# Rigorous proof of basic absolute value theorem?

1. Aug 28, 2008

### scorpion990

Rigorous proof of basic "absolute value" theorem?

Hey :) I'm working through a Real Analysis text, and I came across this theorem and "proof":
http://img352.imageshack.us/img352/6725/proofbx2.png [Broken]

It kind of took me by surprise, because the author of the text is usually very careful about defining every little operation, and took 2-3 pages to show that you can actually "add a constant to both sides" of an equation and equality without changing the solution set.

Except for the sloppy notation, two parts of the "proof" confused me. First, why can you add "plus or minus a" and "plus or minus b" to get "plus or minus (a+b)"? I'm also confused as to how the author jumped from plus or minus (a+b) to |a+b|.

Maybe I'm just slow. Any thoughts?

Last edited by a moderator: May 3, 2017
2. Aug 28, 2008

### quasar987

Re: Rigorous proof of basic "absolute value" theorem?

It's not so much that ±a + ±b = ±(a+b) as much as the fact that if ±a<=c and ±b<=d, then a+b<=c+d and -a-b<=c+d; that is to say, ±(a+b) <= c+d.

As to why ±(a+b)<=|a|+|b| implies |a+b|<=|a|+|b|, it follows from the definition of the absolute value that states that |x|=x if x>=0 and -x if x<0. Here, if (a+b)>=0, then (a+b)=|a+b| and we have |a+b|= (a+b) <=|a|+|b|. In the other case, (a+b)<0, we have -(a+b)=|a+b|, and here again, |a+b|=-(a+b)<=|a|+|b|.

3. Aug 29, 2008

### scorpion990

Re: Rigorous proof of basic "absolute value" theorem?

Oh.. I see.
The author did prove the necessary theorems for me to understand why:
$$\pm$$(a+b)$$\leq$$|a|+|b| $$\forall$$a,b$$\in$$R.

Now, if I understand this correctly, this can really be broken down into 5 cases:
1. +(a+b)<|a| + |b|, where a+b is positive.
2. -(a+b)<|a| + |b|, where a+b is positive.
3. +(a+b)<|a| + |b|, where a+b is negative.
4. -(a+b)<|a| + |b|, where a+b is negative.
5. (a+b)<|a| + |b|, where a+b is 0.

For |a+b| < |a| + |b| to be true, you stated that the only three conditions that need to be true are:
1. +(a+b)<|a| + |b|, where a+b is positive.
2. -(a+b)<|a| + |b|, where a+b is negative.
3. (a+b)<|a| + |b|, where a+b is 0.

...which are all true. So, in a way, |a+b| < |a| + |b| is just a special case of a more general theorem. Is my logic here correct?

4. Jan 28, 2009

### MackBlanch

Re: Rigorous proof of basic "absolute value" theorem?

I would do something like this. You could be a little more rigorous though.

1. A and B are real numbers.

2. A real number is less than or equal to its magnitude. (by def of absolute value)

3. A + B is a real number (by closure)

4. Assume with out loss of generality that A >= B

There are two cases.

Case 1: |A + B| = A + B
Therefore A + B = |A| + |B| = |A + B|

Case 2: |A + B| > A + B
Therefore, A + B < 0
Therefore, A < 0 and B < 0 Therefore |A| + |B| = |A + B|
-------OR, A >= 0 and B < 0 Therefore |A| + |B| >| A + B|

In all cases, |A| + |B| >= |A + B|

Therefore,

|A| + |B| >= |A + B|

Last edited: Jan 28, 2009
5. Jan 29, 2009

### poutsos.A

Re: Rigorous proof of basic "absolute value" theorem?

The author obviously is wrong,particularly with the last line in his proof .The complete proof of what he started is the following:

$$-|a|\leq a\leq |a|$$..................................................1

and $$-|b|\leq b\leq |b|$$..........................................................2

Now by adding (1) and (2) we get:

$$-( |a|+|b|)\leq (a+b)\leq (|a|+|b|)$$ ,which according to the theorem:

.........................$$|x|\leq y\Longleftrightarrow -y\leq x\leq y$$............................

is:

$$|a+b|\leq |a|+|b|$$.

Another way would ,since (a>0 or a=0 or a<0) and (b>0 or b=0 or b<0), be to consider the following cases one by one :

1) a>0 and a>0. in this case a+b>0====> |a+b|=a+b and also a=|a|,b=|b| and thus:

$$|a+b|=|a|+|b|\Longrightarrow |a+b|\leq |a|+|b|$$.

The other cases to consider are:

(a>0 and b=0),(a>0 and b<0), (a=0 and b>0), (a=0 and b=0),(a=0 and b<0),( a<0 and b>0),(a<0 and b=0), (a<0 and b<0) and they all give the same result ,like the one above.

Another way would be the double implications trick.We assume $$|a+b|\leq |a|+|b|$$ to be true and we do calculations that work either way i.e.

We square both sides of $$|a+b|\leq |a|+|b|$$ and we get

$$a^2 +b^2 +2ab\leq |a|^2+|b|^2 +2|a||b|$$,hence

$$|a+b|\leq |a|+|b|\Longrightarrow a^2+b^2 +2ab\leq |a|^2+|b|^2 +2|a||b|$$.

On the other hand if we take the sqrt of the right hand side of the implication we will get the left hand side of the implication ,hence we can write:

$$|a+b|\leq |a|+|b|\Longleftrightarrow a^2+b^2 +2ab\leq |a|^2+|b|^2 +2|a||b|$$.

And since $$|a|^2=a^2,|b|^2=b^2, |a||b|=|ab|$$ we have:

$$a^2+b^2+2ab\leq |a|^2+|b|^2+2|a||b|$$<======>$$ab\leq |ab|$$

Now we can start from $$ab\leq |ab|$$ which is a true fact and thru the implications arrive at the desired result

Last edited by a moderator: May 3, 2017