Ieee 754 floating point converter

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IEEE-754 floating point numbers. IEEE-754 - floating point converter; Addition Subtraction (IEEE-754) Multiplication (IEEE-754) Division (IEEE-754) To convert a number with a

Base Convert: IEEE 754 Floating Point

Find More Calculator ☟ This IEEE 754 single precision calculator converts a hexadecimal representation of a floating-point number into its decimal value, making it a useful tool for programmers and engineers dealing with binary data.Historical BackgroundThe IEEE 754 standard was established to provide a uniform representation of floating-point numbers in computer systems. It ensures consistency across different platforms and programming languages.Calculation FormulaThe conversion from hexadecimal to decimal using IEEE 754 format follows these steps:Convert the hexadecimal to binary.Extract the sign bit, exponent, and mantissa.Calculate the value using the formula:\[\text{Value} = (-1)^{\text{sign}} \times (1 + \text{mantissa}) \times 2^{(\text{exponent} - 127)}\]Example CalculationFor the hexadecimal value C6400000:Convert to binary: 11000110010000000000000000000000Sign bit: 1 (negative)Exponent: 10001100 (binary) = 140 (decimal), so exponent = 140 - 127 = 13Mantissa: 1.100100 (binary) = 1 + 0.5 + 0.125 = 1.5625Value: \(-1 \times 1.5625 \times 2^{13} = -12800\)Importance and Usage ScenariosThis tool is vital in fields such as computer science, engineering, and any area where precise floating-point calculations are necessary.Common FAQsWhat is IEEE 754?IEEE 754 is a standard for floating-point arithmetic that defines how binary representations of real numbers are stored and manipulated.Why is floating-point representation important?It allows for efficient and precise representation of a wide range of values, essential for scientific calculations and simulations.What happens if the hexadecimal value is invalid?The calculator will not perform a conversion and may display an error message or undefined value.

IEEE 754-Style Floating-Point Converter

Is same for (R1 + R2 ).Option 2: R1 = 1100 and R2 = 1010True,R1 = 1 1 0 0 = -(0100)= -4+ R2 = 1 0 1 0 = -(0110)= -6 -------------------------------------------- 0 1 1 0 = = -10 Here Overflow occurred because the sign bit is different for (R1 + R2 ).Option 3: R1 = 0011 and R2 = 0100False,R1 = 0 0 1 1 = +(0011)= +3+ R2 = 0 1 0 0 = +(0100)= +4 -------------------------------------------- 0 1 1 1 = +7 Here No overflow occurred, because the sign bit is the same for (R1 + R2 ).Option 4: R1 = 1001 and R2 = 1111False, R1 = 1 0 0 1 = -(0111) = -7+ R2 = 1 1 1 1 = -(0001) = -1 -------------------------------------------- 1 0 0 0 = = -8Here No overflow occurred, because the sign bit is the same for (R1 + R2 ).Hence the correct answer is R1 = 1100 and R2 = 1010. ‎Ieee 754 Floating Point Number Representation Question 5: If we decide to stay away from IEEE 754 format by making our Exponent field 10 bits wide and our Mantissa field 21 bits wide, then which of the following statement is TRUE? None of the aboveIt will provide less precision as there will be fewer Mantissa bitsIt will provide more precision as there will be fewer Mantissa bitsIt will not change the precisionAnswer (Detailed Solution Below) Option 2 : It will provide less precision as there will be fewer Mantissa bits The correct answer is option 2.Concept:The Institute of Electrical and Electronics Engineers created the IEEE Standard for Floating-Point Arithmetic (IEEE 754) in 1985 as a technical standard for floating-point calculation (IEEE). The standard addressed several issues encountered in various floating-point implementations.IEEE 754 has 3 basic components are Sign, exponent, and Mantissa.Given that,Exponent field =10 bitsMantissa field = 21 bits Sign= 1 bit (represents the positive number or negative number)Bias=Excess=2n-1 where n is the number of bits in the Exponent.To get Number = (-1)S x (1.M) x BE-BiasExplanation:Fewer mantissa bits mean less precision. The smallest change that can be represented in floating-point representation is called precision. In single precision, the mantissa is having 23 bits, and double-precision the mantissa is having 52 bits are required. Hence the less precision as there will be fewer Mantissa bits.Hence the correct answer is It will provide less precision as there will be fewer Mantissa bits. Top ‎Ieee 754 Floating Point Number Representation MCQ Objective Questions Given the following binary number in 32-bit (single precision) IEEE-754 format:00111110011011010000000000000000The decimal value closest to this floating-point number is 1.45 × 1011.45 × 10-12.27 × 10-12.27 × 101Answer (Detailed Solution Below) Option 3 : 2.27

