# Triangle Inequality

1. Jan 5, 2007

### theperthvan

What's so special about the Triangle Inequality?

$$abs(x + y) <= abs(x) + abs(y)$$

I have learnt it in two or three units, but it seems too obvious to be given a special name.

What have I missed?

EDIT: I screwed up the tex here... if someone can fix it that would be cool

2. Jan 5, 2007

### James R

It's more interesting when you apply it to vectors.

i.e.

$$|a + b| \le |a| + |b|$$

where a and b are vectors.

3. Jan 5, 2007

### Hurkyl

Staff Emeritus
It's not so much that it's special, but that it's useful.

Algebraically, it's hard to manipulate an equation where addition happens inside an absolute value, but it's easy to manipulate an equation where the additions are outside.

Geometrically, it's says that a straight line is the shortest distance between two points.

4. Jan 5, 2007

### theperthvan

Alright, thanks.

And James, how did you do the tex?

5. Jan 5, 2007

### HallsofIvy

It is, in fact, one of the defining properties of a distance, i.e. metric function:
1) $d(x,y)\ge 0$
2) d(x,y)= 0 if and only if x= y
3) d(x,y)= d(y,x)
4) $d(x,y)\le d(x,z)+ d(z,y)$ for any point z.

To see the code for any LaTex, just click on it.

6. Jan 5, 2007

### Gokul43201

Staff Emeritus
Halls, is positive definiteness a property of any metric or only when the metric is given by an inner product (as in a Hilbert space)?

Last edited: Jan 5, 2007
7. Jan 5, 2007

### HallsofIvy

If I remember correctly, "positive definiteness", that d(x,y)> 0 except in the case x= y, is a requirement for a metric. Allowing d(x,y)= 0 for x not equal to y, gives what is called a "pseudo-metric". For example, in measure theory, we may define
$$d(f,g)= \int_C |f(x)-g(x)|dx[/itex] where C is some base set. In that case the two d(f,g)= 0 if f and g are equal "almost everywhere" but not necessarily equal. Of course, given any psuedo-metric, we can say that two points are "equivalent" if and only if d(x,y)= 0. Then we can treat the equivalence classes as a metric space with a true metric. 8. Jan 5, 2007 ### Gib Z Hurkyl, could you demonstrate how the Triangle Inequality, geometrically, shows the shortest distance between 2 points is a straight line? Thanks 9. Jan 5, 2007 ### Hurkyl Staff Emeritus |x+y| <= |x| + |y| is the base case. If x and y are the vectors describing two sides of a triangle, then x+y describes the third side, and this inequalty states the fact that the length of one side is no greater than the sum of the lengths of the other two. Thus the name. You can repeat it iteratively, and you get the triangle inequality for finite sums. For example, with 5 terms |a+b+c+d+e| <= |a| + |b| + |c| + |d| + |e| when you take the limit, you get the triangle inequality for infinite sums and for integrals. In particular, if P and Q are two points and c is a curve of length L between them, then: [tex] \vec{Q} - \vec{P} = \int_c d\vec{s}$$

applying the triangle inequality for integrals gives

$$\left|\vec{Q} - \vec{P}\right| = \left|\int_c d\vec{s}\right| \leq \int_c |d\vec{s}| = L$$

If you're not comfortable with ds then, if t is the parameter for the curve c, this is what those integrals mean:

$$\int_c d\vec{s} := \int_0^1 \frac{dc(t)}{dt} \, dt$$

$$\int_c \left|d\vec{s}\right| := \int_0^1 \left|\frac{dc(t)}{dt}\right| \, dt$$

Last edited: Jan 5, 2007
10. Jan 6, 2007

### Gib Z

O thanks, clearly explained. Ty

11. Jan 6, 2007

### Gokul43201

Staff Emeritus
Hmmm...in non-relativistic QM, one of the postulates states that (roughly) the system is described by a state vector that resides in a space with a positive definite metric (which is why I asked the question). This is needed so we can make the probabilistic interpretation.

Mathworld doesn't seem to require this as a property of a metric.

12. Jan 6, 2007

### Hurkyl

Staff Emeritus
Metric: A nonnegative function g(x, y) ... A metric also satisfies ... the condition that g(x, y) = 0 implies x = y.

13. Jan 6, 2007

### Gokul43201

Staff Emeritus
Ouch! For some reason, I start reading only from the third word of each sentence.

So I guess the term "positive definite metric" is either a misnomer, or more likely, a concoction of my tired imagination(?)

14. Jan 6, 2007

### Hurkyl

Staff Emeritus
'When I use a word,' Humpty Dumpty said, in a rather scornful tone,' it means just what I choose it to mean, neither more nor less.' -- Through the Looking-Glass

Physicsts frequently study pseudometrics, generally arising in some way from an inner product. So, it is reasonable that in their language, the word "metric" means what a mathematician would refer to as "pseudometric" or "inner product". The physicist would then have to add the qualification "positive definite" to denote what the mathematician simply calls "metric".

It's actually like that in a lot of fields, even purely within mathematics. For example, if I'm thinking about category theory, I might call something a "graph". But if I was thinking about discrete math, I would call the exact same object a "directed multigraph with loops".

Last edited: Jan 6, 2007
15. Jan 19, 2007

### MeJennifer

It is obvious until one realizes that it does not hold in all spaces.

The Minkowski space-time, frequently associated with the theory of relativity, violates this inequality. For instance in a Minkowski space-time not the shortest but the longest distance between two points is a straight line. And the distance between two points can be zero.

Last edited: Jan 19, 2007