Can Structural Induction Prove All Elements of Set T Are Powers of 2?

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Discussion Overview

The discussion revolves around the recursive definition of a set T and the use of Structural Induction to prove that all elements of T are powers of 2. Participants explore the implications of the recursive rules defining T and the nature of its elements.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the proof should start by assuming that for all x in T, there exists an n in N such that x = 2n, and then show this holds for x/2 and x^2.
  • Others argue that the recursive definition of T only generates powers of 2, not any even number, suggesting that the structure of T limits its elements to powers of 2.
  • A participant mentions that examining simple examples can lead to a general argument, noting that starting from 2 and applying the recursive rules consistently yields only powers of 2.
  • There is a discussion about the role of the number 1 in the set, with some suggesting it does not contribute new elements to T beyond those generated from 2.

Areas of Agreement / Disagreement

Participants generally agree that the recursive rules lead to the conclusion that elements of T are powers of 2, but there is some uncertainty regarding the role of the number 1 and the completeness of the proof using Structural Induction.

Contextual Notes

Some participants express confusion about the application of Structural Induction and the implications of substituting values into the recursive definitions. There are unresolved questions about the completeness of the proof and the assumptions made regarding the elements of T.

thy
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Consider the set T defined recursively as follows:

• 2∈T,

• if x∈T and x>1,then x/2 ∈T,

• if x∈T and x>1,then x^2 ∈T,

• T contains no other element.

Use Structural Induction to write a detailed, carefully structured proof that ∀ x ∈ T, ∃ n ∈ N, x = 2n.


I'm not sure how to do this proof.
I assume ∀ x ∈ T, ∃ n ∈ N, x = 2n is true and try to prove then x/2 ∈T and x^2 ∈T?
 
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thy said:
Consider the set T defined recursively as follows:

• 2∈T,

• if x∈T and x>1,then x/2 ∈T,

• if x∈T and x>1,then x^2 ∈T,

• T contains no other element.

Use Structural Induction to write a detailed, carefully structured proof that ∀ x ∈ T, ∃ n ∈ N, x = 2n.


I'm not sure how to do this proof.
I assume ∀ x ∈ T, ∃ n ∈ N, x = 2n is true and try to prove then x/2 ∈T and x^2 ∈T?

You just get powers of 2 from the recursion not any even number.
 
lavinia said:
You just get powers of 2 from the recursion not any even number.

substituting 2n into x? So it becomes (2n) ^2?
I don't really get it...
 
I find with inductions that looking at simple examples leas to the general argument.

So if 2 is in X then 1 is in X (2/2 = 1) so these are both powers of 2. The recursion only allows you to take these two numbers to get the next numbers. But 1 does not give any new numbers so we can forget about it and just look at 2. 2^2 is 4, another power of 2, and 4/2 = 2. So again you only have powers of 2. Form 4 you get 4^2, another power of 2, and dividing by 2 you get 8 and 4 and 2 again. This is the only thing that happens. Forn * and 16 you get more powers of two and so on.
 
lavinia said:
I find with inductions that looking at simple examples leas to the general argument.

So if 2 is in X then 1 is in X (2/2 = 1) so these are both powers of 2. The recursion only allows you to take these two numbers to get the next numbers. But 1 does not give any new numbers so we can forget about it and just look at 2. 2^2 is 4, another power of 2, and 4/2 = 2. So again you only have powers of 2. Form 4 you get 4^2, another power of 2, and dividing by 2 you get 8 and 4 and 2 again. This is the only thing that happens. Forn * and 16 you get more powers of two and so on.


thanks, it helps a lot
 

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