IEEE-754 Floating Point Numbers Converter

Number is in unsigned representation, it's decimal value starts with 0. So Minimum value will be zero.Range of unsigned representation is 0 to 2i - 1.So, the mum value with i bits goes to 2i - 1.Fraction of value is in the form of 2(-i). So, when we take the value of i = 1, 2, 3 … n this range of fractional value goes like, 2-1, 2-2, 2-3, …So, it makes a GP series, with f bit maximum number possible is sum of GP series.Consider a = ½, r = ½Maximum value with f bits possible= \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \ldots \)= \(a \times \frac{{1 - {r^n}}}{{1 - r}}\)\(= \frac{1}{2}\;\frac{{1 - {{\left( {\frac{1}{2}} \right)}^f}}}{{1 - \frac{1}{2}\;}} = 1 - {2^{ - f}}\;\)So, maximum fractional value possible= maximum value with i bits + maximum value with f bits= 2i - 1 + 1 - 2-f= 2i - 2-fSo, require range will be 0 to 2i - 2-f. What is the range of the exponent E in IEEE 754 Double Precision (Binary64) format? -1022 ≤ E ≤ 1022-1023 ≤ E ≤ 1023-1023 ≤ E ≤ 1022-1022 ≤ E ≤ 1023Answer (Detailed Solution Below) Option 4 : -1022 ≤ E ≤ 1023 Option 4) -1022 ≤ E ≤ 1023 is the correct answer.Explanation:- According to the IEEE Std 754-2008 standard, the exponent field width of the binary 64 double-precision floating-point formats is 11 bits, which is compensated by an exponent bias of 1023. The standard also specifies that the maximum exponent is 1023, and the minimum is -1022.Important Points In IEEE 754 floating-point numbers, the exponent is biased in the engineering sense of the word - the value stored is offset from the actual value by the exponent bias, also called a biased exponent. Double precision may be chosen when the range or precision of single-precision would be insufficient. Double-precision floating-point format is a computer number format usually occupying 64 bits in computer memory. It represents a wide dynamic range of numeric values by using a floating radix point.. IEEE-754 floating point numbers. IEEE-754 - floating point converter; Addition Subtraction (IEEE-754) Multiplication (IEEE-754) Division (IEEE-754) To convert a number with a IEEE-754 floating point numbers: floating point numbers converter (IEEE-754) addition subtraction calculator (IEEE-754) multiplication calculator (IEEE-754) division calculator (IEEE-754) Other numeral systems: BCD code converter;

