# Thread: absolute Value aka modulus

1. ## absolute Value aka modulus

mod(a+b) is greater than or equal to mod(mod(a)+mod(b))
i dont get it

2. ## Re: absolute Value aka modulus

Hello, abdulrehmanshah!

mod(a+b) is greater than or equal to mod(mod(a)+mod(b))
i dont get it

Neither do I . . . It is not true.

3. ## Re: absolute Value aka modulus

Originally Posted by abdulrehmanshah
mod(a+b) is greater than or equal to mod(mod(a)+mod(b))
i dont get it
What is true: $$\left| {\left| a \right| - \left| b \right|} \right| \le \left| {a + b} \right|$$

To see this,
\begin{align*} |x|&= |x+y-y| \\ &\le |x+y|+|y|\\|x|-|y|&\le |x+y|\end{align*}

Likewise, $$|y|-|x|\le |y+x|$$. Together we get the result.

4. ## Re: absolute Value aka modulus

how did u get second step the step after mod(x+y-y) how did u get mod(x+y)+mod(y)?

5. ## Re: absolute Value aka modulus

||a|−|b||≤|a+b| this is what i meant

6. ## Re: absolute Value aka modulus

Originally Posted by abdulrehmanshah
how did u get second step the step after mod(x+y-y) how did u get mod(x+y)+mod(y)?
$|a+b| \le |a| + |b|$ is known as the Triangle Inequality. It is fairly simple to prove for real numbers. In this case, Plato used the Triangle Inequality to show that

$|(x+y)+(-y)| \le |x+y|+|-y| = |x+y|+|y|$.

Then, Plato goes on to show that both

$|x|-|y| \le |x+y|$

and

$|y|-|x| \le |x+y|$

so

$\left\lvert \lvert x \rvert - \lvert y \rvert \right\rvert \le |x+y|$

7. ## Re: absolute Value aka modulus

it is solved thanks people