Question

Advanced engineering part(4)

Show more generally that if $\large\rm{\displaystyle\sum_{1}^{m} n_{i} = n}$ , then $\large\rm{\displaystyle\prod_{i=I}^{m} \mathbb{R}^{n} i}$ is isomorphic to $\large\rm{\mathbb{R}^{n}}$​

Answer 1

Answer:

Advanced engineering part(4)

Show more generally that if $\large\rm{\displaystyle\sum_{1}^{m} n_{i} = n}$ , then $\large\rm{\displaystyle\prod_{i=I}^{m} \mathbb{R}^{n} i}$ is isomorphic to $\large\rm{\mathbb{R}^{n}}$

Step-by-step explanation:

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