Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question re proof of irrational nos.

  1. Jun 25, 2006 #1
    Hi everyone,

    I've been reading about the proof of irrational nos. and I often encounter this phrase: "decreasing sequence of positive integers must be finite". What does this actually mean? Can anyone explain or point me to a link.

    Here's a link of one proof I've read re proof of irrational nos: http://mathforum.org/library/drmath/view/55839.html

    Thanks in advance :)
  2. jcsd
  3. Jun 26, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    A decreasing sequence of positive integers must be finite. Let s be any such sequence. Let s(1) denote the first element of the sequence. Well the sequence can then have at most s(1) elements, and if it had exactly s(1) elements, it would be the sequence s = (s(1), s(1)-1, ...., 2, 1). And s(1) is obviously a finite number.
  4. Jun 26, 2006 #3
    So basically, in the context of proving irrational numbers, what that phrase means is that the numerator and denominator just gets smaller and smaller indefinitely (infinitely) and therefore (the numerator and denominator) cannot both be integers as required by rational numbers.

    Is my conclusion right?

    Thanks in advance.
  5. Jun 26, 2006 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    The phrase 'proving irrationals' doesn't mean anything at all.

    That every set of positive integers has a least element is nothing to do with any proofs of any properties of that the irrationals may or may not have on its own.

    I think the idea you're loooking for is reductio ad absurdum: show that if something is true it leads to an infinite strictly decreasing set of positive integers, which is impossible.
  6. Jun 26, 2006 #5


    User Avatar
    Science Advisor

    I really have to ask: what do you mean by "proving rational numbers"??
    Proving that rational numbers exist?? That is, proving that integers alone don't solve all equations that can be stated in terms of integers? What integer satisfies 2x= 1?
  7. Jun 27, 2006 #6
    It's actually proving the existence of irrational numbers. I'm asking what the phrase "decreasing sequence of positive integers must be finite" means in the context of proving the existence of irrational numbers. Please see link that I've posted.
  8. Jun 27, 2006 #7
    What do you mean? I'm asking about "decreasing sequence of positive integers must be finite" not the phrase you just said.

    Isn't it that one proof of irrational numbers is that 2^(1/2) cannot be expressed as a ration of integers i.e. a/b because as the proof in the link I posted has shown, both a and b decreases indefinitely (infinitely) and therefore cannot be integers because "decreasing sequence of positive integers must be finite".

    Yup. Like I've mentioned above.

    Have I understood the proof right?

    Please correct me if I haven't. Thanks! :biggrin:
  9. Jun 27, 2006 #8

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    The phrase 'the proof of irrationals nos' appears in post one. There is no such things as 'the proof of irrational nos'. Some words are missing. In post three you use the phrase 'in the context of proving irrationals'.

    You might have meant

    1. prove that irrational numbers exist
    2. prove that some specific number is irrational.
    Last edited: Jun 27, 2006
  10. Jun 27, 2006 #9


    User Avatar
    Science Advisor

    The set of positive integers is "well ordered"- that is, any set of positive integers contains a smallest member. In particular, "1" is the smallest of all positive integers so if {an} is a decreasing sequence of positive integers, then a0 is the largest number in the sequence and the sequence cannot have more than a0[\sub] members. That's what AKG said in the first response to your post.

    The method of proving that [itex]\sqrt{2}[/itex] is not rational is called "infinite descent". Assuming that [itex]\sqrt{2}= \frac{m}{n}[/itex] we can show that both m and n must be even: m= 2a, n= 2b so that [itex]\sqrt{2}= \frac{a}{b}[/itex]. Of course, a< m, b< n and we can keep doing that- we get an infinite sequence of decreasing positive integers which is impossible.

    Often that hidden by simply asserting that any rational number can be written in "lowest terms"- that m and n have no common factors. Of course, then, showing that both m and n are even is an immediate contradiction.
  11. Jun 27, 2006 #10
    A neat, simple proof of the irrationality of phi follows from it's geometric definition, the ratio A/B that divides a length A+B into lengths A and B such that (A+B)/A = A/B. If one takes (A+B)/A to be in lowest terms, then A/B is in even lower terms, which is a contradiction.
    Last edited: Jun 27, 2006
  12. Jun 27, 2006 #11


    User Avatar
    Homework Helper

    If you just want to prove irrational numbers exist, probably the simplest examples are logs. For example, log3 5 must be irrational because if it were equal to m/n for some positive integers m and n, then 3m/n=5, so 3m=5n, which is clearly absurd (for example, one is divisible by 5 and the other isn't).
  13. Jun 28, 2006 #12
    Yes. What I meant was #1
  14. Jun 28, 2006 #13
    In that case, perhaps you would be interested in the concept of a "Dedekind cut".


    You might also pick up a book on Number Theory -- I recommend Hardy and Wright's "Introduction to the Theory of Numbers", but almost all introductory texts cover irrational numbers.

    By the way, the proof you originally gave in post 1 doesn't actually prove the existence of irrationals -- for example, it could simply be that sqrt(2) isn't actually a number. It is in fact impossible to "prove" the existence of irrational numbers -- the best you can do is show that there exists some quantities that cannot be expressed as a rational numbers. That's not a proof.
    Last edited: Jun 28, 2006
  15. Jun 29, 2006 #14
    Thanks. I'll check that out.

    Technically you're right. You cannot prove irrational numbers but you can prove that there are numbers that are not rational (i.e. cannot be expressed as a ratio of integers) and therefore are irrational.

    Thanks again :biggrin:
  16. Jun 29, 2006 #15

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    "Prove irrationals exist" not "prove irrationals". Sorry to bang on about this, but please try to make your sentences correct, especially after people have pointed out the mistake at length several times - maths is all about clarity and communication.
  17. Jun 29, 2006 #16
    Yup. Thanks for pointing that out. I had "Prove irrationals exist" in mind but expressed it incorrectly. 'Will be careful next time.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook