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|.□