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### Triangle inequality

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Date: Mon, 14 Nov 94 15:27:52 -0400
From: Anonymous
Organization: Yorktown High School
Subject: Math question

I am a high school student who recently (about a week and a half ago)
asked the question if it is possible to prove that the absolute value of
(a-b) is greater than or equal to the absolute value of a minus the
absolute value of b.  I was wondering if you had come up with an answer
or are currently working on it.  Please let me know.

```

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Date: Mon, 14 Nov 1994 14:56:34 -0500 (EST)
From: Dr. Sydney
Subject: Re: Math question

Dear Jonathan:

Thanks for writing Dr. Math!!  I don't remember getting your
question a week and a half ago--maybe something with the email went
today, though!  You ask a very good question, and the answer is that there
is indeed a way to prove that |a-b| is greater than or equal to |a| - |b|.

Before going into the proof, though, we must go over another property of the
absolute value function that is called the triangle inequality.  This says that

|x| + |y| is greater than or equal to |x+y|

Have you seen this before?  If you think about it, it makes sense--the
maximum value |x+y| will ever reach is going to be |x| + |y|.  If you are
curious about how a rigorous proof of this might work, try playing around
with this:

2       2
|x+y| = (x+y)

See if you can figure out a proof of the triangle inequality.

Knowing that the triangle inequality holds, it follows pretty directly that

|a| = |a - b + b|

and |a -b + b| is less than or equal to |a - b| + |b| by the triangle
inequality.

So we have:

|a| is less than or equal to |a - b| + |b|

Subtracting |b| from both sides we get

|a| - |b| is less than or equal to |a - b|,

as we wanted to show.

Does that make sense to you?  Do you think you could prove the triangle
inequality (that is actually more tricky than the above proof, but try it
back with any more questions you might have.

--Sydney, "Dr. Analysis"
```
Associated Topics:
High School Basic Algebra

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