Talk:Triangle inequality

Old unsigned comments

What about |x+y| <= |x|+|y|, on the real line. I've seen that called the triangle inequality. Is it just by analogy to Euclidean space, and from length to absolute value?

For me this is just a poor formulation. More direct is |x|-|y| <= |x+y|<=|x|+|y|, this is the actual "triangle."

Image correction

Image "Vector triangle inequality.PNG" needs to be edited to correct ||x+y|| < ||x|| + ||y|| to this ||x+y|| <= ||x|| + ||y||.(Jalal0 (talk) 08:27, 5 May 2012 (UTC))[reply]

Consequences unclear and Seem Contradictory

On the inverse triangle inequality, it is given | |x| - |y| | <= |x - y| This makes sense to me. But in Consequences, some technical specifications are given and then a seemingly contradictory statement is given. Either the format of this section has gone awry, or something else needs to be clarified. Someone more knowledgable, please? 128.171.31.11 (talk) 09:13, 18 February 2008 (UTC)[reply]

There is no contradiction. I'm not sure how you came to that conclusion - all the statements given in the article are true, although some use sums and some use differences, and some use norms while others use distance functions. In particular, both "| |x| - |y| | <= |x - y|" and "| |x| - |y| | <= |x + y|" are true (in fact | |x| - |y| | is at most the smallest of these two quantities). Dcoetzee 01:33, 21 February 2008 (UTC)[reply]

Why is it true that? "| |x| - |y| | <= |x + y|" —Preceding unsigned comment added by 152.23.215.245 (talk) 21:59, 22 November 2008 (UTC)[reply]

I know this is a year old, but I don't like dangling questions. The answer is that |y| = |-y| for any y (either real, complex, or indeed a vector if we take norms instead). So if you know | |x| - |y| | <= |x - y|, then applying this to some x and -y gives you | |x| - |y| | <= |x + y|. Although this is a simple point, it is actually a fairly common stumbling point, so I think it would be worth adding to the article. Quietbritishjim (talk) 15:19, 6 September 2009 (UTC)[reply]

Intuition

I think that the triangle inequality is also quite intuitive, because it is impossible to draw a triangle if the "base" is longer then the sum of the two other stems. Given that my English is not good enough I am afraid to formulate this is a proper way, but it would be good to add such a section to the article. —The preceding unsigned comment was added by 89.80.147.73 (talk) 10:16, 21 January 2007 (UTC).[reply]

It is intuitive in R³, and other similar vector spaces. However, the result is more general and can be applied in any well defined inner product space - it can be shown from Cauchy Schwarz, and as such applies to many more situations that just a simple triangle in a Euclidean Geometry. It can easily be extended to more abstract versions of the inner product - for example it can be used in quantum mechanics when considering "overlap integrals". The Young Ones (talk) 20:48, 25 April 2008 (UTC)[reply]

Proof

You can not state the triangle inequality without giving a proof. The article is severly lacking and should be deleted. —Preceding unsigned comment added by 65.32.93.17 (talk • contribs) 23:14, 30 July 2007

I wouldn't say it needs to be deleted, but would suggest it is in need of a proof. The Young Ones (talk) 20:48, 25 April 2008 (UTC)[reply]

Have expanded this to include a short 'proof' using Cauchy-Schwarz Inequality. Let me know what you think. The Young Ones (talk) 20:59, 25 April 2008 (UTC)[reply]

The proof seems to have a mistake in it. The relation (x,y)+(y,x) = 2|(x,y)| doesn't hold in the general case (if (x,y) is not non-negative real). It should probably be ||x||²+(x,y)+(y,x)+||y||² = ||x||²+2*Re(x,y) +||y||² <= ||x||²+2*|(x,y)| +||y||²

Well-spotted, thanks. I've made the correction. -- simxp (talk) 11:16, 5 July 2008 (UTC)[reply]

Yes; good call. This probably has to be corrected on the article on the Cauchy Schwarz Inequality - I just lifted the proof from there...without checking it rigorously myself. Sorry for that...I should have gone through it - I just assumed wikipedia would be correct - force of habit, I guess! The Young Ones (talk) 21:32, 5 July 2008 (UTC)[reply]

