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Math Help - absolute Value aka modulus

  1. #1
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    absolute Value aka modulus

    mod(a+b) is greater than or equal to mod(mod(a)+mod(b))
    i dont get it
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  2. #2
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    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.
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  3. #3
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    Re: absolute Value aka modulus

    Quote Originally Posted by abdulrehmanshah View Post
    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.
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  4. #4
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    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)?
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  5. #5
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    Re: absolute Value aka modulus

    ||a|−|b||≤|a+b| this is what i meant
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  6. #6
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    Re: absolute Value aka modulus

    Quote Originally Posted by abdulrehmanshah View Post
    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

    $\left\lvert \lvert x \rvert - \lvert y \rvert \right\rvert \le |x+y|$
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  7. #7
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    Re: absolute Value aka modulus

    it is solved thanks people
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