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!

  • $\begingroup$ You probably mean induction. $\endgroup$
    – Git Gud
    Feb 20, 2016 at 0:04
  • $\begingroup$ Yes, my apologies. You are right. $\endgroup$ Feb 20, 2016 at 0:08

1 Answer 1


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

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


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

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 $n$.

  • $\begingroup$ YOU ARE AMAZING!!!!!!!!!!!!!!!!!!!!!!!!!!!! $\endgroup$ Feb 20, 2016 at 0:24

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