Looking at the Cauch-Swarz article, it must have been corrected seperately since you took the proof. It doesn't have the 2Re(x,y) line, but the third line correctly uses <= rather than =. Since the TE isn't the main subject of the C-S article, there's so no need to have the proof longer than necessary in that article, so I'll probably leave it as it is. -- simxp (talk) 13:30, 7 July 2008 (UTC)[reply]

Shouldn't the penultimate line be = (not <=) to the last line of the proof? It seems straightforward enough, but I haven't made the change in case I missed some nuance. CinchBug (talk) 20:58, 12 July 2008 (UTC)[reply]

Yes, I think so, so I changed it. Thanks. -- Jitse Niesen (talk) 22:34, 12 July 2008 (UTC)[reply]

The penultimate line is equal to the last line. However, the idea is that the left hand side is less than or equal to both the penultimate and the last line. I won't change this back until we have had further consultation, but I'm sure that there should be an inequality in the last line of the proof. It is the Triangle Inequality! I guess this depends on how you wish to notate... The Young Ones (talk) 12:47, 15 July 2008 (UTC)[reply]

Every left hand side is compared to the right hand side. Why should the left hand side of the last line be interpreted as being compared with the left hand side of the previous line? That is inconsistent, and likely to be misleading. Waynariffic (talk) 04:23, 17 December 2012 (UTC)[reply]

Article Deletion

I nominate this article for deletion on the basis that it is extremely incomplete and makes unfounded statements without proof. --anon

There is no need to have proof in an encyclopedia. This is not a book. Oleg Alexandrov (talk) 03:00, 13 August 2007 (UTC)[reply]

Whilst Wikipedia is not a Maths text book, I don't think it is at all unreasonable to request to have a proof in an article about a theorem, especially if the proof is very short as in this case. From a Methematical perspective, the proof is the most important part of the theorem, and, if at all reasonable, as an Encyclopedia we have a duty to document it. So many thanks to User:The Young Ones for adding a proof in yesterday! -- simxp (talk) 14:37, 26 April 2008 (UTC)[reply]

Contrary Definition

The definition given seems to imply the exclusion of colinear vectors and points.

My reading, including the reading of articles in refereed mathematical and physics journals include mathematical analysies that depend on colinear vectors and points being accepted by the triangle inequality theorem.

Riley K. F., Hobson M. P. and Bence S. J., Mathematical Methods For Physics And Engineering, 2nd ed., Cambridge, 2002 , Section 8.1.3 states the triangle inequality theorem as ||a + b||>=||a||+||b|| and includes a proof. I suggest that the statement of the triangle inequality theorem should be ammended to satisfy the triangle inequality theorem stated by Riley et. al. —Preceding unsigned comment added by 130.56.65.25 (talk) 02:05, 10 December 2007 (UTC)[reply]

I agree that the inequality should be

\|a+b\|\leq \|a\|+\|b\|

instead of

\|a+b\|<\|a\|+\|b\|

. But I can't find which part of the Wikipedia article you are having problems with. Could you please be more specific where in the article collinear vectors and points are excluded? -- Jitse Niesen (talk) 13:47, 10 December 2007 (UTC)[reply]

My problem had been with the first paragraph which seems to have been fixed.

The problems that I had yesterday no longer exist. Maybe I was on another planet! My objection was based on the fact that colinear vectors and points are excluded if |a+b| < |a| + |b| but are included with |a+b| =< |a| + |b|. The article now seems quite acceptable to me. Sorry for the confusion. —Preceding unsigned comment added by 130.56.65.25 (talk) 23:47, 10 December 2007 (UTC)[reply]

Improve the illustration

I think the illustration of the triangle inequality in the case of equality should be improved such that differently coloured line segments can be identified. Now they have collapsed into one blurry linewhich makes it impossible to see how the quantities are defined. // Jens Persson (193.10.104.171 (talk) 10:47, 2 September 2008 (UTC))[reply]

I've replaced the figure with one showing three stages of collapse to make clearer what is happening. Brews ohare (talk) 17:10, 1 July 2010 (UTC)[reply]

