# For real numbers $z$ and $w$, $|(1+z)(1+w)-1| \leq (1+|z|)(1+|w|)-1$.

I have written an attempted proof of the theorem on the title, and I need help verifying it. I have used the following theorems to proof the theorem on the title.

Theorem 5.14)a) Let $$x$$ be a real number. $$-|x| \leq x \leq |x|$$.

Theorem 5.14)b) Let $$a \geq 0$$. $$|x| \leq a$$ if and only if $$-a \leq x \leq a$$.

Theorem 5.14)c) Let $$x$$ and $$y$$ be real numbers. $$|x+y| \leq |x| + |y|$$ (The Triangle Inequality).

For real numbers $$z$$ and $$w$$, $$|(1+z)(1+w)-1| \leq (1+|z|)(1+|w|)-1$$.

Proof. From Theorem 5.14)a), $$(1+z) \leq |(1+z)|$$ and $$(1+w) \leq |(1+w)|$$. Multiplying the $$(1+z) \leq |(1+z)|$$ by $$|(1+w)|$$, one obtains

\begin{align} (1+z)|(1+w)| \leq |(1+z)||(1+w)| \end{align}

Since $$(1+w) \leq |(1+w)|$$, it follows that

\begin{align} (1+z)(1+w) \leq |(1+z)||(1+w)| \end{align}

Observe that from The Triangle Inequality,

\begin{align} |(1+z)| \leq 1 + |z|\\ |(1+w)| \leq 1 + |w| \end{align}

Since $$|(1+z)|(1 + |w|) \leq (1 + |z|)(1 + |w|)$$ and $$|(1+w)| \leq (1 + |w|)$$, it follows that

\begin{align} (1+z)(1+w) -1 \leq |(1+z)||(1+w)| -1 \leq (1 + |z|)(1 + |w|) -1 \end{align}

It is clear that $$(1 + |z|)(1 + |w|) -1 \geq 0$$.

In case that $$(1+z)(1+w) -1 \geq 0$$, $$|(1+z)(1+w) -1| = (1+z)(1+w) -1$$.

Hence,

\begin{align} -[(1 + |z|)(1 + |w|) -1] \leq 0 \leq (1+z)(1+w) -1 \leq (1 + |z|)(1 + |w|) -1 \end{align}

From Theorem 5.14)b),

\begin{align} |(1+z)(1+w) -1| \leq (1 + |z|)(1 + |w|) -1 \end{align}

establishing the result for this case.

On the other hand, in case that $$(1+z)(1+w) -1 < 0$$, $$|(1+z)(1+w) -1| = -[(1+z)(1+w) -1]$$.

Hence,

\begin{align} -[(1 + |z|)(1 + |w|) -1] \leq (1+z)(1+w) -1 < 0 \end{align}

Since $$(1 + |z|)(1 + |w|) -1 \geq 0$$,

\begin{align} -[(1 + |z|)(1 + |w|) -1] \leq (1+z)(1+w) -1 \leq (1 + |z|)(1 + |w|) -1 \end{align}

Multiplying the inequalities by $$-1$$, one obtains

\begin{align} (1 + |z|)(1 + |w|) -1 \geq -[(1+z)(1+w) -1] \geq -[(1 + |z|)(1 + |w|) -1] \end{align}

From Theorem 5.14)b),

\begin{align} |(1+z)(1+w) -1| \leq (1 + |z|)(1 + |w|) -1 \end{align}

establishing the result for this case.

Because the result for all cases of $$(1+z)(1+w) -1$$ have been established, it is the case that $$|(1+z)(1+w) -1| \leq (1 + |z|)(1 + |w|) -1$$.

• Were these Theorems and the problem from Daepp & Gorkin's book, "Reading, Writing, and Proving"? I am working through it as well and I believe you can just use the Triangle Inequality twice to prove the result. Dec 30, 2022 at 0:21

We have :

$$(1+z)(1+w)-1 = z+w+zw$$

Thus :

$$|(1+z)(1+w)-1| = |z+w+zw| \le |z| + |w| + |zw| = (1+|z|)(1+|w|) -1$$

QED

• Is there any mistake on my attempted proof? , and thank you for the shorter version of the proof. Dec 1, 2021 at 15:31
• Not that I could spot, but it is quite convoluted Dec 1, 2021 at 15:54

To answer the solution-verification part of the question, this step is wrong.

one obtains

\begin{align} (1+z)|(1+w)| \leq |(1+z)||(1+w)| \end{align}

Since $$(1+w) \leq |(1+w)|$$, it follows that

\begin{align} (1+z)(1+w) \leq |(1+z)||(1+w)| \end{align}

The above is of the form $$\, a \cdot |b| \le c \implies a \cdot b \le c\,$$, which does not hold true in general. For example, if $$\,a=b=-2, \,c=1\,$$ then $$\,a \cdot |b| = -4 \le 1 = c\,$$, but $$\,a \cdot b = 4 \gt 1 = c\,$$. The implication does hold true for $$\,a \ge 0\,$$, but here $$\,a = 1 + z\,$$ which is not necessarily positive.

The correct way to derive the inequality is to use that $$\,|a \cdot b| = |a| \cdot |b|\,$$, then:

$$(1+z)\cdot (1+w) \;\leq\; |(1+z)\cdot(1+w)| \;=\; |1+z|\cdot|1+w|$$

• But in this case $c$ is not any real numbers. It is $|a||b|$ where $a = 1+z$ and $b = 1+w$ in which Theorem 5.14)a) have stated that $ab \leq |ab|$. Hence, why is it not the case that $a\cdot |b| \leq |a||b|$? Dec 6, 2021 at 16:02
• @Approxiz The inequality you get in the end is correct, but the justification as-written is not. "Since $(1+w) \leq |(1+w)|$, it follows that..." sounds like you take $(1+w) \leq |(1+w)|$ and multiply it by $1+z$ to get $(1+z)\cdot(1+w) \color{red}{\leq} (1+z) \cdot |(1+w)| \le |1+z| \cdot |1+w|$ but the first $\color{red}{\le}$ does not hold true in general. The end result is, again, correct, so you can justify that step in a different way - for example as proposed in my answer, or hinted in your comment here.
– dxiv
Dec 6, 2021 at 16:30