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Prove some properties of the absolute value

Prove the following:

  1. |x| = 0 if and only if x = 0.
  2. |-x| = x.
  3. |x-y| = |y-x|.
  4. |x|^2 = x^2.
  5. |x| = \sqrt{x^2}.
  6. |xy| = |x||y|.
  7. If y \neq 0 then |x/y| = |x|/|y|.
  8. |x-y| \leq |x| + |y|.
  9. |x| - |y| \leq |x-y|.
  10. ||x|-|y|| \leq |x-y|.

  1. Proof. (\Rightarrow) If |x| = 0 then -|x| \leq x \leq |x| \implies 0 \leq x \leq 0 \implies x = 0.
    (\Leftarrow) If x = 0, then |x| = x = 0 by definition of |x|. \qquad \blacksquare
  2. Proof. If x \geq 0, then |x| = x and |-x| = -(-x) = x.
    If x < 0, then |x| = -x and |-x| = -x (since x < 0 \implies -x > 0). \qquad \blacksquare
  3. Proof. If x - y > 0, then y - x < 0, so |x-y| = x-y and |y-x| = -(y-x) = x-y.
    If y-x > 0, then x-y < 0, so |x-y| = -(x-y) = y-x and |y-x| = y-x.
    If x-y=0, then y-x= 0 so |x-y| = |y-x| = 0. \qquad \blacksquare
  4. Proof. First, |x|^2 = |x|\cdot |x|. So, if x \geq 0 then |x|=x \implies |x| \cdot |x| = x^2.
    If x < 0 then |x| = -x \implies |x| \cdot |x| = (-x)\cdot(-x) = x^2. \qquad \blacksquare
  5. Proof. Since \sqrt{x^2} is defined to be the unique non-negative square root of x^2, from part (d) we have |x|^2 = x^2, and |x| is non-negative by definition; hence, |x| = \sqrt{x^2}. \qquad \blacksquare
  6. Proof. If x \geq 0, y \geq 0, then xy \geq 0, so |xy| = xy. Further, |x| = x, \ |y| = y, so |x||y| = xy.
    If x < 0 and y \geq 0, then xy < 0 so |xy| = -(xy). Then, |x| = -x and |y| = y, so |x||y| = (-x)y = -(xy).
    If x \geq 0 and y < 0, then xy < 0 so |xy| = -(xy). Then |x| = x and |y| = -y, so |x||y| = x(-y) = -(xy).
    Finally, if x < 0 and y < 0, then xy > 0 so |xy| = xy. Then |x| = -x and |y| = -y, so |x||y| = (-x)(-y) = xy. \qquad \blacksquare
  7. Proof. Using the definition of x/y we have

        \[ \left| \frac{x}{y} \right| = |xy^{-1}| = |x||y^{-1}|, \]

    by part (f). But then since |y||y^{-1}| = |yy^{-1}| = |1| = 1, we have |y^{-1}| = |y|^{-1}. So,

        \[ |x||y^{-1}| = |x|(|y|^{-1}) = \frac{|x|}{|y|}. \qquad \blacksquare \]

  8. Proof. Using the triangle inequality and the fact that |-y| = |y|,

        \[ |x-y| = |x+(-y)| \leq |x| + |-y| = |x|+|y|. \qquad \blacksquare \]

  9. Proof. Here we use the trick of adding and subtracting y, and then the triangle inequality,

        \[ |x| = |(x-y)+y| \leq |x-y| + |y| \implies |x| - |y| \leq |x-y|. \qquad \blacksquare \]

  10. Proof. Using a similar trick to part (i), we have

        \[ |y| = |x+(y-x)| \leq |x| + |y-x| = |x| + |x-y| \implies |y| - |x| \leq |x-y| \implies |x| - |y| \geq -|x-y|. \]

    Then, combining this inequality with the one in part (i) and applying the definition of the absolute value,

        \[ -|x-y| \leq |x| - |y| \leq |x-y| \implies ||x|-|y|| \leq |x-y|. \]

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