Parallelogram Law

I propose that we add a note saying: The triangle inequality should not be confused with the Parallelogram_law, which is based on the Pythagorean_theorem and specifies precise ratios for the sides of a right-triangle in Euclidean_space. 09:22, 18 November 2008 128.187.80.2

Such confusion would be objectionable, but I don't understand how it might arise. Brews ohare (talk) 17:12, 1 July 2010 (UTC)[reply]

Curves

The following is from the lede:

That is, this theorem indicates that in Euclidean geometry “the shortest distance between two points is a straight line”, an observation proved for curved lines rather than straight-line segments using the calculus of variations.[3]

This is easily provable without the calculus of variations: if the curve were shorter by ε then one could make a rectification that is shorter by at least ε/2, and then apply the usual triangle inequality repeatedly to obtain a contradiction.

Moreover, the source provided makes no claim that calculus of variations is important for the proof that the shortest distance is a straight line. The source just says they are presenting such a proof as an example of what can be done with the calculus of variations.

There's no reason to mention the calculus of variations in the lede of this article; it's far from the topic at hand, and there's no reason a reader would need to look it up to understand the triangle inequality. So I am going to remove that mention. — Carl (CBM · talk) 12:27, 2 July 2010 (UTC)[reply]

Converse

In Pythagorean theorem there is a section about the converse of that theorem. But what is the converse of the triangle inequality? The article doesn't state nor presents its proof. 187.107.8.106 (talk) 10:19, 2 November 2013 (UTC)[reply]

I agree. The article should state with proof that if the inequalities hold, then a triangle exists with those sides.208.50.124.65 (talk) 22:10, 8 July 2014 (UTC)[reply]

Definition needs a bit of work

I am going to finesse the first few sentences to address the following issues:

The main (first) definition differs significantly from its referenced sources.
The formula uses x and y without defining them.
The formula is odd if the reader associates x and y with the nearby diagram of the triangle. Think about the meaning of |x+y|<=|x|+|y| if x and y are the sides of a triangle, and the issue will be clear.

This is a heads-up. Any thoughts? Whikie (talk) 16:13, 25 July 2014 (UTC)[reply]

It ended up being a small change - just 3 new sentences tying things together and no significant changes to the surrounding text. It is a bit more "mathy" then I would like in the lead, but it isn't too bad. Whikie (talk) 21:11, 25 July 2014 (UTC)[reply]

Proof of converse

A recent thread in the Math help desk called to attention some issues with the proof of the converse given here. I realize the proof was asked for in a previous comment, but the proof given is unsourced and seems somewhat ORish. Here are some specific issues.

In the second paragraph the proof seems to be assuming the conclusion. It fact it's not but it finds conditions necessary for a point to be the vertex of the required triangle, then finds a point that meets this condition. You would then have to work backwards to show that the point is the vertex of the required triangle. The proof could be rephrased to put it the correct order.
The proof assumes the existence and properties of complex numbers which seems unnecessary since it's a theorem in Euclidean geometry. In fact it's given in The Elements as Book 1 Prop. 22.
The proof assumes Cartesian coordinates which in turn assumes the parallel postulate. Euclid does not use the parallel postulate so presumably the theorem would still be true in hyperbolic geometry.
Most of the proof involves computing the coordinates of the vertex which is the intersection of two circles. But for a geometric construction you really only need to specify the circles and label the point where they meet, assuming of course that conditions are met which guarantees such a point.

As I see it there are several alternatives:

Drop the proof entirely and put it a reference for it; since it's in The Elements it shouldn't be hard to find one.
Find a source for the existing proof and rephrase to put it a more logical form.
Replace the proof with Euclid's. (Euclid's proof has some issues as well, but the necessary interpolations can be found in Heath's commentary.)
Source and rephrase the existing proof and give Euclid's version as an alternative.

Here is my version of Euclid's proof with modifications given (mostly) as in Heath:

Proposition: Given lengths a, b, and c with a+b>c, a+c>b, b+c>a, construct a triangle with sides a, b and c.

