On the real line, define d (x, y ) = sqrt(abs(x − y)) . Prove that d satisfies the axioms of a
metric.
To prove that d(x, y) = sqrt(abs(x − y)) satisfies the axioms of a metric, we need to show that it satisfies three properties: positivity, symmetry, and triangle inequality.
1. Positivity:
The first property is that d(x, y) should always be greater than or equal to zero, and it should only be equal to zero when x = y.
To prove this, we can start by calculating d(x, y) using the definition:
d(x, y) = sqrt(abs(x − y))
Since the absolute value of any number is always greater than or equal to zero, we have abs(x - y) >= 0.
Taking the square root of a non-negative number will also yield a non-negative number or zero.
Therefore, sqrt(abs(x − y)) >= 0.
Furthermore, abs(x - y) = 0 only when x = y. So if x = y, then sqrt(abs(x − y)) = sqrt(0) = 0.
Hence, d(x, y) = 0 if and only if x = y.
Therefore, the first property of positivity is satisfied.
2. Symmetry:
The second property states that d(x, y) should be the same as d(y, x) for any x and y.
Using the definition of d(x, y) and d(y, x), we have:
d(x, y) = sqrt(abs(x − y)) and d(y, x) = sqrt(abs(y − x))
Since the absolute value function is symmetrical, abs(x - y) is the same as abs(y - x).
Therefore, sqrt(abs(x − y)) = sqrt(abs(y − x)), which implies that d(x, y) = d(y, x).
Therefore, the second property of symmetry is satisfied.
3. Triangle Inequality:
The third property states that for any x, y, and z, d(x, z) ≤ d(x, y) + d(y, z).
Let's calculate the values of each side of the equation using the definition of d(x, y):
Left-hand side: d(x, z) = sqrt(abs(x - z))
Right-hand side: d(x, y) + d(y, z) = sqrt(abs(x - y)) + sqrt(abs(y - z))
To prove the triangle inequality, we need to show that sqrt(abs(x - z)) is less than or equal to sqrt(abs(x - y)) + sqrt(abs(y - z)).
To do this, we can square both sides of the equation (squaring preserves the inequality):
abs(x - z) ≤ [sqrt(abs(x - y)) + sqrt(abs(y - z))]^2
Expanding the right-hand side using the identity (a + b)^2 = a^2 + b^2 + 2ab, we get:
abs(x - z) ≤ abs(x - y) + 2 * sqrt(abs(x - y)) * sqrt(abs(y - z)) + abs(y - z)
Now, by the triangle inequality property of the absolute value function, abs(x - z) ≤ abs(x - y) + abs(y - z).
So, if we substitute abs(x - z) with abs(x - y) + abs(y - z), we get:
abs(x - y) + abs(y - z) ≤ abs(x - y) + 2 * sqrt(abs(x - y)) * sqrt(abs(y - z)) + abs(y - z)
Notice that the absolute values cancel out on both sides, leaving us with:
0 ≤ 2 * sqrt(abs(x - y)) * sqrt(abs(y - z))
Again, since sqrt(abs(x - y)) and sqrt(abs(y - z)) are non-negative, their product is non-negative as well.
Therefore, 2 * sqrt(abs(x - y)) * sqrt(abs(y - z)) is non-negative.
Hence, since 0 is less than or equal to a non-negative number, the inequality 0 ≤ 2 * sqrt(abs(x - y)) * sqrt(abs(y - z)) holds.
Therefore, it follows that abs(x - y) + abs(y - z) ≤ abs(x - y) + 2 * sqrt(abs(x - y)) * sqrt(abs(y - z)) + abs(y - z).
Thus, the triangle inequality is satisfied.
Since d(x, y) satisfies all three properties - positivity, symmetry, and triangle inequality, we conclude that d(x, y) = sqrt(abs(x − y)) is indeed a metric on the real line.