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Topic title: What does IB consider to be binary "normalization"?
Topic summary:
Created on: 22/03/2006 07:30:49 AM
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22/03/2006 07:30:49 AM
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Gerry Donaldson

Posts: 67
Joined: 20/03/2003

Binary normalization looks like scientific notation such as 1.01001 x 2^3 for example.

IEEE drops the "1." as redundant since, in binary, there will always be a "1.", thus the above example would be stored as "01001" rather than "1.01001"?

The quandry is this: IEEE does set the international standard for such things, but I have not seen IB documentation using the IEEE standard on this. In the May papers, which convention should students assume?

-------------------------
 Gerry Donaldson
Computer Science Teacher
Sir Winston Churchill High School
Calgary, Alberta, Canada
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28/03/2006 03:39:44 AM
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Richard Jones

Posts: 25
Joined: 06/08/2003

Oh Jerry

Here is a potential can of worms.

The Subject Guide does not specify IEEE or indeed any other system of storing floating point binary. For me, what you are talking about here is a specific "representation" and not "normalisation".

Normalisation refers to setting the fractional part such that it is stored with the most possible precision, thus the first bit after the (imaginary) point is always a 1. In IEEE representation this bit is never stored 'cos it will always be 1) as you say.

However, your 1.01001 x 2^3 seems to me to be a negative number because the MSB is a 1, so I would have said it was already normalised (since in 2's comp the 0 is really a 1). I'm pretty sure IEEE uses 2's comp for the mantissa and excess-n notation (not in syllabus) for the exponent - but I'm prepared to be corrected.

If it is positive, for me, normalised it would look like: 0.101001 x 2^4. IEEE would then represent it as 001001 plus a bunch more zeroes and the exponent.

Light blue touch paper and retire to safe distance :-)

Richard

-------------------------
 Richard Jones
Between occupations :-)
richard@ib-computing.com
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28/03/2006 11:11:39 PM
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MADHUMITA ROYCHOWDHURY

Posts: 2
Joined: 17/03/2006

In conventional method the normalissation is to shift the point so as to make the number between 1 and 2 ,ie the mantissa should be in the form   1.*********** and while the mantissa is written, only the digits after the point is written ( as always the integer part will be 1)
eg   1111.01
=    1.11101  x 2^3

and mantissa is 11101 exponent is 3 (11)

But it is mentioned in IBO that  the normalissation should be done in the method where the number should be made  less than 1
that means    1111.01

=  .111101 x 2^4

where the mantissa is 111101   and exponent is 4 (100).


I think the above examples anwer you question.

Thanks
madhumita R.C
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