Wikipedia:Reference desk/Archives/Mathematics/2017 January 15

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January 15

Triangle inequality

I'm thinking of studying maths at uni so to see what it's like I've been looking at some textbooks. In the following textbook (http://ramanujan.math.trinity.edu/wtrench/texts/TRENCH_REAL_ANALYSIS.PDF) on page 3 there is a proof of the triangle inequality which I don't understand. So can someone help me understand this proof? And also should I be put off studying maths as I'm already confused by the first proof in a textbook? --178.208.207.19 (talk) 18:25, 15 January 2017 (UTC)

Have you read our article Triangle inequality, and did it help? Rojomoke (talk) 18:45, 15 January 2017 (UTC)

Not really as I have read the Wikipedia article as well as other articles on the web dealing but they all deal with different forms of the inequality and different proofs which I don't understand much either. --178.208.207.19 (talk) 18:51, 15 January 2017 (UTC)

Which part of the proof confuses you ? One way of approaching a proof that you do not understand is to try to think of your own proof instead. The text splits the proof into four cases, but maybe it is simpler to think of two cases initially. If a and b have the same sign (both positive or both negative) then can you see that |a + b| = |a| + |b| ? If a and b have opposite signs then can you see that |a + b| < max(|a|, |b|) ? For completeness you also have to consider the cases where a or b (or both) is 0.

In answer to your second question I would say do not be put off. Reading a maths textbook for the first time is often confusing. You have to get used to the author's style and approach. In this text there seems to be very little motivation for proving the triangle inequality. You are not told why it is important or what would follow if the triangle inequality were not true. Try reading another real analysis text in parallel - topics that are obscure in one text may be clearer in another. Gandalf61 (talk) 18:54, 15 January 2017 (UTC)

Different parts confuse me for different reasons. I think my problem is that I'm trying to follow the authors thinking exactly. I can see for example why as Gandalf shows it works where a and b are positive but I don't see why it was important for the author to point out that "so a + b > 0". Then also assuming I understood cases a, b, c, and d they just seem to prove equality with an = not less than or equal to. --178.208.207.19 (talk) 19:05, 15 January 2017 (UTC)

The proof given isn't as clear as it could be. There are really six cases which the author is disguising as four:

1) a≥0, b≥0

2) a≤0, b≤0

3a) a≥0, b≤0, a≥−b

3b) a≥0, b≤0, a≤−b

4a) a≤0, b≥0, −a≥b

4b) a≤0, b≥0, −a≤b

Try working out the left and right sides of the inequality in each case.

One thing you learn if you try to teach yourself math is that a single book is usually not enough. Get access to another book or two on the same subject, that way if you're confused by one book you can try another one. There are many on-line resources now: course lectures on YouTube, free PDF's of textbooks, etc. It helps if you rewrite the proof in your own words, trying to guess what each next step will be before seeing it the text. There is a whole grammar of proofs that you learn at about college sophomore level, basically how to construct proofs of if-then's, either-or's, iff's, cases, induction, etc. If you've never been to exposed to it then it will be hard to follow a proof where the author is assuming you have. --RDBury (talk) 21:04, 15 January 2017 (UTC)

The reason the author points out that $a+b>0$ is to deduce that $|a+b|=a+b$ (by the definition of absolute value, if $x>0$ then $|x|=x$.)

In the author's cases a and b, he shows that $|a+b|$ is necessarily equal to $|a|+|b|$. In cases c and d he shows the (weak) inequality - $|a+b|$ is shown to be equal to something which itself is less than or equal to $|a|+|b|$.

Anyway, I don't think you should be put off by this. This is not a textbook designed for people who are just starting out, it assumes some background knowledge and mathematical maturity and skims over topics without much of an explanation. Try finding a more beginner-friendly textbook. -- Meni Rosenfeld (talk) 21:19, 15 January 2017 (UTC)

Actually, I would say that in cases c and d aren't really proved. The proof just asserts what |a+b| is equal to without showing why this is correct. If you are familiar with real numbers and absolute values then it should be obvious that these steps are valid on real numbers, but just using them without proof is like asserting that "these operations on any ordered field work the way they do on real numbers". Really they should be proved—and you might find it a valuable exercise to do that. --69.159.60.210 (talk) 21:33, 15 January 2017 (UTC)

I disagree, the proof is complete but you need to read it properly. The part with the black square should be applied to each of the cases c and d separately and then the proof follows directly from the definition (the one which appears just before the theorem, and says $|x|=x$ if $x\geq 0$ and $|x|=-x$ if $x\leq 0$.)

For example, part (c) says: If $a\geq 0$ and $b\leq 0$, then by definition $|a|=a$ and $|b|=-b$ so $a+b=|a|-|b|$. Now, either $|a|\geq |b|$ so $|a|-|b|\geq 0$ so by definition $||a|-|b||=|a|-|b|$ so $|a+b|=||a|-|b||=|a|-|b|$, or $|b|\geq |a|$ so $|a|-|b|\leq 0$ so by definition $||a|-|b||=-(|a|-|b|)=|b|-|a|$, so $|a+b|=||a|-|b||=|b|-|a|$. So $|a+b|$ is either $|a|-|b|$ or $|b|-|a|$, and in either case is $\leq |a|+|b|$ (since $|a|\geq 0$ and $|b|\geq 0$).

For part (d) the proof is the same.

The proof in the book says the exact same thing I did here, the only thing missing is pasting "by the definition of absolute value" everywhere, which is of course unneeded for a non-beginner text. -- Meni Rosenfeld (talk) 22:37, 15 January 2017 (UTC)

First, I didn't mean to suggest that it was defective; only that it skipped steps (as you say, the way a non-beginner text will). Beside the "by definition" step, there's also the bit where you wrote "$|a|\geq |b|$ so $|a|-|b|\geq 0$". That's true because in a field you can construct $-|b|$ and subtract it from both sides, but that's still a step in the full proof. --69.159.60.210 (talk) 08:08, 16 January 2017 (UTC)

Well, you can always take any human-readable proof and deconstruct it to smaller and smaller steps until you end up with a computer-verifiable proof. It doesn't detract from the original proof. $|a|\geq |b|\Rightarrow |a|-|b|\geq 0$ is as obvious as steps go. -- Meni Rosenfeld (talk) 11:43, 16 January 2017 (UTC)