Number System Practice Questions

The floating point format consists of a $1$-bit sign, $7$-bit exponent (excess-$64$), and $8$-bit mantissa. For the number $+0.239 \times 2^{13}$:

1. Sign bit: Positive $\implies 0$.
2. Exponent: Excess-$64$ representation of $13$ is $13 + 64 = 77_{10} = 1001101_2$.
3. Mantissa: Convert $0.239_{10}$ to binary $\implies 0.00111101_2$.
4. Concatenation: Sign ($0$) + Exponent ($1001101$) + Mantissa ($00111101$) $= 0100110100111101_2$.
5. Hex conversion: Grouping $4$ bits $\to (0100)_2=4, (1101)_2=D, (0011)_2=3, (1101)_2=D$.
6. The result is $4D3D$.

The conversion of the numbers to base 10 yields the equation $7x + 3 = 5y + 4$, which simplifies to $7x - 5y = 1$. Additionally, the digits must be smaller than the bases, implying $x > 7$ and $y > 5$. Testing the given options against the linear equation:

• For option D ($x=8, y=11$): $7(8) - 5(11) = 56 - 55 = 1$.
  This pair satisfies both the linear equation $7x - 5y = 1$ and the required base constraints ($8 > 7$ and $11 > 5$). Other options do not result in a value of 1.

In three-digit floating-point arithmetic, operations are performed sequentially, with rounding applied to maintain three significant digits.

For $(113. + -111.) + 7.51$:
First, $113. + (-111.) = 2.00$.
Then, $2.00 + 7.51 = 9.51$.

For $113. + (-111. + 7.51)$:
First, $-111. + 7.51 = -103.49$, which rounds to $-103$ when limited to three significant digits.
Then, $113. + (-103.) = 10.0$.

Thus, the results are $9.51$ and $10.0$ respectively.

A represents a negative 8-bit 2's complement number: $1111\ 1010_2 = -2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^1 = -128 + 122 = -6_{10}$.
B represents a positive 8-bit number: $0000\ 1010_2 = 2^3 + 2^1 = 10_{10}$.
The product is $(-6) \times 10 = -60_{10}$.
To convert $-60_{10}$ to 8-bit 2's complement, first identify $+60_{10} = 0011\ 1100_2$.
Performing the 2's complement operation by inverting the bits ($1100\ 0011_2$) and adding $1$ results in $1100\ 0100_2$.

To convert $(123456)_8$ to base 16 and base 4, first convert the octal digits to their 3-bit binary equivalents:
$(123456)_8 = (001\ 010\ 011\ 100\ 101\ 110)_2 = (1010011100101110)_2$.
For base 16, group the bits into sets of 4 from the right:
$(1010)_2 = A$, $(0111)_2 = 7$, $(0010)_2 = 2$, and $(1110)_2 = E$, yielding $(A72E)_{16}$.
For base 4, group the bits into sets of 2 from the right:
$(10)_2=2, (10)_2=2, (01)_2=1, (11)_2=3, (00)_2=0, (10)_2=2, (11)_2=3, (10)_2=2$, resulting in $(22130232)_4$.
Both conversions confirm that Option A is correct.

In 4-bit 2's complement arithmetic, the representable range is $[-8, +7]$. An overflow occurs if the addition result falls outside this range or if the sign of the result is incorrect.

I. $1100 (-4) + 1100 (-4) = 1000 (-8)$, which is within the range; no overflow.
II. $0011 (+3) + 0111 (+7) = 1010_2 (-6)$, which exceeds $+7$ and produces a negative result, indicating overflow.
III. $1111 (-1) + 0111 (+7) = 0110_2 (+6)$, which is within the range; no overflow.

Thus, only condition II results in an overflow.

Based on the provided tables, we observe that $a \times x = x$ and $a \times y = y$ (the first row/column of multiplication acts as an identity for $a$). Thus, the first equation $(a \times x) + (a \times y) = c$ simplifies to $x + y = c$.

From the addition table, the pairs $(x, y)$ satisfying $x + y = c$ are $(a, c)$, $(b, c)$, and $(c, b)$. We test these in the second equation $(b \times x) + (c \times y) = c$:

1. For $(a, c)$: $(b \times a) + (c \times c) = b + b = b \neq c$.
2. For $(b, c)$: $(b \times b) + (c \times c) = c + b = c$. (Valid)
3. For $(c, b)$: $(b \times c) + (c \times b) = a + c = c$. (Valid)

There are exactly 2 solutions: $(b, c)$ and $(c, b)$.

