# Thread: Triangle inequality induction problem

1. ## Triangle inequality induction problem

Prove by induction that:

$|z_{1}+z_{2}+...+z_{n}| \leq |z_{1}|+|z_{2}|+...+|z_{n}|$

$n \in \mathbb{N}$ and $n\geq 1$

I seem to understand the components of this problem as I feel comfortable with the inductive proof process and I can solve for the modulus of complex numbers, e.g., if $z = 6+2i$ then $|z| = \sqrt40$. I also see that the two shorter sides of a triangle must be at least the length of the longest side and, if they are not parallel with it, then their sum must be greater than the longest side etc. However, I am not sure how to start with this inductive proof. How do I test the base case if z represents a complex number and I don't know the formula for it? How do I sub in a value for it?

How do I start? where is the gap in my understanding of these topics.

Thanks.

2. ## Re: Triangle inequality induction problem

Originally Posted by terrorsquid
Prove by induction that:
$|z_{1}+z_{2}+...+z_{n}| \leq |z_{1}|+|z_{2}|+...+|z_{n}|$
$n \in \mathbb{N}$ and $n\geq 1$
I am not sure how to start with this inductive proof. How do I test the base case if z represents a complex number and I don't know the formula for it? How do I sub in a value for it?
How do I start? where is the gap in my understanding of these topics.
The real difficulty here is proving the base case: $|z_1+z_2|\le |z_1|+|z_2|.$
That is the triangle inequality. If have done that, the rest is easy.
Proving that requires showing $|z_1+z_2|^2\le \left|\,|z_1|+|z_2|\,\right|^2$.

Once you have that, then assume an inductive step.
NOTATION: $A_n = \sum\limits_{k = 1}^n {z_k } \;\& \,B_n = \sum\limits_{k = 1}^n {\left| {z_k } \right|}$
If we know that $|A_N|\le B_N$ then $|A_{N+1}|=|A_N+z_{n+1}|\le B_N+|z_{N+1}|=B_{N+1}$.

3. ## Re: Triangle inequality induction problem

I am a little unsure of the notation I guess; how do I treat the $z_{1}, z_{2}, ...$ expressions? Do I let them equal to $x + iy$ or something? I don't see how I could test the base case if they are just variables and they are non negative numbers (natural).

4. ## Re: Triangle inequality induction problem

Originally Posted by terrorsquid
I am a little unsure of the notation I guess; how do I treat the $z_{1}, z_{2}, ...$ expressions? Do I let them equal to $x + iy$ or something? I don't see how I could test the base case if they are just variables and they are non negative numbers (natural).
What part of the notation do you not understand?
Surely you know that $\sum\limits_{k = 1}^{N + 1} {z_k } = \left( {\sum\limits_{k = 1}^N {z_k } } \right) + z_{N + 1}$

Once again the really difficult part of the problem is the proof of the base case: $|z+w|\le |z|+|w|$.

5. ## Re: Triangle inequality induction problem

I wasn't talking about your notation. I meant the $z_{n}$ terms. If they are just symbols then squaring both sides would give $z_{1}^2+2z_{1}z_{2}+z_{2}^2$ on both sides so then they are equal and the base case has held true? I don't know what they represent.

I can see that it should be true as when $z_{n}$ is positive and $z_{n+1}$ is negative, then the LHS would be the absolute value of the difference and the RHS would still be the sum of the two absolute values and therefore the RHS would be greater. Obviously if both terms are positive then they are equal. But, again, I don't know what the terms represent in this example.

6. ## Re: Triangle inequality induction problem

Originally Posted by terrorsquid
I wasn't talking about your notation. I meant the $z_{n}$ terms. If they are just symbols then squaring both sides would give $z_{1}^2+2z_{1}z_{2}+z_{2}^2$ on both sides so then they are equal and the base case has held true? I don't know what they represent. I can see that it should be true as when $z_{n}$ is positive and $z_{n+1}$ is negative, then the LHS would be the absolute value of the difference and the RHS would still be the sum of the two absolute values and therefore the RHS would be greater. Obviously if both terms are positive then they are equal. But, again, I don't know what the terms represent in this example.
I am really confused now.
Are you proving for complex numbers or not?
If $z$ is complex then it is neither positive nor negative
If these are real numbers the proof is easy.
If they are complex then the base case is difficult.
To avoid subscripts suppose that $z~\&~w$ are complex.
$|z+w|^2=(z+w)(\overline{z}+\overline{w})$. Expand that.

7. ## Re: Triangle inequality induction problem

Originally Posted by Plato
I am really confused now.
Are you proving for complex numbers or not?
If $z$ is complex then it is neither positive nor negative
If these are real numbers the proof is easy.
If they are complex then the base case is difficult.
To avoid subscripts suppose that $z~\&~w$ are complex.
$|z+w|^2=(z+w)(\overline{z}+\overline{w})$. Expand that.
Sorry to confuse you :S I am confused myself, as Im sure you have noticed I suspect it is complex considering all the other questions I did were complex... Its mainly that z is arbitrary. I was fine solving equations where z was defined as $5-3i$ or something but now they are just z's and I don't know what to do with them.

$z\overline{z}+z\overline{w}+w\overline{z}+w \overline{w}$

$= |z|^2+|w|^2+z\overline{w}+w\overline{z}$

8. ## Re: Triangle inequality induction problem

Originally Posted by terrorsquid
$z\overline{z}+z\overline{w}+w\overline{z}+w \overline{w}$
$=|z|^2+|w|^2+z\overline{w}+w\overline{z}$
You need to know that
$z\overline{w}+w\overline{z}=2\text{Re}(z\overline{ w})$
and
$\text{Re}(z\overline{w})\le|z||w|$

9. ## Re: Triangle inequality induction problem

So I have:

$|z|^2+|w|^2+2Re(z\overline{w})\le |z|^2+|w|^2+2|z||w|$

$Re(z\overline{w})\le |z||w|$ which is true.

I haven't come across that identity yet. What is it called so I can learn more about it?

Thanks so much for the help. Excellent as always.

10. ## Re: Triangle inequality induction problem

Originally Posted by terrorsquid
$Re(z\overline{w})\le |z||w|$ which is true.
I haven't come across that identity yet. What is it called so I can learn more about it?
Just prove it. Let $z=a+b\mathif{i}~\&~w=c+d\mathif{i}$

Is it true that $Re(z)\le |z|~?$