Solving a Problem with Sets: x+y <xy, then y>0

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The discussion revolves around proving that if x and y are integers with x > 0 and x + y < xy, then y must be greater than 0. Participants express confusion about how to approach the problem and emphasize the need to understand the properties of integers (Z) in this context. A suggestion is made to use proof by contradiction, assuming y is less than or equal to 0 and deriving a contradiction from the inequality. The conversation highlights the importance of manipulating the given inequality to explore possible contradictions. The focus remains on clarifying the proof strategy and understanding the implications of integer properties in the problem.
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Hi, I'm having a lot of trouble with the following question:

Homework Statement



(a) Let x,y ∈ Z. Prove that if x>0 and x+y <xy, then y>0

Homework Equations


x+y <xy, then y>0


The Attempt at a Solution



I am very confused with this problem, and am not even sure on how to start. Any tips/suggestions to help me get started would be greatly appreciated.
 
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What properties does Z have ? Is it an ordered field ? A commutative ring , a subset of R etc. Without this information I do not see how we can help you.
 
╔(σ_σ)╝ said:
What properties does Z have ? Is it an ordered field ? A commutative ring , a subset of R etc. Without this information I do not see how we can help you.
Z is just the set of integers.
 
I think this might be a way to prove it, using a proof by contradiction.

Assume that x and y are in Z, x + y < xy, and y <= 0.

Since by assumption, y <= 0, then x + y <= x.
Then (x + y)2 <= x2
From the above, it follows that y(2x + y) <= 0.

Now, work with that inequality to try to get a contradiction, keeping in mind that x and y can only be integer values, and that x > 0 and y <= 0.
 

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