To determine divisibility, convert the numbers from 2's complement to decimal. The divisor $11111011_2$ corresponds to $-5$ (\text{invert } 11111011 + 1 = 00000101 = 5$). Option A ($11100111_2$) corresponds to$-25$(\text{invert } 11100111 + 1 = 00011001 = 25$). Since $-25 \div -5 = 5$, Option A is divisible by the divisor. Calculating the other options results in $-28$, $-41$, and $-37$, none of which are divisible by $-5$.

The floating-point value is defined as $V = (-1)^{s}(1 + m \cdot 2^{-9}) \cdot 2^{e-31}$. The difference $\Delta$ between two successive representable numbers with the same exponent $e$ is determined by the mantissa precision:
$\Delta = ((1 + (m+1) \cdot 2^{-9}) \cdot 2^{e-31}) - ((1 + m \cdot 2^{-9}) \cdot 2^{e-31}) = 2^{-9} \cdot 2^{e-31} = 2^{e-40}$
To maximize this difference, we must choose the maximum possible value for the exponent $e$. Given 6 bits for the exponent ($2^6 = 64$ combinations), the maximum value excluding the special case $111111_2$ (which is $63$) is $111110_2 = 62$. Substituting this maximum exponent $e = 62$ into the difference formula yields:
$\Delta_{max} = 2^{62-40} = 2^{22}$

An $n$-bit 2's complement system represents integers within the range $[-2^{n-1}, 2^{n-1} - 1]$. Overflow occurs only when adding two numbers of the same sign results in a value outside this range. When adding a positive value $P$ ($0 \le P < 2^{n-1}$) and a negative value $N$ ($-2^{n-1} \le N < 0$), the result $S = P + N$ is bounded such that $N \le S < P$. Since both $P$ and $N$ are within the valid representation limits, their sum $S$ is mathematically guaranteed to remain within the range $[-2^{n-1}, 2^{n-1} - 1]$. Consequently, adding numbers with opposite signs can never result in overflow.

To convert a decimal fraction to binary, repeatedly multiply the fractional part by $2$ and record the integer part.
First, $0.25 \times 2 = 0.5$, resulting in an integer part of $0$.
Next, $0.5 \times 2 = 1.0$, resulting in an integer part of $1$.
Since the fractional part is now $0$, the process terminates.
Reading the integer parts in order gives the binary representation $0.01_2$.

To represent $-15$ in 2's complement, we use $5$ bits. First, convert the absolute value $+15$ to a $5$-bit binary string: $01111_{2}$.
Next, find the 1's complement by inverting all the bits: $10000_{2}$.
Then, add $1$ to this value to calculate the 2's complement: $10000_{2} + 1_{2} = 10001_{2}$.
Validating the result: $-1 \cdot 2^{4} + 0 \cdot 2^{3} + 0 \cdot 2^{2} + 0 \cdot 2^{1} + 1 \cdot 2^{0} = -16 + 1 = -15$.
This sequence of operations confirms the representation is $10001$.

Sign extension is the process of replicating the most significant bit (the sign bit) of a two's complement binary number into higher-order bit positions when increasing its bit width. This operation preserves the numerical value of the number, ensuring the sign interpretation remains consistent after expansion. For instance, converting an $8$-bit signed value like $11000000_2$ to $16$-bit requires filling the upper $8$ bits with the sign bit $1$, resulting in $11111111 \ 11000000_2$. Consequently, this technique is fundamentally required when converting a signed integer from a smaller data size to a larger one.

To find the 2's complement representation of $(-539)_{10}$, we first convert the absolute value $|539|_{10}$ to hexadecimal: $539 \div 16 = 33$ remainder $11$ ($B$), $33 \div 16 = 2$ remainder $1$, and $2 \div 16 = 0$ remainder $2$, resulting in $21B_{16}$.
Representing $21B_{16}$ as a 12-bit binary number gives $001000011011_2$.
To find the 2's complement, invert the bits to get $110111100100_2$ and add $1$, resulting in $110111100101_2$.
Convert this binary number back to hexadecimal by grouping every 4 bits: $1101_2 = D$, $1110_2 = E$, and $0101_2 = 5$.
The resulting hexadecimal representation is $DE5_{16}$.