IEEE-754 Floating Point Converter - lokker.net

0x408000000xC08000000x834000000xC85800000Answer (Detailed Solution Below) Option 2 : 0xC0800000 Concept: In IEEE- 754 single precision format, a floating-point number is represented in 32 bits. Sign bit (MSB) Biased Exponent (E’) (8 bits) Normalized Mantissa (M’) (23 bits) Sign bit value 0 means positive number, and 1 means a negative number.The floating-point number can be obtained by formula: ± 1. M × 2(E-127)Data:Content of R1: 0x 42200000 (0x means Hexadecimal notation)Content of R2: 0x C1200000Calculation:Content of R1 in Hex (0x) is 42200000. After converting into binary, it can be represented in IEEE- 754 format as: 0 100 0010 0 010 0000 0000 0000 0000 0000 Sign bit is 0 i.e. the number is positiveBiased Exponent (E’) = 100 0010 0 = 132Normalized Mantissa (M) = 010 0000 0000 0000 0000 0000 = .25Therefore, the number in register R1 = + 1.25 * 2(132-127) = 1.25 × 32 = 40Content of R2 in Hex (0x) is C1200000. After converting into binary, it can be represented in IEEE- 754 format as: 1 100 0001 0 010 0000 0000 0000 0000 0000 Sign bit is 1 i.e. the number is negativeBiased Exponent (E’) = 100 0001 0 = 130Normalized Mantissa (M) = 010 0000 0000 0000 0000 0000 = .25Therefore, the number in register R1 = - 1.25 * 2(130-127) = -1.25 * 8 = -10R3 = R1/R2 = 40/-10 = -4Since the number is negative, Sign bit (MSB) = 1Converting 4 into binary of a floating point gives: (100.0)2Representing it into normalized form gives: (1.000000….) × 22Therefore, Mantissa is 23 bits of all 0sBiased Exponent (E’) = E+ 127 = 2+127 = 129 = (10000001)2It can be represented in IEEE- 754 format as: 1 100 0000 1 000 0000 0000 0000 0000 0000 Converting it into Hex format gives: 0x C0800000 The decimal floating-point number -40.1 represented using IEEE-754 32-bit representation and written in hexadecimal form is _____ 0xC22060000xC20066660xC20060000xC2206666Answer (Detailed Solution Below) Option 4 : 0xC2206666 Concept:32-bit floating-point representation of a binary number in IEEE- 754 is Sign (1 bit) Exponent (8 bit) Mantissa bit (23 bits) In IEEE-754 format, 32-bit (single precision)(-1)s × 1.M × 2E – 127Calculation:Convert: 40.1 to binaryStep 1: convert 40 2 40 2 20 0 2 10 0 2 5 0 2 2 1 2 1 0 0 1 ↑ (40)10 = (101000)2Step 2: convert .1 to binary0.1 × 2 = 0.2 (0)0.2 × 2 = 0.4 (0)0.4 × 0.2 = 0.8 (0)0.8 × 0.2 = 1.6 (1)0.6 × 0.2 = 1.2 (1)0.2 × 0.2 = 0.4 (0) and so onGiven binary number is(40.1)10 = (101000.000110011001100…)2(40.1)10 = 1.0100 0000 1100 1100 … × 25Signed (1 bit) = 1 (given number is negative)Exponent (8 bit) = 5 + 127 =

IEEE 754 Floating Point Converter - hhelwich.github.io

‎Ieee 754 Floating Point Number Representation Question 1: Arrange the following steps in the proper sequence for the process of floating-point representation in a computer system.(A) Normalize the significand to fit within the specified range.(B) Convert the decimal number to binary.(C) Determine the sign bit based on the original number.(D) Calculate the exponent based on the normalized significand.(E) Combine the sign bit, exponent, and significand to form the final floating-point representation.Choose the correct answer from the options given below: (C), (A), (D), (B), (E)(C), (B), (A), (D), (E)(B), (D), (A), (C), (E)(B), (C), (A), (D), (E)Answer (Detailed Solution Below) Option 4 : (B), (C), (A), (D), (E) The correct answer is (B), (C), (A), (D), (E)Key PointsThe correct sequence for the process of floating-point representation is: Convert the decimal number to binary (B) - The first step is to convert the decimal number into its binary equivalent. Determine the sign bit based on the original number (C) - Identify whether the number is positive or negative to set the sign bit. Normalize the significand to fit within the specified range (A) - Adjust the significand so that it fits within the required range for floating-point representation. Calculate the exponent based on the normalized significand (D) - Determine the exponent value that corresponds to the normalized significand. Combine the sign bit, exponent, and significand to form the final floating-point representation (E) - Finally, combine all components to create the complete floating-point representation.Therefore, the correct answer is: 4) (B), (C), (A), (D), (E). ‎Ieee 754 Floating Point Number Representation Question 2: A floating-point (FP) number is said to be normalized, if the most significant bit of the mantissa is 10-12Answer (Detailed Solution Below) Option 1 : 1 A floating-point (FP) number is said to be normalized, if the most significant bit of the mantissa is 1.If the exponent is all zeros, the floating point number is denormalized and the most significant bit of the mantissa is known to be zero.The floating number representation has four parts:1. The first part represents a single fixed point number is called the mantissa.2.The second part designates the position of the decimal point and is called the exponent.The mantissa always a positive number holds the significant digits of the floating point number.The exponent indicates the positive or negative power of the radix that the mantissa and sign should be multiplied by. ‎Ieee 754 Floating Point Number Representation Question 3: Consider three floating-point numbers A, B and C stored in registers RA, RB and RC, respectively as per IEEE-754 single-precision floating point format. The 32-bit content stored in these registers (in hexadecimal form) are as follows. RA= 0xC1400000 RB = 0x42100000 RC = 0x41400000 Which one of the

