Lost on an induction proof my colleague called basic.

Question

I just have no idear what is being asked (like what is set J) and if it is so basic why is it so general (I'm probably an idiot) .

Prove by induction on n that for any finite set I with cardinality n with $\forall i \in I$ and $\forall a_i \in \mathbb{R}$ $$\prod _{i\in I} (1 + a_i) = \sum _{j \in \mathcal{P}(I)} \left ( \prod_{j \in J} a_j \right )$$

I do not even know what $S_0$ would say. I think I am just so unfamiliar with indexed sets in pi notation that I am confusing myself.

I think that on the right the index $j$ in the sum should be $J$. $\Pi$ does for products what $\Sigma$ does for sums. I would start solving this question by writing out both sides in detail for a small explicit example. Make one up for $n=2$. — Ethan Bolker, Mar 15 '17 at 14:27
Everything's trivial once you've seen the explanation. "Obvious", "clearly", "trivial", "basic", etc. generally don't serve any purpose in mathematical exposition beyond pomp, intentional or not. To do mathematics is to struggle; you're not an idiot (or, perhaps more accurately, we all are). :) — Kaj Hansen, Mar 15 '17 at 14:32

score 1 · Accepted Answer · answered Mar 15 '17 at 14:31

1

The theorem is just stating how to expand $(1+a_1)(1+a_2)...(1+a_n)$

P(I) is the power set of I, i.e. the set of all subsets of I.

j is an element of P(I).

Also, the pi notation is similar to the $\Sigma$ notation. Just replace summation by multiplication.

Hope this helps.

answered Mar 15 '17 at 14:31

Udayan Joshi

83

Also note that the empty product ($J = \emptyset$) is by convention equal to $1$. – Umberto P. Mar 15 '17 at 14:32
Yeah I see the gist now thanks and cheers – MaQ Mar 15 '17 at 14:43

Lost on an induction proof my colleague called basic.

1 Answers1