Use induction to prove $|a_{1} + a_{2} + \dots + a_{n} | \leq |a_{1}| + |a_{2}| + \dots + |a_{n}|$.

Question

The base case is obvious, as $|a_{1}| \leq |a_{1}|$ is true.

The inductive case is where I'm having trouble. Here is what I have: Suppose $|a_{1} + a_{2} + \dots + a_{n}| \leq |a_{1}| + |a_{2}| + \dots + |a_{n}|$ is true for $n\in \mathbb{N}$. From this, $|a_{1} + (a_{2} + \dots + a_{n} + a_{n+1})| \leq |a_{1}| + |a_{2} + \dots + a_{n} + a_{n+1}|$.

How can I show that if you keep applying the additive associative property of fields repeatedly to the inside part of the absolute value, then you will eventually show the original proposition?

Any help would be greatly appreciated. Thank you.

Answer 1

$|a_1 + a_2 + ..... + a_n + a_{n+1}| = |(a_1 + a_2 + ..... + a_n) + a_{n+1}|\le$

$|a_1 + a_2 + ..... + a_n)| + |a_{n+1}|$ (as those are only two terms)

$\le (|a_1| + |a_2| + ..... + |a_n|) + |a_{n+1}|$ (as those are $n$ terms)

$= |a_1| + |a_2| + ..... + |a_n| + |a_{n+1}|$

P.S.

"How can I show that if you keep applying the additive associative property of fields repeatedly to the inside part of the absolute value, then you will eventually show the original proposition?"

You don't have to.

$|a_1 + ..... + a_n| \le |a_1| + |a_2| + ..... + |a_n|$ because there are $n$ terms.

So

$|a_2 + ....... + a_{n+1} | \le |a_2| + |a_3|+ ..... +|a_{n+1}|$ because there are $n$ terms.

P.S.S (!!!IMPORTANT!!!)

Because the induction case uses the hypotheses on two terms, the base case that $n = 1$ won't do it.

This leads to all "all horses are the same color" paradox.

Consider $|a_1 + .... a_n| \ge |a_1| + .... + |a_n|$ (Not true).

True for $n = 1$: As $|a_1| \ge |a_1|$.

And True for induction as $|(a_1+ ..... + a_n) + a_{n+1}| \ge |a_1+ .... + a_n| + |a_{n+1}| \ge |a_1|+..... + |a_n| + |a_{n+1}|$

But it isn't true as we never showed it was true for $n = 2$.

So base case is you must show $|a + b| \le |a| + |b|$ can you do that?

Answer 2

Type instead $$|(a_1+a_2+\cdots + a_n)+a_{n+1}|\leq |a_1+\cdots + a_n|+|a_{n+1}|$$ And now use the inductive step to conclude

Answer 3

If your "from this" step is valid, couldn't you do this: $$\begin{align} |(a_{1} + a_{2} + \dots + a_{n}) + a_{n+1}| & \leq |a_{1} + a_{2} + \dots + a_{n}| + |a_{n+1}| \\ &\leq |a_{1}| + |a_{2}| + \dots + |a_{n}| + |a_{n+1}| \end{align}$$ with the last step being from the inductive hypothesis?

Something seems fishy about the first step though, unless you can consider it a case where n=2?