3.1 The absolute value
Before we look at subsets of \(\mathbb{Q}\), we introduce the notion of absolute value.
For \(x\in \mathbb{Q}\), the absolute value or modulus of \(x\), denoted \(|x|\), is defined by \[|x| := \begin{cases}x &\text{ if } x\geq 0;\\ -x &\text{ if } x<0.\end{cases}\]
It is often helpful to think of the absolute value \(|x|\) as the distances between the point \(x\) and the origin \(0\). Likewise, \(|x-y|\) is the distance between the points \(x\) and \(y\).
For any \(x, y \in \mathbb{Q}\)
\(|x|\geq 0\) with \(|x| = 0\) if and only if \(x=0\);
\(|xy| = |x||y|\);
\(|x^2| = |x|^2 = x^2\).
Exercise
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Statement Show that \(a^2+b^2\geq 2ab\) for any \(a,b \in \mathbb{Q}\).
Solution Let \(a,b\in\mathbb{Q}\). We have that \((a-b)^2\geq 0\), so [expanding the bracket] \(a^2-2ab+b^2\geq 0\). Rearranging, this gives \(a^2+b^2 \geq 2ab\).
For all \(x,y \in \mathbb{Q}\) we have \(|x+y|\leq |x|+|y|\).
We prove this by case by case analysis. First note that for all \(x\in\mathbb{Q}\) we have \(x\leq |x|\) and \(-x\leq |x|\). Let \(x,y \in \mathbb{Q}\).
Case 1: Suppose \(x\geq -y\) then \(x+y\geq 0\) and so \(|x+y| = x+y \leq |x|+|y|\).
Case 2: Suppose \(x< -y\) then \(x+y<0\) and so \(|x+y|=-(x+y)=-x+(-y)\leq |x|+|y|\).□