Proof That an Integer is Divisible by 2

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An integer N is divisible by 2 if its units digit x is one of the even digits: 0, 2, 4, 6, or 8. The proof begins by expressing N as N=10k+x, where k is an integer. It establishes that since 10k is always even, N's divisibility by 2 depends solely on x. The discussion also emphasizes that the proof must demonstrate the reverse implication, confirming that if x is even, then N is even as well. Thus, the criteria for divisibility by 2 are valid in both directions.
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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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