Discrete Math Proof: Necessary Condition for Divisibility by 6

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Homework Help Overview

The discussion revolves around a proof concerning divisibility, specifically addressing the necessary condition for an integer to be divisible by 6, which is that it must also be divisible by 2. Participants are exploring the structure and clarity of mathematical proofs in the context of discrete mathematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants share their attempts at constructing a proof, with one original poster expressing doubt about their proof's validity. They present their reasoning and seek feedback on potential flaws. Others engage in discussing the clarity and conciseness of the proof, questioning whether certain statements are necessary.

Discussion Status

The conversation is ongoing, with participants providing feedback on the original proof attempts. Some guidance has been offered regarding the simplification of language and the removal of unnecessary details, indicating a productive exchange of ideas. However, there is no explicit consensus on the proof's overall effectiveness.

Contextual Notes

Participants are navigating the challenges of writing proofs for the first time, which may influence their confidence and the complexity of their explanations. There is a focus on clarity and conciseness in mathematical writing.

in the rye
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Homework Statement


We have JUST started writing proofs recently, and I am a little bit doubtful in my abilities in doing this, so I just want to verify that my proof actually works. I was expecting this one to be a lot longer since the previous 2 were. I don't see any glaring flaws in it, but I'd just like to be sure (writing these feel awkward since this is my first proof base course):

A necessary condition for an integer to be divisible by 6 is that it is divisible by 2.

Homework Equations

The Attempt at a Solution


Assume true.

Pf./
[For all integers n, if n is divisible by 6, then n is divisible by 2]. Assume 6|n, n ∈ ℤ. By definition, n = 6k, k ∈ ℤ. Consider that n = 2(3k). See that 3k ∈ ℤ has closure by multiplication of the set of integers, and let 3k = t, t ∈ ℤ. Notice n is even since n = (3k) = 2t. Therefore, 2|n. QED.
 
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in the rye said:

Homework Statement


We have JUST started writing proofs recently, and I am a little bit doubtful in my abilities in doing this, so I just want to verify that my proof actually works. I was expecting this one to be a lot longer since the previous 2 were. I don't see any glaring flaws in it, but I'd just like to be sure (writing these feel awkward since this is my first proof base course):

A necessary condition for an integer to be divisible by 6 is that it is divisible by 2.

Homework Equations

The Attempt at a Solution


Assume true.

Pf./
[For all integers n, if n is divisible by 6, then n is divisible by 2]. Assume 6|n, n ∈ ℤ. By definition, n = 6k, k ∈ ℤ.
n = 6k = 2*3k, so n is even, hence is divisible by 2. That's really all you need to say.
in the rye said:
Consider that n = 2(3k). See that 3k ∈ ℤ has closure by multiplication of the set of integers, and let 3k = t, t ∈ ℤ. Notice n is even since n = (3k) = 2t. Therefore, 2|n. QED.
 
Mark44 said:
n = 6k = 2*3k, so n is even, hence is divisible by 2. That's really all you need to say.
Thanks. I'm having trouble being too wordy because I feel like I need to cover my basis. Do you have any general tips to proof reading (no pun intended) proofs?
 
in the rye said:
Thanks. I'm having trouble being too wordy because I feel like I need to cover my basis. Do you have any general tips to proof reading (no pun intended) proofs?
"Keep it as simple as possible, but no simpler." I don't know how much help that is, but if you have trouble being too wordy (as you said), see if you have extra baggage in there that isn't needed. In you proof you have this sentence: "See that 3k ∈ ℤ has closure by multiplication of the set of integers, and let 3k = t, t ∈ ℤ." It's completely unnecessary.
 

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