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Operations on irrational numbers

  1. Oct 13, 2011 #1
    Heres two problems from an A Level related paper: prove that if pq is irrational then atleast one of p or q is irrational. Also prove that if if p + q is irrational then atleast one of p or q is irrational. These two proofs are trivial proof by contradiction problems but it got me thinking more about cases where operations on two irrational numbers result in irrational or rational numbers

    Now whats interesting are the inverses of the statements, i.e. if atleast one of p or q is irrational then pq is irrational and if atleast one of p or q is irrational then p+q is irrational.
    I can think of two seemingly counter examples to both but they still leave some questions. To the first clearly anything of the form (sqrt(a))^2 is counterexample and the second something like (a + irrational number)+(a - irrational number).

    The question that is remaining is: Other than the case of square roots squared and any other special case, can two normal different irrational numbers multiply to give a rational number? And can two irrational numbers add or subtract to give a irrational number which does not involve cancellation of the irrational parts?
    The answer seems to be an intuitive "no" because both cant involve cancellation but heres a way of looking at things that just makes everything confusing, consider the special case of like (sqrt2)^2, we can break it into the sum of a rational and irrational part namely

    (1+ir)^2 where ir is the irrational part then
    (1+ir)^2 = 1 +2ir + ir^2 = 2

    Now heres two seemingly irrational numbers 2ir +ir^2 that give a rational, whats going on? What about subtraction and division of irrationals?
     
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  3. Oct 13, 2011 #2

    pwsnafu

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    I think you are talking about http://en.wikipedia.org/wiki/Transcendental_number" [Broken].
    I don't think you know what http://en.wikipedia.org/wiki/Normal_number" [Broken] means.
    In general, given any two transcendental numbers, x and y, then either x+y or xy must be transcendental (or both).

    Lastly consider this: there exists two irrational numbers such that xy is rational. Consider [itex]\sqrt{2}^{\sqrt{2}}[/itex]. If this is rational we are done. If not let this number be x, and let y equal square root 2. So that xy=2. QED.
     
    Last edited by a moderator: May 5, 2017
  4. Oct 14, 2011 #3

    Deveno

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    normally, with any kind of "structure" (i am using the word loosely here), if there is a distinguished element (a 0-object), and you wish to make a subset a "substructure", the subset has to contain the 0-object.

    with addition, the 0-object is 0.

    with multiplication, the 0-object is 1.

    both of these are rational, which effectively kills the chances of the irrationals having any interesting self-contained structure (at least ones involving ordinary arithmetic functions).

    what you might be interested in, is studying fields of the form Q(√k) where k is a positive integer that is not a perfect square. this might answer your question about how certain "special" sets of irrational numbers can interact.
     
    Last edited: Oct 14, 2011
  5. Oct 14, 2011 #4
    I think what you're asking is if there are examples that don't seem contrived. For example, 3 - pi and 3 + pi are irrationals that sum to a rational; but one could say, well, those are obviously contrived examples ... is there a more "random" or natural looking example?

    First, there are quite a lot of examples of pairs of irrationals that sum to a rational. Given any irrational, consider its decimal expansion. Any finite truncation of the expansion expresses that irrational as the sum of a rational and an irrational. For example take pi = 3.14159...

    I can say that pi = 3 + .14159... or pi = 3.1 + .04159... and so forth. In other words every irrational is the sum of a finite truncation of its decimal expansion, plus the tail remaining after the truncation.

    Now, for each of those pair, just combine the irrationals to get a pair of irrationals that sum to a rational. For example pi = 3 + .14159, so pi - .14159... = 3. That's a sum of two irrationals that's rational. So you can see that there are lots and lots of these kinds of examples that appear "normal," but are actually contrived.

    Did you know that it is unknown whether e + pi is rational or not? That's quite amazing. If their sum were to turn out to be rational, that would be as normal or non-contrived example as you can imagine, I think.

    I didn't answer your question, but I hope I have you more to think about. Every irrational is the sum of a rational and an irrational in many different ways; so in the end, it's likely that every "normal" appearing rational sum of irrationals, it's really a contrived example at some level.

    You can pursue similar reasoning for multiplication. In the end, the "contrived versus natural" distinction is artificial and more a matter of psychology than mathematics. Some numbers just look more random than others.
     
  6. Oct 17, 2011 #5
    nothing special,
    in both cases you are not doing any operation: x [+y -y] = x , x [√ ²] = x
    in the first example x = 2, in the second [(ir=√2-1), 2ir+ ir²] x = 1: 1 [+(2√2-2) + (2 -2√2)] = 1
    When you truncate a real to x decimal digits, you can perform division and any other operation.

    P.S. when you say rational + ir, remember also 1/7 is defined as rational
     
    Last edited: Oct 17, 2011
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