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 (Proved Differently)

Question

My Proof:

This problem has mainly four cases, they are as follows:

1) $a, b > 0$

2) $a, b < 0$

3) $a > 0 > b$

4) $a < 0 < b $

Let suppose that the sum of the real numbers $a +b$ equals the third real number $c$ OR $a +b = c$.

Case 1 $a, b > 0$:

$|a +b| = |a| + |b|$

$|c| = a +b$

$c = c$ ..... Proved.

Case 2 $a, b <0$:

$|(-a) +(-b)| = |(-a)| + |(-b)|$

$|-a -b| = |-a| + |-b|$

$|-(a +b)| = -(-a) -(-b)$

$|-(c)| = a +b$

$|-c| = c$

$-(-c) = c$

$c = c$ ..... Proved.

Case 3 $a>0>b$:

$|a +(-b)| < |a| + |(-b)|$

$|a -b| < a + |-b|$

$a -b < a -(-b)$

$a -b < a +b$

Case 4:

$|(-a) +(b)| < $|(-a)| + |b|$

$|-a +b| < |-a| +|b|$

$|b -a| < -(-a) +b$

$ b -a < a +b$

As, $b > a$ therefore their difference will be a positive real number but it will be less than the sum of $a$ and $b$, and Case 3 can also be reasoned in this way.

Please check my proof.

Answer 1

What you're doing isn't logically correct. If you look at Case 1, for example, you're starting off with a statement which you're trying to prove. Treat it like this:

Case 1: $a,b > 0$

$|a + b| = a + b = |a| + |b|$ which follows from the definition of the absolute value of $x$:

$|x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{ if } x < 0 \end{cases}$

The other cases are done in a similar fashion.

Answer 2

Your idea of a proof is wrong. You start from your conclusion and work to get something true. Instead, you should start with something true, and work towards your conclusion. This may sound pedantic, but it is VERY important.

I will now prove $1 = 0$. $$1= 0$$ $$1 \cdot 0 = 0 \cdot 0$$ $$0 = 0$$ $$\mathrm{True}$$

So clearly that's not a valid way to prove things.