Conditional divisibility with if-then

1. A student was asked to prove, for \(p \in \mathbb { N }\), that
   "if \(p ^ { 3 }\) is a multiple of 3 , then \(p\) must be a multiple of 3 "

The start of the student's proof by contradiction is shown in the box below. Assumption:
There exists a number \(p , p \in \mathbb { N }\), such that \(p ^ { 3 }\) is a multiple of 3 , and \(p\) is NOT a multiple of 3 Let \(p = 3 k + 1 , k \in \mathbb { N }\). $$\text { Consider } \begin{aligned} p ^ { 3 } = ( 3 k + 1 ) ^ { 3 } & = 27 k ^ { 3 } + 27 k ^ { 2 } + 9 k + 1 \\ & = 3 \left( 9 k ^ { 3 } + 9 k ^ { 2 } + 3 k \right) + 1 \quad \text { which is not a multiple of } 3 \end{aligned}$$

1. Show the calculations and statements that are required to complete the proof.
2. Hence prove, by contradiction, that \(\sqrt [ 3 ] { 3 }\) is an irrational number.

9. (i) Relative to a fixed origin \(O\), the points \(A , B\) and \(C\) have position vectors \(\mathbf { a } , \mathbf { b }\) and \(\mathbf { c }\) respectively. Points \(A , B\) and \(C\) lie in a straight line, with \(B\) lying between \(A\) and \(C\).
Given \(A B : A C = 1 : 3\) show that $$\mathbf { c } = 3 \mathbf { b } - 2 \mathbf { a }$$ (ii) Given that \(n \in \mathbb { N }\), prove by contradiction that if \(n ^ { 2 }\) is a multiple of 3 then \(n\) is a multiple of 3

16. Prove by contradiction that if \(n ^ { 2 }\) is a multiple of \(3 , n\) is a multiple of 3 .