For all Real Numbers $a, b \in \mathbb R$, $|a + b|^2 = (a + b)^2$

Question

I want to show that for all real numbers $a, b \in \mathbb R$, $|a + b|^2 = (a + b)^2$. I have the following axioms for the inequality relation "$\leq$":

(i) For every $a,b \in \mathbb R$, either $a \leq b$ or $b \leq a$.

(ii) If $a \leq b$ and $b \leq a$, then $a=b$.

(iii) If $a \leq b$ and $b \leq c$, then $a \leq c$.

(iv) If $a \leq b$ and $c \in \mathbb R$, then $a+c \leq b+c$.

(v)If $a \leq b$ and $0 \leq c$, then $ac \leq bc$.

I also have been given the following rules about absolute values:

(i) For all $a \in \mathbb R$, $-|a| \leq a \leq |a|$ and $a = |a|$ if, and only if, $0 \leq a$.

(ii)For all $a, b \in \mathbb R$, $|ab| = |a||b|$.

Now, I want to show that for all $a, b \in \mathbb R$, $|a + b| \leq |a| + |b|$. But in part of a proof I'm reading in a textbook, it says that "$|a + b|^2 = (a+b)^2$". I understand it intuitively, but how do you prove exactly that only using the axioms and the rules above? I also know that $0 \leq |a|$ for all $a \in \mathbb R$, so that implies that $0 \leq |a + b|$, and every number squared gives a positive number. But I don't understand how to show that these two are thus equal to each other.

Thanks in advance.

Answer 1

Note that, for every real $x$,

$$\vert x \vert \stackrel{\text{def}}{=} \begin{cases} x &,\ x \geq 0 \\ -x &,\ x < 0 .\end{cases}$$

Now look at the two cases:

• If $ a + b \geq 0$, we have $\vert a + b \vert = a + b$, thus $\vert a + b \vert^2 = (a + b)^2$.
• Otherwise, $\vert a + b \vert = - (a + b)$, therefore it follows $\vert a + b \vert^2 = (- (a+b))^2 = (a+b)^2$, since $(xy)^2 = x^2y^2$ for real $x,y$ and $(-1)^2 = 1$.

Answer 2

Why should we care about those inequalities properties for this problem? All we need are two basic properties of absolute value:

$$\begin{cases}1)\;\; |a|^n=|a^n|\,&\text{for all}\,\,\,n\in\Bbb N\\{}\\2)\;\;|a|=a &\text{forall}\,\,a\ge0\end{cases}$$

In the above, $\;a\in\Bbb R\;$ .

Answer 3

$$|a+b|^2=|a+b||a+b|=|(a+b)(a+b)|=|(a+b)^2|$$ by $(\textrm{ii})$ of the absolute value.

Now, prove that $(a+b)^2\geq 0$ for all $a$ and $b$.

Answer 4

Show that $$|c|^2=c^2$$ for all $c\in\Bbb R$ and let $c=a+b$.

Answer 5

Note that $|x|$ is either $x$ or $-x$

Thus for any real number $x$ we have $$|x|^2=(\pm x)^2=x^2$$

Therefore for $x=(a+b)$ we have $$ |(a+b)|^2=(a+b)^2$$

Answer 6

According to (i), for any $c\in\mathbb R$, either $c\le 0$ or $0\le c$.

In the case $c\le 0$, according to (iv), $c+(-c) \le 0 + -c$, that is, $0\le -c$.

So for all $c\in \mathbb R$, either $0\le c$, or $0\le -c$.

Now if $0\le c$, then according to (v), $0c\le c^2$, that is, $0\le c^2$.

And if $0\le -c$, then according to (v), $0(-c) \le (-c)^2$, that is, $0 \le c^2$ again.

So since either $0\le c$ or $0\le -c$, and in both cases it follows that $0\le c^2$, we have $0\le c^2$ for all $c\in\mathbb R$.

Now apply the rules for absolute values.