First, relabel if necessary so that c is at least as great as a and b. (This step is my own but it saves having to go into cases as in Heath's version.) Then a, b ≤ c < a+b. On line DE, starting at D, mark off DF=a, FG=b and GH=c. Draw circle S with center F and radius a, and circle T with center G and radius c. From G, mark of GM=c in the direction of D. At this point the diagram looks something like:

  D  M F      G        H  E
--*--*-*------*--------*---
  |  a |   b  |    c   |

The point M lies on T and b ≤ c < a+b so GF ≤ GM < GD and M is on the segment DF. Then a+b>c=GM=GF+FM=b+FM so FM<a and M lies inside circle S. The point H is also on T and FH=b+c≥b+a>a so H lies outside the circle S. Since S has points both inside and outside of T, it must have a point on T, say K. (Here Heath invokes what he calls the "Principle of Continuity", not to be confused with Leibniz' Law of Continuity.) But FK = a, FG=b and GK=c so FGK is the required triangle.

Euclid's original proof is much shorter since it simply assumes that the circles S and T will intersect; in this way the proof does not seem to use the necessary hypotheses.

Anyway, what options to people prefer? Is there a source for the proof given or a similar modern style proof which can be given? --RDBury (talk) 06:11, 18 February 2017 (UTC)[reply]

I think you should resequence the current proof more to your satisfaction. I thought I fixed it yesterday with a slight rewording, but if you don't think it's okay now please go ahead and resequence it.
I don't see where the current proof assumes the existence and properties of complex numbers. It just says that h exists because we are not taking the square root of a negative number.
The proof says it uses the Cartesian plane but doesn't really. I'll replace it with Euclidean plane, and get rid of the superfluous coordinates.
The proof doesn't involve computing any coordinates, just the length h. (His notation (d, h) need not be taken as coordinate notation, just as a pair of numbers being computed.)
I like the current proof (and particularly how it factors out the radicand), and think it should be retained. I'll look for a source, but my access to references is very limited so I'm not too optimistic about finding one. If you like and can put in a good diagram, you could put in your proof as well.

Loraof (talk) 19:05, 18 February 2017 (UTC)[reply]

I did find a source for a similar proof, see [1]. The factorization of the radicand is interesting and should be mentioned, but the same factorization appears in in Heron's formula. Maybe it makes more sense to mention that it's the triangle inequality which guarantees that the radicand is positive. My issue with the use of coordinates is that they require the parallel postulate, but even it you remove them you would still be using the Pythagorean theorem which also uses the postulate, and it seems significant that the converse would hold in the non-Euclidean plane. Also, if you remove coordinates you have to contend with the possibility that d is negative, which means you'd need to use signed lengths or split into cases.

In any case, I gather that you'd like to see both proofs in the article, which I don't have a problem with if the issues are fixed. The proof above should probably be trimmed some before it goes in the article; I'll think about how best to do that. --RDBury (talk) 07:12, 19 February 2017 (UTC)[reply]

hypervolume and hyperarea

@JayBeeEll: thank you for your recent edits. One quibble, though. I would like to refer to the measure of an $(n - 1)$ -dimensional facet as a "hyperarea" rather than a "hypervolume". Specifically, I use "hypervolume" as the general term for the measure of a "face" of a structure (e.g., a simplex), where a face can be of any number of dimensions relative to the number of dimensions of the structure. However, a "facet" is defined to be a face of one less dimension than the full structure. When one is referring specifically to this number of dimensions, it is nice to have this more-specific word to use. Likewise, with "facet" comes "hyperarea". When referring specifically to this number of dimensions, it is nice to have this more-specific word to use.

