g) Use the truth table to show that the proportion is a tautology $$ [p \Lambda(p \rightarrow q] \rightarrow q $$. (4mks) h) Prove that $$ \forall \mu \in \mathbb{N} n^{3}-n $$ is divisible by 3 . i) Given that $$ f(x)=3 x^{2}+4 $$ find i. $$ f(2) $$ ii. $$ f^{-1}(7) $$

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a) To show that the proportion $$ [p \wedge (p \rightarrow q)] \rightarrow q $$ is a tautology using a truth table, we need to construct a truth table for the proposition and show that the proposition is always true regardless of the truth values of $$ p $$ and $$ q $$.

Let's construct the truth table for the proposition $$ [p \wedge (p \rightarrow q)] \rightarrow q $$:

| $$ p $$ | $$ q $$ | $$ p \wedge (p \rightarrow q) $$ | $$ [p \wedge (p \rightarrow q)] \rightarrow q $$ |
|--------|--------|----------------------------------|---------------------------------------------|
| T      | T      | T                                | T                                           |
| T      | F      | F                                | T                                           |
| F      | T      | F                                | T                                           |
| F      | F      | T                                | T                                           |

From the truth table, we can see that the proposition $$ [p \wedge (p \rightarrow q)] \rightarrow q $$ is always true, regardless of the truth values of $$ p $$ and $$ q $$. Therefore, the proportion is a tautology.

b) To prove that $$ \forall \mu \in \mathbb{N} n^{3}-n $$ is divisible by 3, we need to show that for any natural number $$ n $$, $$ n^{3}-n $$ is divisible by 3.

Let's prove this by mathematical induction:

Base case: For $$ n = 1 $$, $$ 1^{3}-1 = 0 $$, which is divisible by 3.

Inductive step: Assume that for some natural number $$ k $$, $$ k^{3}-k $$ is divisible by 3. We need to show that $$ (k+1)^{3}-(k+1) $$ is also divisible by 3.

Expanding $$ (k+1)^{3}-(k+1) $$:
$$ (k+1)^{3}-(k+1) = k^{3}+3k^{2}+3k+1-k-1 = k^{3}+3k^{2}+2k $$

Since $$ k^{3}+3k^{2}+2k $$ is divisible by 3 by the inductive hypothesis, we have shown that $$ (k+1)^{3}-(k+1) $$ is divisible by 3.

Therefore, by mathematical induction, $$ \forall \mu \in \mathbb{N} n^{3}-n $$ is divisible by 3.

c) Given that $$ f(x)=3x^{2}+4 $$, we need to find:
i. $$ f(2) $$
ii. $$ f^{-1}(7) $$

i. To find $$ f(2) $$, we substitute $$ x = 2 $$ into the function $$ f(x) $$:
$$ f(2) = 3(2)^{2}+4 = 3(4)+4 = 12+4 = 16 $$

ii. To find $$ f^{-1}(7) $$, we need to find the value of $$ x $$ such that $$ f(x) = 7 $$. This means solving the equation $$ 3x^{2}+4 = 7 $$ for $$ x $$.

Solving the equation:
$$ 3x^{2}+4 = 7 $$
$$ 3x^{2} = 3 $$
$$ x^{2} = 1 $$
$$ x = \pm 1 $$

Therefore, $$ f^{-1}(7) = \pm 1 $$.
a) The proportion $$ [p \wedge (p \rightarrow q)] \rightarrow q $$ is a tautology. b) $$ \forall \mu \in \mathbb{N} n^{3}-n $$ is divisible by 3. c) i. $$ f(2) = 16 $$; ii. $$ f^{-1}(7) = \pm 1 $$.

g) Use the truth table to show that the proportion is a tautology \( [p \Lambda(p \rightarrow q] \rightarrow q \). (4mks) h) Prove that \( \forall \mu \in \mathbb{N} n^{3}-n \) is divisible by 3 . i) Given that \( f(x)=3 x^{2}+4 \) find i. \( f(2) \) ii. \( f^{-1}(7) \)

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