Absolute Value and Inequality
Date: 8/23/96 at 23:21:14 From: Anonymous Subject: Absolute Value and Inequality Hi - I have an answer for the following question, but I'm not sure if it's correct. "Show that |a+b| = |a| + |b| iff ab is greater than or equal to 0." My answer to this question is: Suppose a=2 and b=3 then |2+3| = |2| + |3| = 5 But if a = -2 and b = 3 then |-2+3| is not equal to |-2|+|3| The result is that the value of ab needs to be positive or negative, but it won't work if one is negative and the other is positive. Do you understand my conclusion? Or if you think that you can explain it better, please tell me.
Date: 8/26/96 at 12:8:17 From: Doctor Mike Subject: Re: Absolute Value and Inequality Hi to you, It certainly is true that |-2+3| is not equal to |-2|+|3| , BUT (-2)*(3) is not positive or zero either. We need to focus on that strange word *iff* which is a shorthand for *if and only if*. Anytime you see X iff Y, it means exactly the same as "If X is true then Y is true, and, if Y is true then X is true." In other words, X is true under exactly the same circumstances as Y. Now back to what this means for your problem. You are being asked to show that |a+b| = |a|+|b| is true under exactly the same circumstances as when the product ab is positive or zero. You are not being asked to show that |a+b| = |a| + |b| is always true - far from it - you are only being asked to show it when the product a*b is positive or zero. I think you were seeing the absolute value facts correctly, but were misinterpreting exactly what you were being asked to do. So, what are the 2 situations or circumstances here? 1. If it is true that a times b is positive, then either a and b are both positive, or a and b are both negative. If a times b is zero then one or both of them are zero. It is in any one or all of these situations that |a+b|=|a| + |b| is true. 2. If it is false that a times b is positive or 0, then one of them is positive and the other is negative. |a+b| = |a| + |b| is false in both of these situations. I hope my different (and longer!) explanation helps. -Doctor Mike, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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