If d(x, y) = max (|x| , |y|), x, y ∈ R, then show that d is not a metric on R.

Solution:

To show that d is not a metric on R, we need to find a counterexample to at least one of the three properties that a metric must satisfy:

1.    Non-negativity: d(x, y) ≥ 0 for all x, y ∈ R, and d(x, y) = 0 if and only if x = y.

2.    Symmetry: d(x, y) = d(y, x) for all x, y ∈ R.

3.    Triangle inequality: d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ R.

Let's first verify that property 1 is satisfied:

Non-negativity: For any x, y ∈ R, we have d(x, y) = max(|x|, |y|) ≥ 0 since |x| and |y| are both non-negative. Moreover, d(x, y) = 0 if and only if both |x| = 0 and |y| = 0, which implies that x = y = 0.

Now let's try to find a counterexample to either symmetry or the triangle inequality.

For symmetry, let's consider two different values of x and y, say x = 1 and y = -2. Then:

d(x, y) = max(|x|, |y|) = max(1, 2) = 2

d(y, x) = max(|y|, |x|) = max(2, 1) = 2

So d(x, y) = d(y, x), which satisfies the symmetry property.

Next, let's examine the triangle inequality. Suppose we choose x = 1, y = 2, and z = 3. Then:

d(x, y) = max(|x|, |y|) = max(1, 2) = 2

d(y, z) = max(|y|, |z|) = max(2, 3) = 3

d(x, z) = max(|x|, |z|) = max(1, 3) = 3

Now, we need to verify that the triangle inequality holds:

d(x, z) = 3 ≤ d(x, y) + d(y, z) = 2 + 3 = 5

Since 3 is not less than or equal to 5, we have found a counterexample to the triangle inequality, and thus, d is not a metric on R.