# Triangular Inequality using Induction

The triangle inequality for absolute value that for all real numbers a and b,

$\left&space;|&space;a+b&space;\right&space;|\leq&space;\left&space;|&space;a&space;\right&space;|+\left&space;|&space;b&space;\right&space;|$

Use the recursive definition of summation, the triangle inequality, the definition of absolute value, and mathematical induction to prove that for all integers n, if

$a_{1},a_{2},...,a_{n}$

are real numbers, then

$\left&space;\right&space;|\sum_{i=1}^{n}&space;a_{i}&space;|&space;\leq&space;\right&space;|\sum_{i=1}^{n}&space;\left&space;\right&space;|a_{i}|$

Please help. I am extremely lost and have no idea where to begin. Any hints will be great. Thank you so very much!

• You probably mean induction. Feb 20 '16 at 0:04
• Yes, my apologies. You are right. Feb 20 '16 at 0:08

Basic step: $n = 1$

$|a_1| = |a_1|$. Done.

Induction step:

Assume $|\sum_{i=1}^k a_i|\le \sum_{i=1}^k|a_i|$. Prove $|\sum_{i=1}^{k+1} a_i|\le \sum_{i=1}^{k+1}|a_i|$

$|\sum_{i=1}^{k+1} a_i| = |(\sum_{i=1}^k a_i) + a_{k+1}|$

By the triangle inequality:

$|(\sum_{i=1}^k a_i) + a_{k+1}| \le |(\sum_{i=1}^k a_i)| + |a_{k+1}|$

we are presuming

$|\sum_{i=1}^k a_i|\le \sum_{i=1}^k|a_i|$

so

$|(\sum_{i=1}^k a_i) | + |a_{k+1}| \le \sum_{i=1}^k|a_i| + |a_{k+1}| = \sum_{i=1}^{k+1}|a_i|$.

So $|\sum_{i=1}^{k+1} a_i|\le \sum_{i=1}^{k+1}|a_i|$

So we have shown $|\sum_{i=1}^{n} a_i|\le \sum_{i=1}^{n}|a_i|$ is true for $n = 1$.

We have shown that if $|\sum_{i=1}^n a_i|\le \sum_{i=1}^n|a_i|$ is true for $n =k$ then it is true for $n = k+1$.

So by induction it is true for all natural numbers $n$.

• YOU ARE AMAZING!!!!!!!!!!!!!!!!!!!!!!!!!!!! Feb 20 '16 at 0:24