Consider three floating point numbers $A$, $B$ and $C$ stored in registers $\text{R}_\text{A}$, $\text{R}_\text{B}$ and $\text{R}_\text{C}$, respectively as per IEEE-754 single precision floating point format. The 32-bit content stored in these registers (in hexadecimal form) are as follows. Which one of the following is FALSE?

In IEEE-754 single-precision format, the 32 bits are divided as follows:

• 1 sign bit ($S$)

• 8 exponent bits ($E$) with bias $127$

• 23 mantissa/fraction bits ($M$)

The value is given by $(-1)^S \times 1.M \times 2^{E - 127}$.

1. For register $\text{R}_\text{A} = \text{0xC1400000}$:

   • In binary: 1 10000010 10000000000000000000000

   • Sign bit $S = 1$ (negative)

   • Exponent $E = (10000010)_2 = 130_{10} \implies E - 127 = 3$

   • Mantissa $M = 0.1_2 = 0.5_{10} \implies 1.M = 1.5$

   • Value $A = -1 \times 1.5 \times 2^3 = -1.5 \times 8 = -12$

2. For register $\text{R}_\text{B} = \text{0x42100000}$:

   • In binary: 0 10000100 00100000000000000000000

   • Sign bit $S = 0$ (positive)

   • Exponent $E = (10000100)_2 = 132_{10} \implies E - 127 = 5$

   • Mantissa $M = 0.001_2 = 0.125_{10} \implies 1.M = 1.125$

   • Value $B = +1 \times 1.125 \times 2^5 = 1.125 \times 32 = 36$

3. For register $\text{R}_\text{C} = \text{0x41400000}$:

   • In binary: 0 10000010 10000000000000000000000

   • Sign bit $S = 0$ (positive)

   • Exponent $E = (10000010)_2 = 130_{10} \implies E - 127 = 3$

   • Mantissa $M = 0.1_2 = 0.5_{10} \implies 1.M = 1.5$

   • Value $C = +1 \times 1.5 \times 2^3 = 1.5 \times 8 = 12$

Evaluating the options:

• (A) $A + C = -12 + 12 = 0$ (True)

• (B) $C = A + B \implies 12 = -12 + 36 = 24$, which is false ($12 \neq 24$)

• (C) $B = 3C \implies 36 = 3 \times 12 = 36$ (True)

• (D) $(B - C) = 36 - 12 = 24 > 0$ (True)

Therefore, statement B is FALSE.