Inductive Proof (+linear equation in four variables)

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Homework Help Overview

The discussion revolves around proving by induction that for all integers n greater than or equal to 5, there exist natural numbers m1 and m2 such that n can be expressed as a linear combination of 2 and 3. The problem involves understanding the conditions under which such combinations can be formed.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the validity of using induction for this proof, with some suggesting that a proof by cases may be more straightforward. Others share their attempts at the inductive step and express curiosity about solving for the variables involved.

Discussion Status

The discussion includes various perspectives on the proof method, with some participants having successfully completed the induction while others are still exploring their understanding of the problem. Guidance has been offered regarding the inductive step, but no consensus has been reached on the best approach.

Contextual Notes

Some participants question the necessity of induction for this problem, considering the simplicity of the cases based on the parity of n. There is also mention of additional base cases that may facilitate the proof.

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Homework Statement



I'm trying to prove by induction that \forall n \geq 5, \exists m_1, m_2 \in \mathbb{N} such that n = 2m_1 +3m_2.

Homework Equations



(none, really)

The Attempt at a Solution



I've done the base case, and the inductive step boils down to showing that \exists m_1 \prime m_2 \prime such that 2m_1 +3m_2 +1 = 2m_1 \prime +3m_2 \prime. Maybe I'm forgetting something from grade school algebra, but I have no idea how to solve for m_1 \prime, m_2 \prime. I've plugged it into wolfram alpha [http://www.wolframalpha.com/input/?i=2*x_1+3*x_2+++1+=+2*y_1+++3*y_2] and got solutions (all I care about is the case n =1 for the integer solutions wolfram gives) but I want to know how to arrive there.
 
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Is this really something that you need induction for?

There are really only two cases for any n. Either 2|n, or it does not. Both cases are pretty easy to see what m1 and m2 need to be for any given n.

I'll think about how to proceed with an inductive proof for a little while (until I give up, at least). But, if it were me, I'd choose a proof by cases.
 
I managed to prove this using induction, actually. It's a lot easier if you prove an additional base case (n=6) and then prove it for n >=7. I am still curious as to how to solve for those variables though.
 
In case you're curious, for the inductive step just add one to both sides of your induction hypothesis, then factor out 2's to obtain the desired expression.
 

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