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Irrational + irrational = rational

  1. Mar 13, 2005 #1
    Can someone prove that there exists x and y which are elements of the reals such that x and y are irrational but x+y is rational? Certainly, there are an infinite number of examples (pi/4 + -pi/4 for example) to show this, but how would you prove the general case?
    Last edited: Mar 13, 2005
  2. jcsd
  3. Mar 13, 2005 #2
    [tex]\frac{\pi}{4} + \frac{3\pi}{4} = \pi[/tex]

    This is not a rational number. I don't believe you can add two irrational numbers and get a rational result. The only case I could think of would be something like [tex]\sqrt{5} - \sqrt{5} = 0[/tex].

    I know you can multiply two irrational numbers to get a rational one, but like I said, I don't think addition can do this.

  4. Mar 13, 2005 #3
    ofcourse you can,
    how about
    (In6)/(In3) and (In1.5)/(In3) they add to give 2.
  5. Mar 13, 2005 #4
    Hmmm... good example tongos. Can you give an example though that doesn't have any division to express the irrational number?
  6. Mar 13, 2005 #5
    how about 2 sqrts that add to equal one.

    y=1-sqrtx+x if x is rational, and not a perfect square then y cant be rational.
    sqrty is irrational
    Last edited: Mar 13, 2005
  7. Mar 13, 2005 #6

    What about [tex](\pi -1)+1[/tex]?

    Although I don't know if [tex](\pi -1)[/tex] can be considered as an irrational number, or a "number".
  8. Mar 13, 2005 #7
    Pi is irrational.

    Tongos - I don't understand what you mean... please explain and give an example.
  9. Mar 13, 2005 #8
    I wrote it wrong. I meant [tex](1-\pi)+\pi[/tex]
  10. Mar 13, 2005 #9
    Yeah, I agree. If you look at my first post I gave an example like that. You're basically just cancelling out the irrational number... can you show me one where you add them and they don't result in zero?
  11. Mar 13, 2005 #10

    The question is whether [tex](1-\pi)[/tex] can be considered as an irrational number.
  12. Mar 13, 2005 #11
    That is an irrational number and I agree it is a case that proves the inital topic. But what you're doing is saying [tex] a + C -a = C[/tex]

    This is where "a" is the irrational number and C is any rational constant. Can you show me an example where you don't simply subtract the irrational number to get a rational one?
  13. Mar 13, 2005 #12
    if there is no pattern in the numbers after the decimal point in x,
    then when you add it to y to obtain a rational number, y would also have to be irrational, or have a no pattern in its numbers.
  14. Mar 13, 2005 #13
    Very true.
  15. Mar 13, 2005 #14
    jameson, my example about the sqrts shows it
  16. Mar 13, 2005 #15
    Yeah, I get it now. Thank you for explaining.
  17. Mar 13, 2005 #16
    What about this for a general case?

    Assume that the difference between a rational number and an irrational number is an irrational number. The sum of any such two irrational numbers would be a rational number.

    All we have to do now is prove the assumption, which I think is easier.

    edit: I just saw tongos' proposition, which I think is better than mine.
  18. Mar 13, 2005 #17


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    this was a homerwork problem freshman year in about my first homework. My idea was to take any decimal with non repeating entries, hence irrational, and then add to it thed ecimal whose entries were 9-the entry in the firstd ecimal. that gives .9999999 .... =1 a rational number. finally i realixzed that this mjust emant thnat for any rational number r and irrational number, we have r-x and x both irrational, hence (r-x) + x = r is rational.
  19. Mar 13, 2005 #18
    I assume we can all agree with

    [tex] \forall x, y \in \mathbb{R}, \ \exists z \in \mathbb{R} \ \mbox{s.t.} \ x + z = y [/tex]

    or, equivalently

    [tex] x, y \in \mathbb{R} \Longrightarrow x-y \in \mathbb{R}[/tex]

    Then let [tex] \mathbb{I}[/tex] denote the set of irrationals. Clearly

    [tex] a \in \mathbb{I}, \ b \in \mathbb{Q} \Longrightarrow a + b \in \mathbb{I} [/tex]

    since otherwise, we assume

    [tex] a + b = d \in \mathbb{Q} \Longrightarrow \exists q, p, n, m \in \mathbb{Z} \ \mbox{s.t.} \ a + \frac{n}{m} = \frac{q}{p} \Longrightarrow a = \frac{qm - np}{pm} \in \mathbb{Q}[/tex]

    which is a contradiction.

    Basically, the difference between a rational number and an irrational number is ALWAYS irrational.
    Last edited: Mar 13, 2005
  20. Mar 14, 2005 #19
    Although it does seem like circular logic:

    "The sum of two irrational numbers is rational iff the difference between the sum and one of the irrational numbers is irrational."
  21. Mar 14, 2005 #20

    matt grime

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    But how can it be any other way? If x+y=r with x, y, irrational and r rational then of course x=r-y, it isn't circulary logic, it's just bleedin' obvious.
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