Convert to IEEE 754 floating point - YouTube

0 = +(0100)= +4 -------------------------------------------- 0 1 1 1 = +7 Here No overflow occurred, because the sign bit is the same for (R1 + R2 ).Option 4: R1 = 1001 and R2 = 1111False, R1 = 1 0 0 1 = -(0111) = -7+ R2 = 1 1 1 1 = -(0001) = -1 -------------------------------------------- 1 0 0 0 = = -8Here No overflow occurred, because the sign bit is the same for (R1 + R2 ).Hence the correct answer is R1 = 1100 and R2 = 1010. Consider three floating-point numbers A, B and C stored in registers RA, RB and RC, respectively as per IEEE-754 single-precision floating point format. The 32-bit content stored in these registers (in hexadecimal form) are as follows. RA= 0xC1400000 RB = 0x42100000 RC = 0x41400000 Which one of the following is FALSE? A + C = 0C = A + BB = 3C(B - C) > 0Answer (Detailed Solution Below) Option 2 : C = A + B The correct answer is option 2.Concept:IEEE single-precision floating-point:IEEE single-precision floating-point computer numbering format is a binary computing format that takes up 4 bytes (32 bits) of memory. Binary32 is the official name for the 32-bit base 2 formats in IEEE 754-2008. IEEE 754-1985 referred to it as single.IEEE single-precision format:Explanation:The given data,Decimal value =(-1)s x 1.M x 2Base Exponent -BiasBias value in IEEE single-precision format is 127RA = 1100 0001 0100 0000 0000 0000 0000 0000RA sign= 1RA Base Exponent =100 0001 0 = 130RA Mantisa = 100 0000 0000 0000 0000 0000 = 1.100 0000 0000.....Decimal value = (-1)1 x1.1 x2130-127 =-1.1x23= -1100 = (-12)10A=-12RB = 0100 0010 0001 0000 0000 0000 0000 0000RA sign= 0RA Base Exponent =100 0010 0= 132RA Mantisa = 001 0000 0000 0000 0000 0000 = 1.001 000000.....Decimal value = (-1)0 x1.001 x2132-127 =+1.001x25= + 100100 = (+36)10B=+36RC = 0100 0001 0100 0000 0000 0000 0000 0000RA sign= 0RA Base Exponent =100 0001 0= 130RA Mantisa =100 0000 0000 0000 0000 0000= 1.100 0000.....Decimal value = (-1)0 x1.1 x2130-127 =+1.1x23= + 1100 = (+12)10C=+12Option 1: A + C = 0True, A+C= -12+12=0Hence it is true.Option 2: C = A + BFalse, A+B= -12+36=+24it not equal to C. Hence it is false.Option 3: B = 3CTrue, B=3C =3x+12 =36 =Bit equal to B. Hence it is true.Option 4: (B - C) > 0True, (B-C) >0=(36-12)=24>0Hence it is true.Hence the correct answer is C = A + B. Consider three registers R1, R2 and R3 that store numbers in IEEE-754 single precision floating point format. Assume that R1 and R2 contain the values (in hexadecimal notation) 0x42200000 and 0xC1200000, respectively.If R3 \(= \frac{{R1}}{{R2}},\) what is the value stored in R3?

