Induction with more than one variable

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Discussion Overview

The discussion revolves around the use of mathematical induction to prove statements involving two variables, specifically whether it is sufficient to prove a statement for one variable while leaving the other arbitrary, or if both variables must be treated through induction. Participants explore different approaches to induction in the context of proving statements like the binomial theorem.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that it may be sufficient to define Q(n) or R(m) and prove one of these by induction to establish the truth of P(n, m).
  • Others argue that it is necessary to prove both P(n, m) => P(n+1, m) and P(n, m) => P(n, m+1) to show that P(n, m) holds for all arbitrary n and m.
  • A participant mentions that proving P(n, m) => P(n, m+1) while leaving n arbitrary could suffice, suggesting that this approach might be valid in certain cases.
  • Another participant raises a concern about the application of induction to non-integer variables, citing the binomial theorem as an example where induction is not applied to arbitrary x and y.
  • One participant suggests that if P(n, 1) is trivially true for all n, it might be possible to prove P(n, m) without needing to perform induction on n.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of proving statements for both variables through induction. There is no consensus on whether proving one variable while leaving the other arbitrary is sufficient.

Contextual Notes

Participants note that the treatment of variables as arbitrary may depend on their nature (e.g., integers vs. real numbers) and the specific context of the statements being proved.

µ³
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Suppose we have a statement (say an equation) P(n, m), and we want to prove it works for arbitrary n and m.

Either

(1) It is sufficient to define Q(n) = P(n,m) or R(m) = P(n,m) and prove either Q(n) or R(m) by induction

(2) It is necessary to prove that P(n,m) => P(n+1,m) and P(n, m) => P(n, m+1), in addition to the base cases, to show that it works for arbitrary n and m.

Which one is correct !?
 
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µ³ said:
Suppose we have a statement (say an equation) P(n, m), and we want to prove it works for arbitrary n and m.

Either

(1) It is sufficient to define Q(n) = P(n,m) or R(m) = P(n,m) and prove either Q(n) or R(m) by induction

(2) It is necessary to prove that P(n,m) => P(n+1,m) and P(n, m) => P(n, m+1), in addition to the base cases, to show that it works for arbitrary n and m.

Which one is correct !?
If in (1) you prove Q(n) for ALL n or prove R(n) for ALL m, that would be sufficient. That would, in most cases involve proving P(n,m)=> P(n+1, m) to prove R(m) for a specific m, and then do another induction, proving R(m)=> R(m+1) to prove it true for ALL m. In other words, the two are really the same!
 
HallsofIvy said:
If in (1) you prove Q(n) for ALL n or prove R(n) for ALL m, that would be sufficient. That would, in most cases involve proving P(n,m)=> P(n+1, m) to prove R(m) for a specific m, and then do another induction, proving R(m)=> R(m+1) to prove it true for ALL m. In other words, the two are really the same!
The thing that confuses me is that for example to prove the binomial theorem, which is a statement P(x,y,n), we can simply prove that P(x,y,n) => P(x,y, n+1) and we don't have to prove it for all x and y, as we can just leave them arbitrary. Similarly why can't we prove, P(n,m) by proving P(n,m) => P(n, m+1), leaving n arbitrary and then assuming since we left n arbitrary that it will work for all n?
 
µ³ said:
The thing that confuses me is that for example to prove the binomial theorem, which is a statement P(x,y,n), we can simply prove that P(x,y,n) => P(x,y, n+1) and we don't have to prove it for all x and y, as we can just leave them arbitrary. Similarly why can't we prove, P(n,m) by proving P(n,m) => P(n, m+1), leaving n arbitrary and then assuming since we left n arbitrary that it will work for all n?
??But x and y are not integers so you wouldn't use "induction" on them anyway! Leaving x and y arbitrary- so that they can be any real numbers- does prove it for all x and y.
 
So what if P(n, 1) is trivially true for arbitrary n, then surely we can prove P(n, m) by showing P(n, m) => P(n, m+1) without having to do induction on n?
 
IF you can prove it for arbitrary n, yes.
 

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