Arithmetic – Are these ways helpful for the complement implementation of a BCD digit for 9?

I have two options for adding 9 to the BCD number.

The first way I tried is as follows:
Here
What I did is if we have 4-bit digit T as abcd:

1001 (~ 9) - abcd (in BCD) = 1001 + (2's complement of abcd) = 9's complement

the image has chart implementation.


The next way is as follows:
Here
in which I showed a truth table with a 4-bit point T (BCD abcd) as the input and a nine's complement of T as the output, I can also use the circuit with K-map.

My question is whether these paths are correct for nine supplements or not.

Floating point – proof that a guard digit has limited the deduction error

I have read what every computer scientist should know about floating point arithmetic, which is extremely interesting. However, I have some difficulty understanding the proof of Theorem 9 (page 33).

First a pretty trivial question. If the formula (15) $ say:
$$ y – bar {y} lt ( beta – 1) ( beta ^ {- p} + dots + beta ^ {- p-k}) $$
Should not it be? $ le $ instead of $ lt $or did I miss something?

More importantly, I do not understand why it says so if $ x- bar {y} lt $ 1, then $ delta = 0 $, How can there be no rounding error?

Then it means:
$$ x – y ge 1.0 – 0. overbrace {0 dots 0} ^ k overbrace { rho dots rho} ^ k ; textrm {with} ; rho = beta – 1 $$

Why is that? The difference can not be arbitrarily small or even $ 0 $?

And overall, correct me if I'm wrong, but the guard digit just saves the day when the input floats $ x $ and $ y $ are the exact numbers we want to subtract. If they are the well-rounded result of another calculation, there can still be a catastrophic cancellation. The guard digit would only bring the calculated value into the correct range.

Discrete Mathematics – Compute the least significant digit of a large number with the exponent

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Return the precision of the first non-zero digit in a number representing the measurement error representation

I want to return the precision of the first non-zero digit of a number. The motivation is that I can quickly build up my error notation, which is the parenthesized notation, eg.
$ x = $ 1,234 with associated error $ Delta x = $ 0.0015 I would represent the measurement as $ 1.23 (2), I want to automate this so I:

  • Determine the accuracy of the error. Round up with the next digit. So in pseudo-code:

    DeltaX = 0.0015
    DeterminePrecissionValue[DeltaX] (* Would come back: 3 *)
    NumberForm[Numb, {DeltaX, 3}]
    NumberForm[Numb, {X, 3}]

  • Based on the above result, I would round up the reading
    the accuracy of the error result and then format accordingly.

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