If 2|3x then 2|x
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D
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I would have done it the same way.
I cannot say whether the proof is "complete" or not - but looks satisfactory to me.
HallsofIvy
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Suppose it had been "6x" rather than "3x". Would writing 2x= 6*n, x= 6(n/2) prove that "if 2 divides 6x then 2 divides x"?
At some point you will have to use the fact that 2 and 3 are prime numbers!
At some point you will have to use the fact that 2 and 3 are prime numbers!
D
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That would certainly be applicable. However, I think those two numbers only need to be "relatively" prime.
HallsofIvy
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Yes, but I was specifically referring to this problem.
Steven G
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There is a fact that says is a | b and a | bc, then a|c
Proof: a|bc means bc=ma for some integer m
a| b means b = ma + k, where 0<k<a
So a|bc = (ma+k)*c. Now ma+k = n, where n is an integer and is not a multiple of a. So a|bc = nc. Now we have that a|c.
Just use this fact for your problem.
2|3x and (we know) 2| 3. Hence 2|x.
Proof: a|bc means bc=ma for some integer m
a
So a|bc = (ma+k)*c. Now ma+k = n, where n is an integer and is not a multiple of a. So a|bc = nc. Now we have that a|c.
Just use this fact for your problem.
2|3x and (we know) 2
Steven G
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Maybe I am missing something but your proof is only valid if n/3 is an integer! Why is n/3 an integer?
It seems that you are saying if 7|5*2 then 5*2 = 7n and 2 = 7*(n/2). Hence 7|2. This is not true since for 7*(n/2) to equal 2, n/2 can't be an integer.