Initial values are $A = 2.0 \times 10^{30}$, $B = -2.0 \times 10^{30}$, and $C = 1.0$.
First, $X := A + B = 2.0 \times 10^{30} + (-2.0 \times 10^{30}) = 0.0$.
Next, $Y := A + C = 2.0 \times 10^{30} + 1.0$. Due to 32-bit floating point precision, $C$ is below the machine epsilon relative to $A$, so $A + C = A = 2.0 \times 10^{30}$.
Then, $X := X + C = 0.0 + 1.0 = 1.0$.
Finally, $Y := Y + B = 2.0 \times 10^{30} + (-2.0 \times 10^{30}) = 0.0$.
Thus, the final values are $X = 1.0$ and $Y = 0.0$.

To represent decimal $43$ in binary, we express it as the sum of powers of $2$: $43 = 32 + 8 + 2 + 1 = 2^5 + 2^3 + 2^1 + 2^0$, resulting in $101011_2$.
To store this in an $8$-bit signed format, we pad the leading bits with $0$s to signify a positive number.
The resulting $8$-bit binary representation is $00101011$.
For positive integers, the $2$'s complement representation is equivalent to the binary representation, confirming the correct option.

In a system with $n$ bits, the Sign-magnitude representation uses the most significant bit as a sign, resulting in $+0$ ($00\dots0$) and $-0$ ($10\dots0$). Similarly, in 1's complement, negative numbers are obtained by inverting bits, yielding $+0$ ($00\dots0$) and $-0$ ($11\dots1$). In contrast, 2's complement calculates $-0$ by adding $1$ to the inverted bits of $+0$, resulting in an overflow that simplifies to a unique $00\dots0$. Consequently, both Sign-magnitude and 1's complement possess two distinct representations for zero.

[CORRECT_OPTION: A, C]

The decimal number $57$ in 8-bit binary is $00111001_2$, so the $2$'s complement representation of $-57$ is $11000111_2$.
To perform Booth's encoding, append an implicit bit $y_{-1} = 0$ to the right, yielding $110001110$.
We apply the encoding rule $y_{i-1} - y_i$ to each adjacent pair $(y_i, y_{i-1})$ from the least significant bit (LSB) to the most significant bit (MSB):
Pair $(y_0, y_{-1}): 0-1 = -1$, $(y_1, y_0): 1-1 = 0$, $(y_2, y_1): 1-1 = 0$, $(y_3, y_2): 1-0 = +1$, $(y_4, y_3): 0-0 = 0$, $(y_5, y_4): 0-0 = 0$, $(y_6, y_5): 0-1 = -1$, $(y_7, y_6): 1-1 = 0$.
Arranging these results from MSB to LSB gives $0, -1, 0, 0, 1, 0, 0, -1$, which corresponds to the string $0-100+100-1$.

The octal representation $(342)_8$ converted to binary is $3 \to 011_2$, $4 \to 100_2$, and $2 \to 010_2$, resulting in the 8-bit binary pattern $11100010_2$.
Since the most significant bit is $1$, the value is negative in 2's complement representation.
To find the magnitude, take the 2's complement of $11100010_2$: invert the bits to get $00011101_2$, then add $1$ to obtain $00011110_2$.
Converting $00011110_2$ to decimal gives $2^4 + 2^3 + 2^2 + 2^1 = 16 + 8 + 4 + 2 = 30$.
Including the negative sign, the decimal equivalent is $-30$.

The range of signed four-digit numbers is $-9999$ to $9999$. In character form, storing a number like $-9999$ requires 5 bytes (1 byte for the sign and 4 bytes for the digits).
In contrast, these numbers can be stored as 16-bit integers, which require only 2 bytes of storage space.
The percentage of reduction is calculated using the formula:
$\text{Reduction} = \frac{\text{Initial Size} - \text{New Size}}{\text{Initial Size}} \times 100\%$
Substituting the values:
$\text{Reduction} = \frac{5 - 2}{5} \times 100\% = \frac{3}{5} \times 100\% = 60\%$