Problem 12

Prove the following:

(i) $|xy| = |x| \cdot |y|$ \[ \begin{align*} (|xy|)^2 &= (xy)^2 \\ &= x^2y^2 \\ &= (|x|)^2 \cdot (|y|)^2 \\ &= (|x| \cdot |y|)^2 && \text{| exponential can be eliminated} \\ \end{align*} \]

(ii) $|\frac{1}{x}| = \frac{1}{|x|}$, if $x \neq 0$. \[ \begin{align*} |\frac{1}{x}| &= |1x^{-1}| \\ \rightarrow (|1x^{-1}|) &= (|1| \cdot |x^{-1}|) && \text{| based on (i)} \\ &= |1| \cdot (\sqrt{(x^{-1})^2}) \\ &= |1| \cdot (\sqrt{\frac{xx}{x}}) && \text{| no inherent order to exponentials here} \\ &= |1| \cdot (\sqrt{(x^2)^{-1}}) \\ &= |1| \cdot ((x^2)^{-1})^{\frac{1}{2}}) \\ &= |1| \cdot (((x^2)^{\frac{1}{2}})^{-1}) \\ &= |1| \cdot ((\sqrt{x^2})^{-1}) \\ &= |1| \cdot \frac{1}{|x|} \\ \end{align*} \]

(iii)* $\frac{|x|}{|y|} = |\frac{x}{y}|$, if $y \neq 0$ \[ \begin{align*} \frac{|x|}{|y|} &= |x| \cdot |y|^{-1} \\ &= |xy^{-1}| && \text{| based on (ii)} \\ &= |\frac{x}{y}| \\ \end{align*} \]

(iv) $|x - y| \leq |x| + |y|$

When $z = -y$ we get $|x + z| \leq |x| + |z|$ and that is already proven to be true.

(v) $|x| - |y| \leq |x - y|$ \[ \begin{align*} (x - y)^2 &= x^2 - 2xy - y^2 \\ &\geq x^2 - 2|x| \cdot |y| - y^2 \\ &= |x|^2 - 2|x| \cdot |y| - |y|^2 \\ &= (|x| - |y|)^2 && \text{| and then } a^2 \geq b^2 \rightarrow a \geq b \\ \end{align*} \]

(vi) $|(|x| - |y|)| \leq |x - y|$ \[ \begin{align*} |x - y| &= ||x - y|| \\ &\geq |(|x| - |y|)| \\ \end{align*} \]

(vii) $|x + y + z| \leq |x| + |y| + |z|$. Indicate when equality holds, and prove your statement. \[ \begin{align*} |x + y + z| &= |x + a| \\ &\leq |x| + |a| \\ \\ |y + z| &\leq |y| + |z| \\ \\ |x + y + z| &\leq |x| + |y| + |z| \\ \end{align*} \] For this expression to be equal $|y + z| = |y| + |z|$ and based on that $|x + a| = |x| + |a|$ which is only the case when $x, y, z \geq 0$. See: \[ \begin{align*} (|a| + |b|)^2 &= |a|^2 + 2|a||b| + |b|^2 \\ &= a^2 + 2|a||b| + b^2 \\ &= a^2 + 2ab + b^2 && \text{| with } a, b \geq 0 \\ &= (|a + b|)^2 \\ \end{align*} \]

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