Thursday, September 03, 2009 1:50 AM,

The second thing I wanted to provide was an example for you to go over before 
you tackle the new problem 8 on the first problem set is a mechanism for 
taking a decimal fraction and putting it into normalized form.

In problem 8, I am asking you to do this for the value 0.3.

Here I will show you how to do it for the value 0.09.  I will use a method 
very similar to the method I showed you for dealing with integers in class.

Let's say I wanted to form the first six significant binary digits of 0.09.

That is, 0.09 is approximately equal to 0. b(-1) b(-2) b(-3) b(-4) b(-5) b(-6).

I say, approximately equal becauase the actual binary string is infinite.

You know, I hope, that 0.1111111111111111 is less than 1.  That is,
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... is always less than 1.  You probably saw 
the proof of that expressed as: Start on one side of the room, and walk half 
way to the other side (1/2).  Now walk half of the remaining half (1/4).
Now half of the remaining half (1/8).  Keep doing this.  You can never reach 
the other side!  It should be also clear that 0.011111111111 is always less 
than 1/2.

Now we are ready.

We start with: 


  0.09 = b(-1)*2^(-1) + b(-2)*2^(-2) + b(-3)*2^(-3) + b(-4)*2^(-4) + b(-5)*2^(-5) + b(-6)*2^(-6)

Note that 0.09 is less than 1/2.  So, b(-1) must be 0.  

Now remove the b(-1) term, multiply 0.09 by 2, and all terms on the right by 
2, giving us


  0.18 =  b(-2)*2^(-1) + b(-3)*2^(-2) + b(-4)*2^(-3) + b(-5)*2^(-4) + b(-6)*2^(-5)


Note that 0.18 is less than 1/2.  So b(-2) must be 0.

Remove the b(-2) term, and multiply everything by 2, yielding:

  0.36 =  b(-3)*2^(-1) + b(-4)*2^(-2) + b(-5)*2^(-3) + b(-6)*2^(-4)


And, b(-3) = 0.

The same operation yields:


  0.72 =  b(-4)*2^(-1) + b(-5)*2^(-2) + b(-6)*2^(-3)

0.72 is larger than 1/2, so b(-4) must be 1.  Subtract .5 from .72, and 
removing b(-4) gives us:


  0.22 =  b(-5)*2^(-2) + b(-6)*2^(-3)


Multiplying by 2 gives us

  0.44 =  b(-5)*2^(-1) + b(-6)*2^(-2)

Since .44 is less than 1/2, b(-5) =0.  Repeating the operation gives us:


  0.88 =  b(-6)*2^(-1)


Since .88 is greater than 1/2, b(-6) = 1.  

We still have .38 to deal with (once we subtract the .5 due to b(-6).
That will be the contribution due to b(-7), b(-8), b(-9), which thankfully we 
don't have to laboriously go through since I am allowing us to stop here.

The result is obtained by pulling together all that we have done:
That is, 


	0.09 is approximately equal to 0.000101.


If you want to check it, try adding 1/16 + 1/64.  What do you get?

The difference between what you get and 0.09 is due to the fact that we 
stopped at b(-6).  Try a few more digits, and note that you keep getting 
closer to 0.09 (never going above 0.09)!

Finally, as an exercise, let's put this number in normalized form:

We move the binary point 4 places to the right and divide by 2^4.

The result: 1.01 * 2^(-4)

And for one last bit of practice, if I wanted to represent that as a 32-bit 
floating point number:


		0 01111011 01000000000000000000000



And, I am done.

Good luck!
Yale Patt