Question

Use the method of direct proof to prove the following statements. Suppose \(a\) and \(b\) are integers. If \(a \mid b,\) then \(a \mid\left(3 b^{3}-b^{2}+5 b\right)\).

Short answer

The statement is true. If 'a' divides 'b', then 'a' will also divide the expression \(3b^{3}-b^{2}+5b\). The proof of this fact can be done by direct substitution and rearrangement.

Step-by-step solution

Define divisibility

Two integers \(a\) and \(b\) where \(a \neq 0\), 'a' is said to divide 'b' if there is an integer 'k' such that \(b = a \cdot k\). So if we are given that \(a \mid b\), then there exists an integer 'k' such that \(b = a \cdot k\).

Substitute b to the expression

We can substitute \(b = a \cdot k\) to the expression \((3b^{3}-b^{2}+5b)\). After substitution we get: \(3(a \cdot k)^{3}-(a \cdot k)^{2}+5(a \cdot k)\). Which simplifies to: \(3a^{3}k^{3}-a^{2}k^{2}+5ak\).

Rearrange to prove divisibility

To prove the resulting expression is divisible by \(a\), we rearrange the expression so that it becomes more apparent: \(a[3a^{2}k^{3}-a k^{2}+5k]\). This shows that our resulting expression clearly contains 'a' as a factor, so it stands to reason that it is divisible by 'a' provided that \(k\) is an integer. Therefore, if \(a \mid b\), then \(a \mid \left(3b^{3}-b^{2}+5b\right)\).