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Between two rationals there are at least one irrational.

  1. Aug 2, 2006 #1
    in courant's and fritz's calculus text im given the assignment to show the above, but first in the same question im given this task:
    1) a) if a is rational and x is irrational then x+a is irrational and if a isnt 0 then ax is irrational too.
    well this task is ofcourse trivial.
    i thought that these two tasks are connected, although ive seen other proof which do not use taks a.
    even so, here's the proof of mine which does incorporate a.

    let's assume a is rational and x is irrational.
    1) if a>x>0 then a-x>0 and a-x<a. all we need to show is that a-x is irrational and from a) above, we need to show that -x is irrational, assume it's not, then -x=p/q and x=-p/q contradiction to x being irrational.
    2) if x>a>0 then 1>a/x>0 again we need to show that 1/x is irrational, suppose it's not then 1/x=p/q x=q/p and thus a contradiction to x being irrational.

    is this proof valid?

    btw, i know other proofs which are quite different than this, so i wonder if this one is valid also.
     
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  3. Aug 2, 2006 #2

    Office_Shredder

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    I'm a bit confused as to what you're proving. If you're trying to prove that there exists an irrational number between any two rationals (as the title of the thread states), then you really haven't gone anywhere with that.

    If you're trying to prove the question at the top, I don't see how you've proven it, since you never actually look at a+x or ax. Also, you haven't covered every case for both the additive and the multiplicative case. Try writing your ideas out more clearly, it looks like you rushed to type out what you could really fast
     
  4. Aug 2, 2006 #3

    StatusX

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    I don't understand these proofs. Why does it matter if a>x>0? All you need to do is assume that, for example, a+x is rational, and show this implies x is rational, a contradiction. Once you show this is true, you can answer the question in the title of the thread by finding a sequence of irrational numbers that tend to zero.
     
  5. Aug 2, 2006 #4

    HallsofIvy

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    I think that the proof you want is given in the title: "between two rationals there are [is] at least one irrational". It would be a really good idea to actually say that in the text. You have apparently already proved that "if a is rational and x is irrational, then x+ a is irrational" and "if a is rational, not equal to 0, and x is irrational, then ax is irrational" and want to use those statements.

    But what are the given two numbers? You seem to saying that if you are given the rational number a, you can choose any irrational x with a> x> 0 and, e voila', produce a- x, an irrational number between a and 0. ?? Well, aren't you assuming that there exist an irrational number, x, between a and 0 to begin with? In other words, what you have done is prove that, given a positive rational number a, if there exist an irrational number x, 0< x< a, then there exist another irrational number, a- x, between 0 and a. But you have not proved "given two rational numbers, a and b, a< b, there exist an irrational number x with a< x< b.

    ?? Okay, this proves that if there exist an irrational number larger than the given positive rational number a, then there also exist an irrational number less than a. Was that what you intended? You still haven't proven that there exist an irrational number between two given rational numbers.
     
  6. Aug 2, 2006 #5
    okay, as i said proving a) was easy, my question is how do i use a) to prove that between any two rational numbers there's at least one irrational number?
    and halls thanks for the correction, i shouldv'e spotted that when a>x>0 that im assuming what is needed to be proven.
     
  7. Aug 2, 2006 #6
    for example, let's assume r1,r2 are rational numbers and x is irrational number.
    let's assume x>r1>r2>0
    1>r1/x>r2/x>0
    but i think it's the same as my first attempt.
    should i be proving by ad absurudm that:
    assuming that between two rational numbers there arent any irrational numbers, and proving a contradiction, that there actually is an irrational number between two given rational numbers.
     
  8. Aug 2, 2006 #7

    NateTG

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    It would help you a whole lot if you took your time and wrote things out more clearly.

    Compare:
    To:
    Which has a clear line of reasoning, doesn't introduce variables such as p or q without specifying what they represent, features complete sentences, and has a conclusion that explicitly and clearly responds to the prompt.

    Notably, your 'proof' (and this is most likely because of sloppy notation) includes "assume [x is not irrational] then ... contradicts [that x is irrational]" which is circular.
     
  9. Aug 2, 2006 #8
    nate you got confused, i didnt say i needed to prove that a+x and ax are irrational, this i did already, as i said in my first post it is trivial.
    what i need to do apparently is to use this fact and prove that between any two rational numbers there's at least one irrational number.

    the methods i listed here were in this direction.
    i hope you can help me realise how to use this part in the proof of the second question.
     
  10. Aug 2, 2006 #9

    StatusX

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    As I mentioned earlier, if you can find a sequence of irrational numbers that tends to zero, then for any rational numbers r1 and r2, assuming r1<r2, then there is some irrational number x in the sequence with x<r2-r1. Then you can use your result from the first post (although what you have in the post now is not a (understandable) proof).
     
