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. Jonathan Wadler
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 wrong. I am glad you were persistent so we could answer your question 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 for fun!)? It's great you are thinking about this, and feel free to write back with any more questions you might have. --Sydney, "Dr. Analysis"
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum