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\(\forall a \in \mathbf{Z},(a-1) / a\) is not an integer.

Short Answer

Expert verified
To show that for any integer \(a\), the fraction \((a-1) / a\) is not an integer, we consider whether \((a-1)\) is divisible by \(a\). If there existed an integer \(k\) such that \(a-1 = ak\), then we would have \(k = \frac{a-1}{a}\), which would mean that \(\frac{a-1}{a}\) is an integer. However, this contradicts the given statement that \((a-1) / a\) is not an integer. Thus, there is no such \(k\), and we conclude that \((a-1) / a\) is not an integer for all \(a \in \mathbf{Z}\).

Step by step solution

01

Identify the Given Information

We are given the set of integers \(\mathbf{Z}\) and a statement that needs to be proved for all integers \(a\): \((a-1) / a\) is not an integer.
02

Test Divisibility

In order for \((a-1) / a\) to be an integer, the numerator (\(a-1\)) must be divisible by the denominator (\(a\)) with no remainder. We can say that \(a-1\) is divisible by \(a\) if there exists an integer \(k\) such that \(a-1 = ak\).
03

Find the Counterexample

Suppose that there exists an integer \(k\) such that \(a-1 = ak\). We can rearrange the equation to find the value of \(k\): \(k = \frac{a-1}{a}\) Since \(k\) is an integer, \(\frac{a-1}{a}\) must also be an integer. However, this contradicts the given statement in which \((a-1) / a\) is, in fact, not an integer. Therefore, there cannot be such a \(k\), and we can conclude that \((a-1) / a\) is not an integer for all \(a \in \mathbf{Z}\).

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Most popular questions from this chapter

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