If $z_1, z_2, \ldots , z_n$ are complex numbers, then prove that $|z_1 + \cdots + z_n| \le |z_1| + \cdots + |z_n|.$
Hello, I've tried to prove this problem with induction, but I'm not sure if I'm using the induction hypothesis properly. From what I understand, I need to reduce the inductive step down to something that looks "similar" to the induction hypothesis, and use that as a proof.
What I did was first use the $n = 2$ case as a base case:
$|z + w| \le |z| + |w|$ (proof was covered in class, and we did not have to re-prove it).
Inductive Hypothesis:
(I) Assume that $|z_1 + \cdots + z_n| \le |z_1| + \cdots + |z_n|.$
Inductive Step:
$$|z_1 + \cdots + z_n + z_{n+1}| \le |z_1| + \cdots + |z_n| + |z_{n+1}|.$$
(II) Here, $z_1 + \cdots + z_n$ is really just some other complex number, $w$. So, we can rewrite the inequality as $|w + z_{n+1}| \le |z_1| + \cdots + |z_n| + |z_{n+1}|.$
(III) From our base case and inductive hypothesis, we know that $|z + w| \le |z| + |w|$ and $w = |z_1 + \cdots + z_n| \le |z_1| + \cdots + |z_n| = |w|.$ (I feel like I'm missing something here. Am I?)
From this, we can rewrite the inductive step as $|w + z_{n+1}| \le |w| + |z_{n+1}|.$ This expression is something that's been proven in the base case, and therefore we know the inequality holds. Therefore, the inductive step holds, and the proof is complete.
Have I properly used induction here? I feel a bit iffy on stating $|z_1| + \cdots + |z_n| = |w|$ in (III). Is this something I'd need to also prove separately? Also, I feel like I'm relying on the base case too much rather than the induction hypothesis. Is this something I'm allowed to do in induction? Thank you for your time and input!
Answer
You have a pretty good foundation here. Some clean-up...
For the inductive step,
$$
|(z_1 + \cdots + z_n) + z_{n+1}| \leq |z_1 + \cdots + z_n| + |z_{n+1}| \leq |z_1|+ \cdots + |z_n|+ |z_{n+1}|
$$
and you're done. The first inequality is your application of the basic fact for two terms, and the last is the usage of the inductive hypothesis.
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