Simple Proof by Induction Problem

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SUMMARY

The discussion centers on a proof by induction problem involving the sets defined as S_0 = {} and S_i = {0} ∪ {s + π | s ∈ S_{i-1}}. The goal is to demonstrate that S_i ⊆ S_{i+1} for all i. The user, Patrick, outlines an approach using nested induction but seeks guidance on progressing from the inductive hypothesis to the conclusion. The solution involves showing that if x is in S_n, then x must also be in S_{n+1}, leveraging the properties of the defined sets.

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  • Understanding of mathematical induction principles
  • Familiarity with set notation and operations
  • Knowledge of sequences and series in mathematics
  • Basic comprehension of the properties of π in mathematical contexts
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  • Study the principles of nested induction in mathematical proofs
  • Explore set theory, focusing on operations involving unions and subsets
  • Review examples of proofs involving sequences defined recursively
  • Investigate the implications of using constants like π in set definitions
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CuppoJava
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Hi,

I devised this problem for myself as part of a bigger question that I'm working on, and am having trouble solving it. I think it involves a nested induction proof but I am not sure how to start. A tip on how to begin would be much appreciated.

Thanks
-Patrick

Homework Statement



<br /> S_0 = \{\}<br />
<br /> S_i = \{0\} \bigcup \{s + \pi | s \in S_{i-1}\}<br />
Show that S_i \subseteq S_{i+1} for all i.

The Attempt at a Solution



I started by converting the problem to the following form:

\forall i &gt; 0, 0 \in S_i
\forall s \in S_i, s + \pi \in S_{i + 1}
Show that \forall s \in S_i, s \in S_{i + 1}

But I don't know what to do next.
 
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Suppose true up to some i = n. Let x be in Sn. You want to show it is in Sn+1. Trivial if x = 0, otherwise x = y+π for some y in Sn-1. By the inductive hypothesis, since y is in Sn-1 it is also in ...?
 
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... therefore y is in S_n according to the assumed hypothesis.
Therefore y + 1 = x is in S_{n+1}.
Therefore S_i \subseteq S_{i+1} by induction.

Thank you!
-Patrick
 

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