How to write an expression in an equivalent form without absolute values?

Question

The question I have in front of me is the very first problem in Trench's Introduction to Real Analysis:

Write the following expression in equivalent form not involving absolute values:

$a+b+|a-b|$

Looking at the provided answer in the back of the book its:

$2\max(a,b)$

In the chapter I can't find any examples with solutions involving the min/max functions, so feel at a complete loss as to how to get to the above answer.

Answer 1

If $a\ge b$ then we can safely say that:$$a+b+|a-b|=a+b+a-b=2a$$If $a\lt b$ then we can safely say that:$$a+b+|a-b|=a+b-(a-b)=a+b-a+b=2b$$Hopefully you reason to the answer from here...

Answer 2

Think about it in two cases:

If $a\geq b$ then $|a-b|=a-b$ and then $a+b+|a-b|=a+b+a-b=2a$.

If $b>a$ then $|a-b|=b-a$ and then $a+b+|a-b|=a+b+b-a=2b$.

In either case, the result is $2\max(a,b)$.

Answer 3

$S = a+b+|a-b| = \begin{cases} a+b+a-b, \text{ if a $\geq$ b} \\ a+b-a+b, \text{ if a $<$ b}\end{cases}= \begin{cases} 2a, \text{ if a $\geq$ b} \\ 2b, \text{ if a $<$ b} \end{cases}= 2\text{ max}(a,b)$

Answer 4

All you have to do is write down what the absolute value means.

The definition we have is $$|x|:=\begin{cases}x&\text{ if }x\ge0\\-x&\text{ if }x<0\end{cases}$$

So now the equation we have means

$$\begin{align} a+b+|a-b|&=\begin{cases}a+b+a-b&\text{ if }a-b\ge0\\a+b-(a-b)&\text{ if }a-b<0\end{cases}\\ \\&=\begin{cases}2a&\text{ if }a\ge b\\2b&\text{ if }a<b\end{cases} \end{align}$$

Now all that is required is to recognize $$\max(a,b)=\begin{cases}a&\text{ if }a\ge b\\b&\text{ if }a<b\end{cases}$$ giving you an answer of $$a+b+|a-b|=2\max(a,b)$$

Answer 5

There is also a trick. $\sqrt{x}$ means normally the positive square root. Thus, we can use $\sqrt{x^2}=|x|$:

$$a+b+|a-b|=a+b+\sqrt{(a-b)^2}$$