Convert IEEE-754 floating-point encoding to floating-point value

132∴ Exponent = (132)10 = (1000 0100)2Mantissa (23 bits ) = 0100 0000 1100 1100 1100 110 Sign (1 bit) Exponent (8 bit) Mantissa bit (23 bits) 1 1000 0100 0100 0000 1100 1100 1100 110 (1100 0010 0010 0000 0110 0110 0110 0110)2 = (C2206666)16(C2206666)16 = 0xC2206666 The format of the single-precision floating-point representation of a real number as per the IEEE 754 standard is as follows:Which one of the following choices is correct with respect to the smallest normalized positive number represented using the standard? exponent = 00000000 and mantissa = 00000000000000000000001exponent = 00000001 and mantissa = 00000000000000000000000exponent = 00000001 and mantissa = 00000000000000000000001​exponent = 00000001 and mantissa = 000000000000000000000001Answer (Detailed Solution Below) Option 2 : exponent = 00000001 and mantissa = 00000000000000000000000 Option 2) is correct answer.Concept:In IEEE- 754 single precision format, a floating-point number is represented in 32 bits. Sign bit (MSB) Biased Exponent (E’) (8 bits) Normalized Mantissa (M’) (23 bits) Sign bit value 0 means a positive number, and 1 means a negative number.The floating-point number can be obtained by formula: (-1)s × 1.M × 2E – 127Explanation: Smallest normalized positive number Sign bit Biased Exponent (E’) Normalized Mantissa (M’) 0 0000 0001 00000000000000000000000 Smallest normalized positive = (-1)0 × 1.00...0 × 21 – 127 = 2-126 ≈1.1755 × 10–38 Consider the following representation of a number in IEEE 754 single-precision floating point format with a bias of 127.S: 1 E: 10000001 F : 11110000000000000000000Here S, E and F denote the sign, exponent and fraction components of the floating point representation.The decimal value corresponding to the above representation (rounded to 2 decimal places) is ______Answer (Detailed Solution Below) -7.75 Concept:In IEEE- 754 single-precision format, a floating-point number is represented in 32 bits. Sign bit (MSB) Biased Exponent (E’) (8 bits) Mantissa (M’) (23 bits) Sign bit value 0 means a positive number, and 1 means a negative number.The floating-point number can be obtained by formula: (-1)s × 1.M × 2E – 127Explanation: Sign bit Biased Exponent (E’) Mantissa (M’) 1 10000001 11110000000000000000000 Exponent = 10000001 = 129decimal value = (-1)1 × 1.1111...0 × 2129 – 127 = (-1)1 ×1.1111 × 22 The decimal value corresponding to the above representation (rounded to 2 decimal places) is -7.75 The n-bit fixed-point representation of an unsigned real number X uses f bits for the fraction part. Let i = n - f. The range of decimal values for X in this representation is. 2-f to 2i2-f to (2i - 2-f) 0 to 2i0 to (2i - 2-f) Answer (Detailed Solution Below) Option 4 : 0 to (2i - 2-f) Diagram:i represents an integral part of the and f represents the fractional part of the number.Since, the n. IEEE-754 floating point numbers. IEEE-754 - floating point converter; Addition Subtraction (IEEE-754) Multiplication (IEEE-754) Division (IEEE-754) To convert a number with a

