Floating-Point Representation and Values

This question explores the behavior of floating-point arithmetic when performing operations on numbers of significantly different magnitudes, using a 32-bit representation (commonly IEEE 754 single-precision). This representation has limitations in precision, which can lead to specific outcomes in calculations.

The given values for the variables are:

Step-by-Step Operation Execution

We will trace the values of X and Y through the given sequence of operations:

1. $X = A + B$
2. $Y = A + C$
3. $X = X + C$
4. $Y = Y + B$

Operation 1: Calculating X = A + B

The first step is to compute $X = A + B$. Substituting the values of $A$ and $B$:

$X = (2.0 \times 10^{30}) + (-2.0 \times 10^{30})$

When a number is added to its exact negative counterpart, the mathematical result is zero. In floating-point arithmetic, especially with numbers of equal magnitude and opposite sign, the result is typically zero. Therefore:

$X = 0.0$

Operation 2: Calculating Y = A + C

The next operation is $Y = A + C$. Here, $A$ ($2.0 \times 10^{30}$) is a very large number, while $C$ ($1.0$) is very small. This difference in magnitude is crucial.

Floating-point addition requires aligning the numbers by their exponents. The number with the smaller exponent must be shifted to match the exponent of the number with the larger exponent. The number $C = 1.0$ has an exponent of $0$ ($1.0 \times 2^0$). The number $A = 2.0 \times 10^{30}$ has a much larger exponent (approximately $100$ in base 2).

When $C$ is shifted to match the scale of $A$, it becomes an extremely small number relative to $A$. This tiny value, when added to the mantissa of $A$, falls beyond the representable precision (23 bits after the binary point in a 32-bit float). This phenomenon is known as loss of significance.

As a result, the addition of $C$ to $A$ does not change the value of $A$ in the floating-point representation. So, $Y$ will effectively be equal to $A$:

$Y = A = 2.0 \times 10^{30}$

Operation 3: Calculating X = X + C

We now update $X$. Its current value is $0.0$ from the first operation.

$X = 0.0 + 1.0$

This is a simple addition of zero and one, which results in:

$X = 1.0$

Operation 4: Calculating Y = Y + B

Finally, we update $Y$. The current value of $Y$ is approximately $2.0 \times 10^{30}$, and $B$ is $-2.0 \times 10^{30}$.

$Y = (2.0 \times 10^{30}) + (-2.0 \times 10^{30})$

Similar to the first operation, adding a number to its exact negative counterpart results in zero. Since $Y$ holds a value very close to $A$, adding $B$ (which is $-A$) results in zero:

$Y = 0.0$

Final Values of X and Y

After completing all the specified operations, the final computed values are:

X = 1.0

Y = 0.0