Proof for a modulus, inequality induction =D (1 Viewer)

b3kh1t · Oct 23, 2011

Hellooooooooo

I thought this may be useful for tomorrow or at least it would not hurt to know it, and it's actually quite simple

well the question and proof is as follows;

Question is prove by mathematical induction $\left | z_{1}+z_{2}+...+z_{n} \right |\leq \left | z_{1} \right | +\left | z_{2} \right |+...+\left| z_{n} \right |$ for $n\geq 1$

Proof:
It is obviously true for n=1, therefore test for n=2
$LHS=\left | z_{1}+z_{2} \right |$ , this is the diagonal of the parallelogram formed by the sides $\left | z_{1} \right |$ and $\left | z_{2} \right |$ . Therefore, by definition of a triangle, the third side of a triangle is always less then the sum of the other two, unless the triangle is flat, meaning the vertices are collinear, this is when $\left | z_{1}+z_{2}\right |= \left | z_{1} \right | +\left | z_{2} \right |$ occurs.
$\therefore \left | z_{1}+z_{2}\right |\leq \left | z_{1} \right | +\left | z_{2} \right |$
Therefore statement is true for n=1 and n=2

Assume statement is true for n=k
$\left | z_{1}+z_{2}+...+z_{k} \right |\leq \left | z_{1} \right | +\left | z_{2} \right |+...+\left| z_{k} \right | -----> \alpha$
and let $Z= z_{1}+z_{2}+...+z_{k}+z_{k+1}$

Prove the statement is true for n=k+1
$\left | z_{1}+z_{2}+...+z_{k}+z_{k+1} \right |\leq \left | z_{1} \right | +\left | z_{2} \right |+...+\left| z_{k} \right |+\left|z_{k+1} \right |$
$LHS=\left | z_{1}+z_{2}+...+z_{k}+z_{k+1} \right |=\left | Z+z_{k+1} \right |$
$LHS\leq |Z|+|z_{k+1}|$ as seen for when proving for n=2
$\therefore LHS\leq |z_{1}|+|z_{2}|+...+|z_{k}|+|z_{k+1}| ----->\alpha$

Therefore statement is true for n=k+1, if it is true for n=k. Therefore statement is true for $n\geq 1$

NB: the equality occurs when for when $0,z_{1},z_{2},...,z_{n}$ are colliner, in other words, when $arg(z_{1})=arg(z_{2})=...=arg(z_{n})$ .

b3kh1t · Oct 27, 2011

Lol I'm not trying to beat a dead horse here but I'm just saying this was the question 8 hahahahaha may have not been exactly but they just changed it a little with the beta and alpha

cutemouse · Nov 1, 2011

Hmm... I might be wrong on this, but I don't think you need to test it for n=2...

EDIT: But it wouldn't hurt doing so because n=1 is trivial.

b3kh1t · Nov 1, 2011

cutemouse said:
Hmm... I might be wrong on this, but I don't think you need to test it for n=2...

EDIT: But it wouldn't hurt doing so because n=1 is trivial.

That's what I thought at first but you do as in the proving step, when you got big Z, you expand the same way |Z+z(k+1)| like you did when you proved for n=2 and you have to state that, and then by continuing the chain you will get the solution.

Proof for a modulus, inequality induction =D (1 Viewer)

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