Question

Use the method of direct proof to prove the following statements. If \(a\) is an integer and \(a^{2} \mid a\), then \(a \in\\{-1,0,1\\}\).

Short answer

The possible integer values such that \(a^{2} \mid a\) are -1, 0, and 1. All the other integer values won't satisfy the condition of \(a^{2} \mid a\).

Step-by-step solution

Understanding the statement

We are asked to prove that for any integer \(a\), if the square of that integer divides the integer, then that integer \(a\) must be -1, 0, or 1. So, we need to consider possible integer values of \(a\) and see if they satisfy \(a^{2} \mid a\).

Testing the dividing property for \(a^{2}\) and \(a\)

To show \(a^{2} \mid a\), we typically need to express \(a\) as a multiple of \(a^{2}\). In term of integers, \(a = xa^{2}\) where \(x\) is any integer. However, we're restricted in this case to \(x\) being 1/a since the only way for \(a^{2}\) to become \(a\), is by dividing by \(a\) itself.

Exploring the integer values of \(a\)

Now, we know that \(x = 1/a\) and it is said that \(x\) is also an integer. Since \(x\) is an integer, \(1/a\) must produce an integer. This limits the possible values of \(a\) to just -1, 0, and 1 since those are the only integers that can make \(1/a\) an integer.

Verifying the results

For case \(a = 1\), \(1 = 1*1\) which is true; for \(a = -1\), \(-1 = 1*(-1)\) which is true; and for \(a = 0\), \(0 = 0*0\) which is also true. Hence, it's proved that \(a\) must be either -1, 0, or 1 if \(a^{2} \mid a\).