Obviously, the more general words could be used instead of "facet" or "hyperarea". However, math is about precision and I'd like to use the more precise "hyperarea" in this case. — $Q$ uantling (talk | contribs) 13:33, 24 October 2021 (UTC)[reply]

I understand your motivation, but as far as I can tell this is a word you just made up for convenience, not one that is represented in reliable sources. In particular, I do not believe there is a meaningful concept to be attached to the phrase "hyperarea of [a thing]". In contrast, length, area, and volume have fixed meanings, and they are all particular cases of hypervolume (in appropriate dimensions). As evidence in favor of the theory that this is not a real thing, the word "hyperarea" is used in only two other articles on Wikipedia: Law of sines (it was added in this edit, with no reference supporting its use) and Polyakov action (where it does not appear to be used with the same meaning). If it were a term with an accepted meaning used by some relevant community, I would expect it to be much better-represented in articles on polyhedra etc. --JBL (talk) 16:57, 24 October 2021 (UTC)[reply]

I don't see much use of "hyperarea" outside of Wikipedia either. (Though, not zero. If I've made it up, so have others!) We should probably do this your way; including correcting the Law of sines Wikipedia page. — $Q$ uantling (talk | contribs) 23:14, 24 October 2021 (UTC)[reply]

Before we put this issue completely to bed, what do you think of "hyper-surface area" (instead of "hypervolume") where we are replacing "hyperarea"? The wording "hyper-surface area" gets a few relevant search engine hits. — $Q$ uantling (talk | contribs) 16:10, 25 October 2021 (UTC)[reply]

Distances, points, vectors

The article is a little sloppy when discussing distances and vectors. Specifically, the triangle inequality for distances is either $z \leq x + y$ for distances $x$ , $y$ , and $z$ or, if the distances are between specified points, something like $d (a, c) \leq d (a, b) + d (b, c)$ . On the other hand, if $u$ is the vector from point $a$ to point $b$ and $v$ is the vector from point $b$ to point $c$ then the triangle inequality for vectors is $| u + v | \leq | u | + | v |$ . Let's have all three of these formulas, properly labeled. — $Q$ uantling (talk | contribs) 20:30, 17 June 2024 (UTC)[reply]

The lead of the article seems very clear; what part are you unhappy about? --JBL (talk) 00:47, 18 June 2024 (UTC)[reply]

I will make an edit to the article. —

Q

uantling (talk | contribs) 12:34, 18 June 2024 (UTC)[reply]

x, y, z vs. a, b, c

The lede uses $x$ , $y$ , and $z$ for distances, but the remainder of the article uses $a$ , $b$ , and $c$ ; shouldn't we be consistent? For vectors the article uses $x$ and $y$ consistently, but that kind of, sort of collides with the notation for lengths; should we make those more distinct? Likewise, when we get to metric spaces, now $x$ , $y$ , and $z$ are points; should we have a third set of letters for these? In summary, we have: $z\leq x+y$ $\|\mathbf {x} +\mathbf {y} \|\leq \|\mathbf {x} \|+\|\mathbf {y} \|$ $d(x,z)\leq d(x,y)+d(y,z)$ but IMHO the following would be better:

{\textstyle c\leq a+b}

for lengths

a

,

b

, and

c

;

{\textstyle \|\mathbf {u} +\mathbf {v} \|\leq \|\mathbf {u} \|+\|\mathbf {v} \|}

for vectors

u

and

v

; and

{\textstyle d(P,R)\leq d(P,Q)+d(Q,R)}

for points

P

,

Q

, and

R

.

I'd make changes myself, but at least one of the figures depends upon the current mishmash of variable names, and I don't have the skills to edit that. — $Q$ uantling (talk | contribs) 12:53, 18 June 2024 (UTC)[reply]

I like

a

,

b

, and

c

for lengths because of their use in the common presentation of the Pythagorean theorem

a 2 + b 2 = c 2

. —

Q

uantling (talk | contribs) 12:56, 18 June 2024 (UTC)[reply]

I edited the article, except using

A

,

B

, and

C

for points. —

Q

uantling (talk | contribs) 16:19, 19 June 2024 (UTC)[reply]

Are you going to change the diagrams? –jacobolus (t) 17:03, 19 June 2024 (UTC)[reply]

No, unfortunately. In that sense, I've changed the article from one inconsistent state to a different inconsistent state :frown:. I think it is now slightly less inconsistent and is less confusing (no reuse of a symbol to mean multiple things) — but updating the diagrams needs to happen too! In the meantime, I left the captions consistent with their figures. —

Q

uantling (talk | contribs) 17:33, 19 June 2024 (UTC)[reply]