  11. Aug 2, 2006 #10
    why there should be an irrational number smaller than r2-r1?

    and to what result do you refer to?
    is it that 'if a is rational and x irrational then x+a is irrational, and for a different than 0 ax is irrational'?
    or do you refer to something else?
    if you refer to the former, then as i said what that i proved isn't posted here, as nate said you assume a+x is rational and get a contradiction and this way also with ax.
    but i didnt asked you to help me on this, for heaven's sake i said it was trivial, which means i solved it already.

    edit: what i asked is for this theorem:'between two rational there's at least one irrational' to solve with help of this:'if a is rational and x is irrational then x+a is irrational...'.
     
    Last edited: Aug 2, 2006
  12. Aug 2, 2006 #11

    StatusX

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    I'm saying there are arbitrarily small irrational numbers, which is to say there is a positive irrational number smaller than e for any e>0. So if r2-r1=e, there is an irrational number smaller than this, and so if you add it to r1, the result will be between r1 and r2.
     
  13. Aug 2, 2006 #12

    NateTG

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    Well...
    Can you show that there is an irrational number between 0 and 1?
    How about between 1 and 2?
    Or between any two integers?
    Or between any two fractions with the same denominator?
     
  14. Aug 2, 2006 #13
    status i think i know your apparoach, i think i have it my notes.
    tell me if this what meant:
    let n be a natural number and sqrt2/n<b-a where a and b are rationals, we will choose the last m for which (m*sqrt2)/n<=a then sqrt2(m+1)/n>a and sqrt2(m+1)/n?<=a+sqrt2/n<a+(b-a)=b, and sqrt2(m+1)/n is irrational.

    is this what you meant?
     
  15. Aug 2, 2006 #14
    ok, thanks status, i get it now.
     
  16. Aug 2, 2006 #15

    benorin

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    Given any two rational numbers r and s, where [tex]r < s,[/tex] there exists an irrational number q such that [tex]r < q < s[/tex].

    Proof: [tex]r < s \Rightarrow 0 < s-r [/tex] and since [tex]1<\sqrt{2}[/tex] we also have [tex]0 < s-r <\sqrt{2}(s-r) \Rightarrow 0 < \frac{s-r}{\sqrt{2}} <s-r \Rightarrow r < r+\frac{s-r}{\sqrt{2}} <s[/tex]

    can you finish?
     
  17. Aug 2, 2006 #16
    but ben two things:
    1)when you state 0<(s-r)/sqrt2<s-r you already have an irrational number between two rational numbers, between s-r and 0 ( you onle need to prove it's irrational which is trivial).

    2) when you assume sqrt2>1 you also assume that sqrt2<2 and thus that there's an irrational number sqrt2 between 2 and 1 which are rationals.

    status, just one thing occured to me, if you say that x<r2-r1
    and you say that if you add r1 to it, then r1<r1+x<r2 so you're assuming already that 0<x<r1-r2 and thus that there's an irrational number between two rational numbers, is it not?
     
  18. Aug 2, 2006 #17
    ok i get it now, it should be between two given rationals, and not between any two rationals.
     
  19. Aug 2, 2006 #18
    uh, you want that [itex]r_2-r_1[/itex] on top :tongue2:. And that's exactly what benorin did!

    There's no difference there.

    It's easy to prove that [itex]\sqrt{2}[/itex] is irrational. He also never needed the assumption that [itex]\sqrt{2} < 2[/itex], and even if he had, the existence of an irrational between 1 and 2 does not immediately imply the existence of an irrational between any two rationals (well, actually it does, and you don't need the [itex] < 2[/itex] part, as benorin's proof shows - but it's not "obvious" why it does :smile:). And [itex]1 < \sqrt{2} < 2[/itex] is pretty clear anyways.
     
  20. Aug 2, 2006 #19

    mathwonk

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    you have to prove there are really short irrationals.


    try pi/n.
     
  21. Aug 3, 2006 #20
    i know already what to do, i choose for example x=sqrt2/n for n natural, it's as status say smaller than a given e=r1-r2, and because r1-r2>x then by adding r2, we get what we wanted.

    btw, data, i think there's a difference between 'any' and 'given' rationals, by the first if it's any rationals then it suffice to show that there's an irrational number smaller than 1 which is positive, from this you ofcourse can conclude that between any two rationals such as 1+r1 and r1-1 there's at least one irrational.
    for example if x is irrational greater than a given rational number a, than x>a, if a>0 than a/x<1 and you can conclude that between 0 and 1 there's at least one irrational number.
    but if it's between given two rationals, than i should show that indeed there's between them an irrational number, and not between any rationals such as 1 and 0.
    at least that's what i think.
     
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