Is this true for $p >1, a_i \epsilon R$: $$(\sum_{i=1}^n |ra_i|^p)^\frac{1}{p} = |r|(\sum|a_i|^p)^\frac{1}{p}$$
It seems to work for $p = 1$: $$(\sum_{i=1}^n |ra_i|^1)^\frac{1}{1} = \sum_{i=1}^n |r||a_i| = |r|\sum_{i=1}^n|a_i| = |r|(\sum_{i=1}^n |a_i|^1)^\frac{1}{1} $$
For $p+1$ it gets a little tricky:
$$(\sum_{i=1}^n |ra_i|^{p+1})^\frac{1}{p+1} = $$ $$. . .$$ $$ |r|(\sum|a_i|^{p + 1})^\frac{1}{p + 1} $$
Any next step ideas to show it is also true for $p+1$?
Let $p$ be a given real number such that $p \geq 1$.
Then for any real numbers $r, a_1, \ldots, a_n$, we obtain $$ \begin{align} \left( \sum_{i=1}^n \left\lvert r a_i \right\rvert^p \right)^{\frac{1}{p}} &= \left( \sum_{i=1}^n \left( \left\lvert r a_i \right\rvert \right)^p \right)^{\frac{1}{p}} \\ &= \left( \sum_{i=1}^n \left( \lvert r \rvert \times \left\lvert a_i \right\rvert \right)^p \right)^{\frac{1}{p}} \\ &= \left( \sum_{i=1}^n \left( \lvert r \rvert^p \times \left\lvert a_i \right\rvert^p \right) \right)^{\frac{1}{p}} \\ &= \left( \lvert r \rvert^p \times \sum_{i=1}^n \left( \left\lvert a_i \right\rvert^p \right) \right)^{\frac{1}{p}} \\ &= \left( \lvert r \rvert^p \right)^{\frac{1}{p}} \times \left( \sum_{i=1}^n \left( \left\lvert a_i \right\rvert^p \right) \right)^{\frac{1}{p}} \\ &= \lvert r \rvert \left( \sum_{i=1}^n \left\lvert a_i \right\rvert^p \right)^{\frac{1}{p}} \\ \end{align} $$