floating point - Convert signed IEEE 754 float to hexadecimal

× 10-1 Concept:32-bit floating-point representation of a binary number in IEEE- 754 is Sign (1 bit) Exponent (8 bit) Mantissa bit (23 bits) Calculation:Given binary number is00111110011011010000000000000000Here, sign bit is 0. So, number is positive. 0 01111100 11011010000000000000000 Exponent bits = E = 01111100 = 124 (in decimal)Mantissa bits M = 11011010000000000000000In IEEE-754 format, 32-bit (single precision) (-1)s × 1.M × 2E – 127 = (-1)0 × 1.1101101 × 2124 – 127= 1.1101101 × 2-3= (1 + 2-1 + 2-2 + 2-4 + 2-5 + 2-7) × 2-3= 0.231 = 2.31 × 10-1 ≈ 2.27 × 10-1 In IEEE floating point representation, the hexadecimal number 0xC0000000 corresponds to –3.0–1.0–4.0–2.0Answer (Detailed Solution Below) Option 4 : –2.0 Concept:32-bit floating-point representation of a binary number in IEEE- 754 is Sign (1 bit) Exponent (8 bit) Mantissa bit (23 bits) Calculation:Binary number is0xC0000000 = (11000000000000000000000000000000)2Here, the sign bit is 1. So, the number is negative. 1 10000000 00000000000000000000000 Exponent bits = E = 10000000 = 128 (in decimal)Mantissa bits M = 00000000000000000000000In IEEE-754 format, 32-bit (single precision)(-1)s × 1.M × 2E – 127= (-1)1 × 1. 0 × 2128 – 127= -1 × 1.0 × 2= -2In IEEE floating-point representation, the hexadecimal number 0xC0000000 corresponds to -2. Let R1 and R2 be two 4-bit registers that store numbers in 2’s complement form. For the operation R1 + R2, which one of the following values of R1 and R2 gives an arithmetic overflow? R1 = 1011 and R2 = 1110R1 = 1100 and R2 = 1010R1 = 0011 and R2 = 0100R1 = 1001 and R2 = 1111Answer (Detailed Solution Below) Option 2 : R1 = 1100 and R2 = 1010 The correct answer is option 2.Concept:Stored numbers in registers R1 and R2 are in 2's complement form. Register size is 4 bits. The range of numbers in 2's complement form is -8 to +7. If R1 + R2, the result is out of the above range, then it is overflow.The given data,Given two four-bit registers R1 and R2.Option 1: R1 = 1011 and R2 = 1110False, R1 = 1 0 1 1 = -(0101)= -5+ R2 = 1 1 1 0 = -(0010)= -2----------------------------------------------- 1 0 0 1 = = -7 Here No overflow occurred, because sign bit is same for (R1 + R2 ).Option 2: R1 = 1100 and R2 = 1010True,R1 = 1 1 0 0 = -(0100)= -4+ R2 = 1 0 1 0 = -(0110)= -6 -------------------------------------------- 0 1 1 0 = = -10 Here Overflow occurred because the sign bit is different for (R1 + R2 ).Option 3: R1 = 0011 and R2 = 0100False,R1 = 0 0 1 1 = +(0011)= +3+ R2 = 0 1 0

IEEE 754 floating point - tools.timodenk.com

Unnaturally saturated orange hues. We might attempt to correct the white and orange areas by making the image darker. However, if the original image has been stored in a low-dynamic-range file format, for example, JPEG, darkening produces a rather ugly image. The areas around the flames become gray, as in Figure 26-2. If the original image has been stored in an HDR file format such as OpenEXR, which preserves bright pixel values rather than clipping them at 1.0, then darkening produces an image that still looks natural. See Figure 26-3. Figure 26-3 High-Dynamic-Range Version of , Made Darker 26.1.2 A "Half" Format Early in 2003, ILM released a new HDR file format with 16-bit floating-point color-component values. Because the IEEE 754 floating-point specification does not define a 16-bit format, ILM created a half format that matches NVIDIA's 16-bit format. The half type provides an excellent storage structure for high-dynamic-range image content. This type is directly supported in the OpenEXR format. The 16-bit, or "half-precision," floating-point format is modeled after the IEEE 754 single-precision and double-precision formats. A half-precision number consists of a sign bit, a 5-bit exponent, and a 10-bit mantissa. The smallest and largest possible exponent values are reserved for representing zero, denormalized numbers, infinities, and NaNs. In OpenEXR's C++ implementation, numbers of type half generally behave like the built-in C++ floating-point types, float and double. The half, float, and double types can be mixed freely in arithmetic expressions. Here are a few examples: half a(3.5);float b(a + sqrt(a));a. IEEE-754 floating point numbers. IEEE-754 - floating point converter; Addition Subtraction (IEEE-754) Multiplication (IEEE-754) Division (IEEE-754) To convert a number with a IEEE-754 floating point numbers: floating point numbers converter (IEEE-754) addition subtraction calculator (IEEE-754) multiplication calculator (IEEE-754) division calculator (IEEE-754) Other numeral systems: BCD code converter;

