Prove That $|a +b| = |a| +|b|$ if $a$ and $b$ Have Same Signs, And $|a +b| < |a| + |b|$ if $a$ and $b$ Have Opposite Signs

Question

My Proof:

$|a +b| = |a| +|b|$ ..... $(i)$

$|a +b| < |a| + |b|$ ..... $(i)$

If $'a'$ and $'b'$ have same signs:

Let $a$ and $b$ be equal to $-x$. Replacing $a$ and $b$ with $-x$ in the equation $(i)$, we get the following result:

$|-x +(-x)| = |-x| + |-x|$

$|-x -x| = -(-x) -(-x)$

$|-2x| = x +x$

$-(-2x) = 2x$

$2x = 2x$ ..... Proved

If $'a'$ and $'b'$ have opposite signs:

Let $a = x$ and $b = -x$. Then, by replacing $a$ and $b$ with $x$ and $-x$ respectively in the inequality $(i)$, we get the following result:

$|x +(-x)| < |x| + |-x|$

$|x -x| < x -(-x)$

$|0| < x +x$

$0 < 2x$ ..... Proved

Is this way valid for proving what has been asked to prove in the problem?

Answer 1

You may want to use the definition of absolute value, i.e.

$|x| := \left\{\begin{array}{l}x, \mbox{if } x > 0 \\ -x, \mbox{if } x < 0 \end{array}\right.$

Next consider four cases:

Case 1. $a, b > 0$

Then $|a| = a$ and $|b| = b$. Further, $|a + b| = a + b$. Can you finish the proof for this?

Case 2. $a, b < 0$

Then $|a| = -a$ and $|b| = -b$. More so, $|a + b| = -(a + b)$. And so on...

Case 3. $a < 0 < b$.

Then $|a| = -a > a$ and $|b| = b$...

Case 4. $b < 0 < a$.

Similar to the previous case.

p.s. Of course one can reduce further the cases, but I think for now it is best that you consider all cases...

Answer 2

No it's not, you assumed your equation was true.

However if you first prove that your equation is true for $a$ and $b$ positive, then using your "$(-x)$ method" would be the right way to show it for negative $a$ and $b$ (considering you use $(-x)$ and $(-y)$ instead of $(-x)$ alone).

Same thing for the other equation ;)

Answer 3

Your proof is incomplete. $a$ and $b$ having the same sign does not imply $a=b$, nor does $a$ and $b$ having opposite signs imply $a=-b$.

Try considering the following cases separately:

1. $a>0$ and $b>0$
2. $a>0$ and $b<0$
3. $a<0$ and $b>0$
4. $a<0$ and $b<0$

Answer 4

Use Real number line .[Motivation from your previous question]

• $a, b$ be two points on opposite side of zero. Then $|a+b|$= distance between $a$ and $(-b)$ .Now as they are of opposite sign then $a$, $(-b)$ are on same sides of $0$. As $|x|$ represent the distance from origin. You get $|a+b|=||a|-|b||$

• Otherwise you get reverse.$|a+b|=|a|+|b|$. if $a,b$ are on same sides of real line.

Answer 5

In fact, this problem has eight cases. $(a+b)$ could be positive or negative, $a$ could be positive or negative, and $b$ could be positive or negative.

However:

1. If $a,b$ are the same sign, then $(a+b)$ is that same sign. This eliminates two of the cases.

2. If $a$ is positive and $b$ is negative (regardless of what $a+b$ is), we may swap the roles of $a$ and $b$; so this case is identical to when $a$ is negative and $b$ is positive. This eliminates two more cases.

This leaves four cases to consider; two more are derived from these, and two are impossible.

1. $a>0, b>0, a+b>0$

2. $a<0, b<0, a+b<0$

3. $a>0, b<0, a+b>0$

4. $a>0, b<0, a+b<0$

With a bit of care one could combine cases 3 and 4 as well, but the important thing is that in these cases we may eliminate all absolute value symbols.