Given below are two statements: One is labelled as Assertion A and the other labelled as reason R.
Assertion A: During the boiling of water having temporary hardness, Mg(HCO$$_3$$)$$_2$$ is converted to MgCO$$_3$$.
Reason R: The solubility product of Mg(OH)$$_2$$ is greater than that of MgCO$$_3$$.
In the light of the above statements, choose the most appropriate answer from the options given below:

When water with temporary hardness (containing Mg(HCO$$_3$$)$$_2$$) is boiled, Mg(HCO$$_3$$)$$_2$$ decomposes. However, unlike Ca(HCO$$_3$$)$$_2$$ which forms CaCO$$_3$$ precipitate, Mg(HCO$$_3$$)$$_2$$ on boiling actually produces Mg(OH)$$_2$$ rather than MgCO$$_3$$. This is because in hot water, Mg(OH)$$_2$$ is less soluble and precipitates preferentially. So Assertion A (that Mg(HCO$$_3$$)$$_2$$ is converted to MgCO$$_3$$) is false.

For Reason R: the solubility product of MgCO$$_3$$ ($$K_{sp} \approx 6.8 \times 10^{-6}$$) is much larger than that of Mg(OH)$$_2$$ ($$K_{sp} \approx 5.6 \times 10^{-12}$$). So the statement that $$K_{sp}$$ of Mg(OH)$$_2$$ is greater than that of MgCO$$_3$$ is false.

Since both A and R are false, the correct option is (D).

The answer is $$\boxed{\text{A and R both are false}}$$.