Prove the equation $2x - 6y = 3$ has no integer solution to $x$ and $y$.

I need to verify my proof I think I did it correctly, but am not fully sure since I don't have solutions in my book. I basically proved by contradiction and assumed there was an integer solution for x or y. I then solved for $x $ and $y$ in $2x - 6y = 3$ getting $x = 3y + 3/2$ and $y = x/3 - 1/2$ .since both $x,y$ are not integers I said it contradicts that $x$ or $y$ had an integer solution, meaning the original statement was correct. Did I prove this right, or should I redo?

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    $\begingroup$ You're on the right track, but why is $x = 3y + 3/2$ not an integer? Also, your statement that "both $x, y$ are not integers" is not correct: $(3/2, 0)$ is a solution with $y$ integral. $\endgroup$ – user296602 May 29 '19 at 5:33
  • $\begingroup$ @T.Bongers I think since adding 3/2 will make it a rational number $\endgroup$ – user675497 May 29 '19 at 5:35
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    $\begingroup$ Some tips for friendly math formatting: math.meta.stackexchange.com/questions/5020/…. $\endgroup$ – uniquesolution May 29 '19 at 5:37
  • $\begingroup$ ggg.2(x-3y)=3; x-3y integer.Means: Left hand side is even.Hence? $\endgroup$ – Peter Szilas May 29 '19 at 5:45

Rewind to the point where you say $x=3y+3/2$. We rearrange this to $x-3y=3/2$, then note that since we have taken $x$ and $y$ to be integers, $x-3y$ is also an integer. But $3/2$ is not an integer, a contradiction.


Alternatively, you may write $$2x - 6y = 2(x-3y).$$ Since $x$ and $y$ are integers, so must be $x-3y$. So $2(x-3y)$ must be an even integer, clearly being divisible by $2$. But $3$ is odd.


Your proof, as it stands, is not correct. Some particular issues:

  • You get to a point where $x = 3y + 3/2$ and $y = x/3 - 1/2$ and state that "since both $x,y$ are not integers..."; this assumes the conclusion that you're aiming for.

  • It's not true that both $x$ and $y$ are not integers. You could have one of the two be an integer, and the other not be an integer. For example, $(0, -1/2)$ and $(3/2, 0)$ are both solutions with $x$ or $y$ integral, but not both.

  • Its not convincing that $3y + 3/2$ is not an integer (and the point immediately above shows that that claim is not true!).

  • Stylistically, your assumptions are not explicitly stated and there is no introduction to the proof. You haven't stated that you're assuming $(x, y)$ to be a pair of integers solving a particular equation. Therefore, it's not clear how the contradiction is reached; at a minimum, you need to form the negation of the statement and clearly include assumptions.

So unfortunately, this is not a properly written proof. But it can be fixed without too much work; here's an outline to follow:

1) Introduce the players. Say "We proceed by contradiction. Assume that $x, y$ are integers such that $2x - 6y = 3.$

2) Isolate one of the variables and get the contradiction. Perhaps "Then $x = 3y + \frac 3 2$. Since $3y$ is an integer (why?), the sum $3y + \frac 3 2$ is not an integer."

3) Tell the reader why this is a problem. "This contradicts the assumption that $x$ is an integer."

  • $\begingroup$ yea I know, my actual written proof was much more detailed and better formatted. Im just lazy and quickly wrote a summary haha. But, thanks for pointing out the issues, they are really helpful $\endgroup$ – user675497 May 29 '19 at 5:43
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    $\begingroup$ @ggg If you're asking a sincere proof-verification question, then it's important to include the actual proof that you want to verify... rather than a "lazy [...] summary." Please don't waste people's time by doing that. $\endgroup$ – user296602 May 29 '19 at 5:44
  • $\begingroup$ Exactly what made you thinking that this question is not a homework? $\endgroup$ – peterh - Reinstate Monica Jun 3 '19 at 10:45
  • $\begingroup$ @peterh If you have a problem with my comments on meta, then discuss it there or flag them. Please do not harass me on main with irrelevant nonsense. $\endgroup$ – user296602 Jun 3 '19 at 10:52
  • $\begingroup$ @T.Bongers I did not harass you, I am just surprised that you seem answering a no-context PSQ. What if the question would be deleted, would it make you happier? $\endgroup$ – peterh - Reinstate Monica Jun 3 '19 at 10:56

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