The triangle inequality for absolute value that for all real numbers a and b,
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
are real numbers, then
Please help. I am extremely lost and have no idea where to begin. Any hints will be great. Thank you so very much!
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 $n$.