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Proof (x + y + xy = 36) -> x,y are not N

  1. Apr 14, 2014 #1
    Hi guys, I've been struggling with this proof for some quite time now. I have to prove that

    x + y + xy = 36; has no solution in Natural numbers.

    I've started with proving that both x and y have to be even numbers (which is easy to prove) but now I'm going in circles. The problem is, that there are even numbers which are solution to this equation ( -2 and -38, if I recall correctly). Any ideas here?
     
  2. jcsd
  3. Apr 14, 2014 #2
    You could always brute-force it, though that seems inelegant. Let x and y each run from 1:36 (or let y run from 1:x, since the expression is symmetric).
     
  4. Apr 14, 2014 #3

    Borek

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    If they are both even, x'+y'+x'y'=18, and there are less cases to consider when using brute force approach.

    And the story doesn't end here.
     
  5. Apr 14, 2014 #4

    pasmith

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    Consider a rectangle with sides of length [itex]x + 1[/itex] and [itex]y + 1[/itex], which are both natural numbers if [itex]x[/itex] and [itex]y[/itex] are.

    What's the area of this rectangle when [itex]x + y + xy = 36[/itex]?
     
  6. Apr 16, 2014 #5

    Curious3141

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    Very neat. :approve:
     
  7. Apr 16, 2014 #6

    Mentallic

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    Nice! I did not expect to be so profoundly enlightened in such a short sentence :smile:
     
  8. Apr 16, 2014 #7

    Borek

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    But if so (and if I understand it correctly), answer depends on whether we consider zero to be a natural number?
     
  9. Apr 16, 2014 #8

    Dick

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    I think the majority of people would agree that 0 is not a natural number. Some wouldn't. But if 0 were a natural number that makes the problem dead easy. x=0, y=36. Hardly worth asking.
     
  10. Apr 16, 2014 #9
    Actually, I would say that this is a 50/50 question. I myself don't agree that N does not include 0. I know that in many cases it makes no sense to say it does, but N* should be used to represent {1, 2, 3, ...}.
    Agreed.
    But the way it is written "Proof (x + y + xy = 36) -> x,y are not N" sounds really imperative. I wouldn't even expect myself to be able to prove that x, y are N*.
    Or y = 0, x = 36.
     
  11. Apr 17, 2014 #10
    Let's start with your (correct) realization that both x and y must be even. This is easily shown.

    Group the equation as follows:

    (X + Y) + XY = 36

    Dividing each term by (X + Y) gives

    1 + XY/(X+Y) = 36/(X+Y)

    Both XY/(X + Y) and 36/(X + Y) must be even.

    But adding 1 to an even number cannot yield an even sum.

    This contradicts the fact that X and Y must be even.

    Check my algebra and logic.
     
  12. Apr 17, 2014 #11
    I really do like this solution.
     
  13. Apr 17, 2014 #12
    I'm a little slow today. Why must XY/(X + Y) and 36/(X + Y) be even? Why indeed must they be natural numbers at all?

    I hope you can show me, because the proof is otherwise delightful.
     
  14. Apr 17, 2014 #13

    Borek

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    Assuming x and y are even, xy is divisible by 4, and x+y is even.
     
  15. Apr 17, 2014 #14
    Right, but that doesn't mean they're even natural numbers at all!

    I mean, if X=6 and Y=4, they are both even, but neither XY/(X + Y) nor 36/(X + Y) is even an integer (they are 2.4 and 3.6 respectively).

    Also, an even number divided by an even number isn't necessarily even.

    Am I missing something, or does the proof fall apart at that step?
     
  16. Apr 17, 2014 #15
    Chogg is correct. My proof stinks. Let me take advantage of pasmith's suggestion.

    (X + 1)(Y + 1) = X + Y + XY + 1 = 36 + 1 = 37

    Since 37 is prime, (X + 1) and (Y + 1) must be 1 and 37 (or vice versa)

    That means X and Y must be 0 and 36 (or vice versa)

    Back to the argument about zero being a natural number!
     
  17. Apr 17, 2014 #16

    LCKurtz

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    What happened to the rule to not provide complete solutions?
     
  18. Apr 18, 2014 #17
    Acoording to the definition it's very elusive whether 0 is a natural number or not. Neither is incorrect, however, if 0 were a natural number, the solution of this assignment is painfully obvious.

    Since the assignment states that proof is needed such that there is no solution among natural numbers, we kind of have to assume right off the bat that 0 is Not a natural number.

    It is relatively easy to prove, one can show that the values for x and y are fractions, fractions that at no time of the day can be natural values.
     
    Last edited: Apr 18, 2014
  19. Apr 18, 2014 #18

    Curious3141

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    You don't have to do any division - if you do, you have to justify that the resulting quotients are all integers, as you yourself have realised.

    Actually, it's very easy to see why x and y must both be even. Just use these well-worn properties:

    odd + odd = even
    even + even = even
    odd + even = odd
    odd*even = even
    even*even = even

    Those should be obvious, but they're trivially proven.

    There are only 3 possibilites: x and y are both odd, exactly one of x and y is odd, or both x and y are even.

    First case. You get odd + odd + odd = even + odd = odd. That can't equal 36.

    Second case. Say x is odd, y is even. The relation is symmetric, so it doesn't matter which you choose to be odd. So you get odd + even + even = odd + even = odd. Again it can't equal 36.

    The final case is the only one that is admissible. Even + even + even = even. There's no contraindication equating that to 36.

    Done.
     
  20. Apr 18, 2014 #19

    Curious3141

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    I agree that enforcement of this has been fairly arbitrary. I've been issued an infraction before for providing an alternate solution *after* the original poster had solved the problem.

    Then again, I don't know if what was provided in this thread constitutes a full solution. Pasmith's post provides a very neat solution and the large insight has already been revealed. But the small insight of recognising that 37 is prime is still required.

    As to the question of whether 0 is a natural number, there is no standard convention. That makes this a dumb question. It could easily have been rectified by saying there are no solutions in the positive integers.
     
  21. Apr 18, 2014 #20

    LCKurtz

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    I say that 0 is not a natural number! That settles it, doesn't it? :smile:
     
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