Search Results

The two's complement of 000 is 000. It is formed by complementing all bits and adding 1 to the result. The one's complement of 000 is indeed 111, but it is not used in computing. The ten's complement …

Hint: The four values are $25/100 = 1/4$ and its multiples. If all you want to do is store an exact amount of currency, you can use an integer which represents the amount of cents. (This is known as …

The second exponent is FF, which signals NaN according to Wikipedia, and so the highest exponent is only FE. This is verified by the following C code snippet: unsigned int i = 0x7FFFFFFF; float f = * …

The ASCII encodings for digits 0 to 9 are 48 to 57 in decimal, or 30 to 39 in hexadecimal.

The reason we get a larger range is denormalized numbers. Generally speaking, floating point numbers have three physical parts: sign (1 bit), mantissa $M$ and exponent $e$. Most of the time we think o …

You are asking two questions: How can I prove that negative numbers are stored as pseudopositive? Why are negative numbers stored as pseudopositive? The answer to the first question is that no pro …

The idea is that a (not denormalized) floating point number is stored as $2^x \times 1.m_0 \ldots m_{k-1}$ (in binary), where $x$ is the exponent and $m_0 \ldots m_{k-1}$ is the mantissa of length $k$ …

If you drop the leading zeroes then $3,5,9,17$ and so on will all have the same representation. Don't forget that the number being represented need not be an integer.

It's not true that when all components of the exponent are 0 then the number must represent 0. A (normalized) floating point number is composed of two parts: sign, exponent, and mantissa. The value of …

It seems that WolframAlpha is using fixed-point representation. In effect, it is doubling your number and then representing it as two's-complement. In more detail, WolframAlpha is guessing that you a …

The fractional part is in binary: $(0.01010101\ldots)_2 = (0.333333\ldots)_{10}$. To convince yourself of this, try multiplying the binary $0.0101010101\ldots$ by $3$: you will get $0.111111\ldots = 1 …

The other answers explain why you're not getting 1.12 exactly (in brief, this is because floating point numbers can only exactly represent rationals whose denominator is a power of 2). The result of …

Suppose that a set of functions $F_0,\ldots,F_{N-1}$ satisfies the constraints, generating the sequence $x_0 = 0^N, x_1, \ldots, x_{2^{N-1}-1}$. When does a function $F_N$ satisfy the constraints? If …

Ignoring denormalized numbers, infinities and the like, floating point formats can store numbers of the form $$ \pm \left(1 + \frac{x}{2^N}\right) \times 2^y, $$ where $0 \leq x < 2^N-1$ and $-M \leq …

Two's complement makes sense when the two entities in question have the same "units" and the same "width". By width I mean that, say, if you're adding an N bit number and an M bit number, where N and …