6+ 2^5+2^4+2^2+2^1\) (Notice again that the power is the position of '1' in the binary number, minus 1)

Now to convert to hex, I think it's better you have a table of conversion with you, because:
Binary | Hex
0000 = 0
0001 = 1
0010 = 2
0011 = 3
0100 = 4
0101 = 5
0110 = 6
0111 = 7
1000 = 8
1001 = 9
1010 = A
1011 = B
1100 = C
1101 = D
1110 = E
1111 = F

You can always re-create that table, it's easy if you notice the pattern. BUt don't forget that it ends at F, or remember that F is the 6th letter and hex means 6, or just remember that it goes up to 1111 (which by the way means \(\displaystyle 2^3+2^2+2^1+2^0 = 15\))

So, let's take the binary we got earlier:

101001110110

Break it down in groups of 4 digits for the conversion: 1010 0111 0110

Look at your table and you should get 'A76'.

 

[SAMUELK]

I have just started a new class, I have never done this before I will give an example of the problems. Can someone please help me understand how to do this?

Thanks

Decimal

2678 need to convert this to binary and hexadecimal

I think the simplest way to look at it is: Conversion to base X is repeated division by X --

2678 / 2 = 1339, rem 0. So 0 is your last (rightmost) digit.
1339 / 2 = 669, rem 1. 1 is your next (to the left) digit.
etc.
until your quotient is zero.

Now, about base X bigger than 10, such as 16? Do the same thing, exactly. But what if the remainder is bigger than 9? Simply create new symbols for the 'digits'. The usual is A = 10,....,F = 15.

 

[soroban]

Hello, weller7503!

Convert 2678 to binary and hexadecimal.

SamuelK is right . . . There is a faster procedure.

To convert to base-2:
. . (1) Divide the number by 2; note the remainder.
. . (2) Divide the quotient by 2; note the remainger.
. . (3) Repeat step (2) until we get a zero quotient.
. . (4) Read up the remainders.

      2 ) 2 6 7 8     Remainders
        ----------
      2 ) 1 3 3 9         0
        ----------
        2 ) 6 6 9         1
          --------
        2 ) 3 3 4         1
          --------
        2 ) 1 6 7         0
          --------
          2 ) 8 3         1
            ------
          2 ) 4 1         1
            ------
          2 ) 2 0         1
            ------
          2 ) 1 0         0
            ------
            2 ) 5         0
              ----
            2 ) 2         1
              ----
            2 ) 1         0
              ----
                0         1

Therefore: .\(\displaystyle 2678 \;=\;101,\!001,\!110,\!110_2\)



To convert to base-16:
. . (1) Divide the number by 16; note the remainder.
. . (2) Divide the quotent by 16; note the remainder.
. . (3) Repeat step (2) until we get a zero quotient.
. . (4) Read up the remainders.

      1 6 ) 2 6 7 8     Remainders
          ----------
        1 6 ) 1 6 7         6
            --------
          1 6 ) 1 0         7
              ------
                  0        10

The remainders are 10, 7, 6.

For "digits" greater than 9, use: .A = ten, B = eleven, C = twelve, . . .

Therefore: .\(\displaystyle 2678 \;=\;A76_{16}\)