IEEE-754 Floating Point Converter - h-schmidt.net

Following is FALSE? A + C = 0C = A + BB = 3C(B - C) > 0Answer (Detailed Solution Below) Option 2 : C = A + B The correct answer is option 2.Concept:IEEE single-precision floating-point:IEEE single-precision floating-point computer numbering format is a binary computing format that takes up 4 bytes (32 bits) of memory. Binary32 is the official name for the 32-bit base 2 formats in IEEE 754-2008. IEEE 754-1985 referred to it as single.IEEE single-precision format:Explanation:The given data,Decimal value =(-1)s x 1.M x 2Base Exponent -BiasBias value in IEEE single-precision format is 127RA = 1100 0001 0100 0000 0000 0000 0000 0000RA sign= 1RA Base Exponent =100 0001 0 = 130RA Mantisa = 100 0000 0000 0000 0000 0000 = 1.100 0000 0000.....Decimal value = (-1)1 x1.1 x2130-127 =-1.1x23= -1100 = (-12)10A=-12RB = 0100 0010 0001 0000 0000 0000 0000 0000RA sign= 0RA Base Exponent =100 0010 0= 132RA Mantisa = 001 0000 0000 0000 0000 0000 = 1.001 000000.....Decimal value = (-1)0 x1.001 x2132-127 =+1.001x25= + 100100 = (+36)10B=+36RC = 0100 0001 0100 0000 0000 0000 0000 0000RA sign= 0RA Base Exponent =100 0001 0= 130RA Mantisa =100 0000 0000 0000 0000 0000= 1.100 0000.....Decimal value = (-1)0 x1.1 x2130-127 =+1.1x23= + 1100 = (+12)10C=+12Option 1: A + C = 0True, A+C= -12+12=0Hence it is true.Option 2: C = A + BFalse, A+B= -12+36=+24it not equal to C. Hence it is false.Option 3: B = 3CTrue, B=3C =3x+12 =36 =Bit equal to B. Hence it is true.Option 4: (B - C) > 0True, (B-C) >0=(36-12)=24>0Hence it is true.Hence the correct answer is C = A + B. ‎Ieee 754 Floating Point Number Representation Question 4: Let R1 and R2 be two 4-bit registers that store numbers in 2’s complement form. For the operation R1 + R2, which one of the following values of R1 and R2 gives an arithmetic overflow? R1 = 1011 and R2 = 1110R1 = 1100 and R2 = 1010R1 = 0011 and R2 = 0100R1 = 1001 and R2 = 1111Answer (Detailed Solution Below) Option 2 : R1 = 1100 and R2 = 1010 The correct answer is option 2.Concept:Stored numbers in registers R1 and R2 are in 2's complement form. Register size is 4 bits. The range of numbers in 2's complement form is -8 to +7. If R1 + R2, the result is out of the above range, then it is overflow.The given data,Given two four-bit registers R1 and R2.Option 1: R1 = 1011 and R2 = 1110False, R1 = 1 0 1 1 = -(0101)= -5+ R2 = 1 1 1 0 = -(0010)= -2----------------------------------------------- 1 0 0 1 = = -7 Here No overflow occurred, because sign bit