Roundoff Error by a machine number

Question

What is the roundoff error when we represent $2^{-1} + 2^{-25}$ by a machine number? (Note: this refers to absolute error, not relative)

Please, help

$0$, when we're using IEEE754 doubles. What do you know about the representation of "machine numbers"? — Daniel Fischer, Sep 17 '13 at 09:20
it has the form of $sign$ $exponent$ $mantissa$ ? Like $0$ $00000000$ $00000000000000000000000$ for 0 in decimal ? Could you explain your answer? — John Lennon, Sep 17 '13 at 09:22
How many bits for exponent and mantissa? How are the bits to be interpreted? (Yes, I'm pretty sure it's the IEEE754 format, but do you know what that means?) — Daniel Fischer, Sep 17 '13 at 09:26
There are 24 bits for the mantissa. The leading one is implied so this number can just fit without loss of precision. — Warren Hill, Sep 17 '13 at 10:05
Yes, but how are the bits to be interpreted, what does the bit pattern 0x30400000, with sign bit $0$ exponent bits $0110000$ and mantissa bits $10000000000000000000000$ mean? — Daniel Fischer, Sep 17 '13 at 10:06
@WarrenHill No, this number needs $25$ bits of precision, so it gets rounded to $\frac12$. — Daniel Fischer, Sep 17 '13 at 10:08

score 2 · Answer 1 · answered Sep 17 '13 at 12:05

Assuming a Single precision IEE754 number.

You have a sign bit: 0 is positive, 1 is negative

8 bits for the exponent with a -127 offset (range -127 to +128)

A 24 bit mantissa. The top bit is always assumed to be 1 so only 23 bits are stored.

For your example you need 25 bits of mantissa so $2^{-25}$ is too small to fit and the closest you can get is $2^{-1} = \frac{1}{2}$

The absolute error is therefore $2^{-25}$.

Note. With double precision numbers the absolute error would be zero.

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