Talk:Triangle inequality/Archive 1

Archive 1

On the real line

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? — Preceding unsigned comment added by 85.250.168.135 (talk) 08:27, 2 November 2005 (UTC)

For me this is just a poor formulation. More direct is |x|-|y| <= |x+y|<=|x|+|y|, this is the actual "triangle." — Preceding unsigned comment added by 207.216.132.146 (talk) 06:14, 13 July 2006 (UTC)

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).

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)

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)

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)

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)

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)

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)

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)

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

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)

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)

Article Deletion

I nominate this article for deletion on the basis that it is extremely incomplete and makes unfounded statements without proof. — Preceding unsigned comment added by 97.97.70.127 (talk) 23:55, 12 August 2007 (UTC)

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

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)

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)

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)

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)

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)

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)

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)

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)

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))

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)

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)

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)

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))

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)

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)

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)

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)