Proof That an Integer is Divisible by 2

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Homework Help Overview

The discussion revolves around establishing a proof for the divisibility of an integer by 2, specifically focusing on the relationship between an integer's units digit and its divisibility by 2.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the proof that an integer is divisible by 2 if its units digit is one of {0, 2, 4, 6, 8}. Questions arise regarding the completeness of the proof, particularly the need to demonstrate the converse relationship.

Discussion Status

Some participants have provided guidance on the proof structure, suggesting that both directions of the divisibility condition should be shown. There is an ongoing exploration of how to formally express the equivalence in the proof.

Contextual Notes

Participants note the importance of addressing both implications in the proof, indicating a focus on rigorous mathematical reasoning and completeness in the argument.

Math100
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Homework Statement
Establish the following divisibility criteria:
An integer is divisible by ## 2 ## if and only if its units digit is ## 0, 2, 4, 6 ##, or ## 8 ##.
Relevant Equations
None.
Proof:

Suppose ## N ## is the integer and ## x ## is the units digit of ## N ##.
Then ## N=10k+x ## for some ## k\in\mathbb{Z} ## where ## x={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} ##.
Note that ## 10k\equiv 0\pmod {2}\implies N\equiv x\pmod {2} ##.
Thus ## 2\mid N\implies N\equiv 0\pmod {2}\implies x\equiv 0\pmod {2}\implies x\in{0, 2, 4, 6, 8} ##.
Therefore, an integer is divisible by ## 2 ## if and only if its units digit is ## 0, 2, 4, 6 ##, or ## 8 ##.
 
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Math100 said:
Homework Statement:: Establish the following divisibility criteria:
An integer is divisible by ## 2 ## if and only if its units digit is ## 0, 2, 4, 6 ##, or ## 8 ##.
Relevant Equations:: None.

Proof:

Suppose ## N ## is the integer and ## x ## is the units digit of ## N ##.
Then ## N=10k+x ## for some ## k\in\mathbb{Z} ## where ## x={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} ##.
Note that ## 10k\equiv 0\pmod {2}\implies N\equiv x\pmod {2} ##.
Thus ## 2\mid N\implies N\equiv 0\pmod {2}\implies x\equiv 0\pmod {2}\implies x\in{0, 2, 4, 6, 8} ##.
Therefore, an integer is divisible by ## 2 ## if and only if its units digit is ## 0, 2, 4, 6 ##, or ## 8 ##.
Where is the other direction? You have shown ##2\,|\,N \Longrightarrow x\in \{0,2,4,6,8\}##. You must also show that all integers that end on an even digit are even themselves. Of course, you simply could replace all ##\Longrightarrow ## by ##\Longleftrightarrow##.
 
fresh_42 said:
Where is the other direction? You have shown ##2\,|\,N \Longrightarrow x\in \{0,2,4,6,8\}##. You must also show that all integers that end on an even digit is even itself. Of course you simply could replace all ##\Longrightarrow ## by ##\Longleftrightarrow##.
So should I put "Thus ## 2\mid N\Leftrightarrow N\equiv 0\pmod {2}\Leftrightarrow x\equiv 0\pmod {2}\Leftrightarrow x\in{0, 2, 4, 6, 8} ##.?
 
Math100 said:
So should I put "Thus ## 2\mid N\Leftrightarrow N\equiv 0\pmod {2}\Leftrightarrow x\equiv 0\pmod {2}\Leftrightarrow x\in{0, 2, 4, 6, 8} ##.?
Yes, the conclusions work in both directions because ##10k \equiv 0 \pmod 2## as you correctly said.
 
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