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I was cruising through my college algebra book, when I encountered this equation, everything is straight-forward, only, I have no idea what they did in regards to the grouping symbol, and the negative operation in front of it.

It looked like this: $$\mathbf{6-|5-8|}+3(4-1) \quad\to\quad \mathbf{6-|{-3}|}+3(3) \quad\to\quad \mathbf{6-3}+9$$

Is the $-3$ supposed to be positive, or is there a reason the grouping symbol for an absolute value is an exception?

Thank's in advance!

Misha Lavrov
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bitbyter
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2 Answers2

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Absolute value is not only a grouping symbol, it is also a function. There are rules for what happens when you pull numbers in and out of it. We have $$|ax|=|a||x|$$ Notice the absolute value symbol around both values. So that means we have $$|-3|=|-1||3|=3$$ but $$-|3|=-3\neq|-3|$$ Furthermore, $$|a|=|-a|$$ for all $a$, so a "distributed" minus sign does nothing.

Matt Samuel
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    I would argue that absolute value is a grouping symbol, same as parentheses, brackets, radicals, and division vinculums. Of course those latter examples are also functions, like absolute value. – ziggurism Sep 06 '23 at 19:20
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    @ziggurism I decided I agree so I edited my post. – Matt Samuel Sep 06 '23 at 19:22
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Welcome, bitbyter! The focus of this site is usually more advanced than this, but let's see what we can do.

When you write that the terms inside are multiplied by the negative outside, I'm guessing that you mean something like $$ 6 - ( 5 - 8 ) = 6 - 5 + 8 \text , $$ with the minus distributed to both the $ 5 $ (making it $ - 5 $) and the $ - 8 $ (making it $ 8 $); is that what you mean?

If so, this is a theorem about algebra, not a definition, and it doesn't apply if there are other operations (like an absolute value) in the way. You can't write $ 6 - \lvert 5 - 8 \rvert = 6 - \lvert 5 \rvert + \lvert 8 \rvert $, or $ 6 - \sqrt { 5 - 8 } = 6 - \sqrt 5 + \sqrt 8 $, or anything like that. Or maybe you're thinking $ 6 - \lvert 5 - 8 \rvert = 6 + \lvert - 5 + 8 \rvert $, or $ 6 - \sqrt { 5 - 8 } = 6 + \sqrt { - 5 + 8 } $, etc; but those aren't correct either. Even though the absolute value and the square root provide grouping, they also involve an operation, and that interferes.

Even $ 6 - ( 5 - 8 ) = 6 - 5 + 8 $, while true, is not how you would evaluate the left-hand side directly; to do it directly, you do $ 5 - 8 = - 3 $ first, and then $ 6 - ( - 3 ) = 6 + 3 = 9 $. Since $$ a - ( b - c ) = a - b + c $$ is an algebraic identity, you can do $ 6 - ( 5 - 8 ) = 6 - 5 + 8 = 1 + 8 = 9 $ instead, but that's an indirect method of evaluation, and not how you'd normally go about it.

So if you have $ 6 - \lvert 5 - 8 \rvert $, first you do $ 5 - 8 = - 3 $, then (since the absolute value went around this) $ \lvert { - 3 } \rvert = 3 $, and then finally $ 6 - ( 3 ) = 3 $. And this time, there's no alternative order to do it in.

The only time you really need the identity $ a - ( b - c ) = a - b + c $ is when you have variable expressions instead of specific numbers. So $ 6 - ( 5 - x ) $ can be simplified to $ 6 - 5 + x = 1 + x = x + 1 $, but $ 6 - \lvert 5 - x \rvert $ can't be simplified at all. (There is an identity $ \lvert a - b \rvert = \lvert b - a \rvert $ that you can use if you want, so you could write $ 6 - \lvert 5 - x \rvert $ as $ 6 - \lvert x - 5 \rvert $ if that looks nicer, but that has nothing to do with the minus outside the absolute value, and the expression still can't get any simpler.)

Toby Bartels
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