# Questions tagged [floating-point]

Approximate representation of numbers as a fixed number of digits multiplied by a logarithmic scale.

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### Adding two numbers in half-precision IEEE754 standart

I have to add two numbers using IEEE754: -14.1875 and -7.4375. I managed to convert them to half-precision numbers in IEEE754: 1 10010 1100011000 (-14.1875) ...
83 views

### Unit conversion - Better to divide by an integer or multiply by a double?

I currently have a long timestamp measured in units of 100ns elapsed since January 1st, 1900. I need to convert it to milliseconds. I have the choice of either ...
60 views

### Half precision floating point question -- smallest non-zero number

There's a floating point question that popped up and I'm confused about the solution. It states that IEEE 754-2008 introduces half precision, which is a binary floating-point representation that uses ...
32 views

### Floating Point Arithmetic with 3 bits mantissa

Find all values of $x ∈ R$ such that x + 1 = 1 in floating point arithmetic with 3 bits mantissa. How do we represent number 1 in floating point arithmetic with 3 bits mantissa I wonder? After that, ...
89 views

### Python versus Matlab on the quantity 1/0

Python and Matlab seem to disagree on the division by 0. Python: ...
52 views

### Negative Numbers in 32 bit Floating Point IEEE Numbers

So I understand the logic behind converting positive decimal numbers to IEEE 32 bit floating numbers but I'm not completely sure behind the negative one's. If for example we have a decimal number say -...
33 views

### Adding two numbers in base 2(floating point) vs Multiplying two numbers in base 2(floating point)

Is it true that adding two numbers in base 2 is more complex than multiplying them? If so can someone please explain why this is the case?
138 views

### Prove that $1^\text{nan} = 1.00$

I know that for most computation involve nan (not a number) the result is a nan itself except for some cases. For example, $1^{\text{nan}} = 1.00$ which proven by mathematicians to be true. I tried to ...