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Irrational numbers - incomprehnsible?

  1. Mar 23, 2010 #1
    Hi, I am having an issue with irrational numbers and the term irrational.

    Main Entry: 1ir·ra·tio·nal
    Pronunciation: \i-?ra-sh(?-)n?l, ?i(r)-\
    Function: adjective
    Etymology: Middle English, from Latin irrationalis, from in- + rationalis rational
    Date: 14th century

    : not rational: as a (1) : not endowed with reason or understanding (2) : lacking usual or normal mental clarity or coherence b : not governed by or according to reason <irrational fears> c Greek & Latin prosody (1) of a syllable : having a quantity other than that required by the meter (2) of a foot : containing such a syllable d (1) : being an irrational number <an irrational root of an equation> (2) : having a numerical value that is an irrational number <a length that is irrational>





    Are irrational deductions, such as the square root of 2 completely irrational in the sense that it is lacking normal coherence and are incomprehensible?



    For example, no matter how many decimals we expand the square root of 2, when we multiply itself it is always 1.9999 ... (with some random numbers at the end depending on the exact expansion)

    So how could the square root of 2 be comprehended, how can it exist logically and conceptually in the mind except on a superficial level that it is the number that when multiplies by itself equals two?

    Are not irrational numbers irrational in the fullest sense of the word? There is no number, or integer that when multiplied by itself that will equal two, stating an infinite expansion will not suffice because that seems illogical, yes? An infinite expansion is not possible .

    Thank you,
     
  2. jcsd
  3. Mar 23, 2010 #2
    The sqrt of 2 times by itself is indeed 2. Irritational in maths terms just means can not be expressed in a form such as a/b. Where a and b are integers. i.e. an ending or repeating decimal. Just because it can't accurately be expressed as a decimal doesn't make it any less of a real number than a rational number. Just that we(the vast majority) think in decimals.

    Hope it has helped somewhat.
     
    Last edited: Mar 23, 2010
  4. Mar 23, 2010 #3
    How is it possible to think any other way though? (√2)(√2) = 2 will only blind us to the true dilemma that a number that can be multiplied by itself will not be able to equal 2? Its uncountable to me. Why do math rules apply it as something countable? This is my dilemma. I think it is unreasonable, aside from its practical applications.
     
  5. Mar 23, 2010 #4
    If you drew a perfect square, with sides of length exactly 1, the diagonal would be of length exactly sqrt(2).

    So, there's a line length that you can comprehend.
     
  6. Mar 23, 2010 #5
    I still can't comprehend that because it still seems like an approximation (although as close as we can get without knowing any better).

    Am I stupid?
     
  7. Mar 23, 2010 #6
    You are using an definition of language and applying it to a mathematical idea. You should be using a definition of mathematics and applying it to a mathematical idea (see how that works?).

    You can find a mathematical definition here: http://mathworld.wolfram.com/IrrationalNumber.html
     
  8. Mar 23, 2010 #7

    Hurkyl

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    No; they are irrational only in the technical mathematical sense.
     
  9. Mar 23, 2010 #8
    What's approximate? If you have a right triangle with equal length sides, the length of the hypotenuse is sqrt(2) times the length of those sides. Exactly. We can even write down the value: [tex] \sqrt{2} [/tex]

    I guess I just don't see what the problem is.
     
  10. Mar 23, 2010 #9
    square root of 2 is a number that when multiplied by itself = 2, although I cannot find that number precisely base 10 somehow it exists in conceptual geometry and is always an approximation based on what we are measuring. If a triangle or pyramid were to exist in physical gauageble reality, we would always have to magnify the end of the hypotenuse towards an infinite magnification that will in turn give us another decimal point due to the irrational value of sqrt of 2, for ever and ever, (amen.)
     
  11. Mar 23, 2010 #10
    "Irrational" refers to its inability to be expressed as a "ratio", I.E. fraction. Furthermore, mathematics is strewn with terms that, to an ordinary human being, would be incomprehensible or misconstrued, such as the infamous "imaginary" number with very real applications and uses. Why, if we were to make mathematics adhere to ordinary language, many things would have to change (the word sine, for instance).
     
  12. Mar 23, 2010 #11
    The question is not really mathematical but physical. It's true that if you measured the diagonal of a square in real life you have (1) an uncertainty, and even if you didn't you (2) wouldn't have enough paper or ink to write down the result of your measurement.

    Re (1): the uncertainty is a physical thing and mathematics isn't concerned in this context
    Re (2): even rational numbers have infinite representations (eg. 2 = 2.0 = 2.0000...), we just shorten them for practicality. Just as we write 3.131313... = 3.13 with dots above the last two digits, we write the square root of 2 as sqrt(2). We were always taught it is fine to leave surds in our answers, because you can't get a more accurate representation of that number on a finite piece of paper.

    If you think a number like sqrt(2) (diagonal of a unit square) is troublesome try pi (circumference of a unit diameter circle). Both are irrational but sqrt(2) is at least algebraic, meaning it is the solution of a polynomial with rational coefficients (x^2 - 2 = 0). pi is transcendental (cannot be the solution of any such equation), and there are infinitely more transcendental numbers than algebraic ones.


    I hope I got all that summary correct but it kind of blows my mind. It does seem impossible to actually imagine irrational numbers, and the number of them as well!
     
  13. Mar 23, 2010 #12
    I think your questions are good. There is a book called "Measure and the Integral" by Henri Lebesgue (don't worry, it's not high level, doesn't even assume you know what "multiplication" or a "number" are, very readable) that I highly recommend you get. It's available on amazon used for $10 or $20 and deeply gets at the heart of what's puzzling you. (Again, don't let the author's name intimidate you, if you've heard of it, it requires next to no background in math for the 1st hundred or so pages).
     
  14. Mar 23, 2010 #13
    Yes so wouldn't it be irrational to consider the square root of two to actually exist? I don't see how a number that when multiplied itself actually equals 2! It doesn't make sense to me.
     
  15. Mar 23, 2010 #14

    CRGreathouse

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    No. It is perfectly rational to consider the square root of two to exist.

    Why does it not make sense? What tells you that there is no number like that? And what about numbers like 555444804887496890116; do their square roots exist?
     
  16. Mar 23, 2010 #15
    It does not make sense to me because no matter how many decimals the square root of 2 is expanded to, it will always be 1.999 with some varying string of numbers at the tail dependent upon how many decimals expanded to, through infinity. Which is why at best it is 1.99...to infinity, not 2.

    I don't know about that large number, but the same applies to finding the square root of any number that results in an irrational number, which as I understand is not precise and infinite.

    As I understand it also, finding the numerical value in decimals is the most precise method to evaluate the square root of 2. If there is another more precise method can you please let me know? Thanks,
     
  17. Mar 23, 2010 #16
    May I add that perhaps your troubles stem from identifying a number, is this case [itex]\sqrt{2}[/itex], with its representation relative to a base? To "understand" [itex]\sqrt{2}[/itex] you don't need to refer to its decimal expansion.
     
  18. Mar 23, 2010 #17
    @WW III ANGRY: does this help: 2^(1/2) times 2^(1/2)= 2?
     
  19. Mar 24, 2010 #18
    If p is any positive real number, there is a positive real number x such that x^2=p.

    The existence of such numbers is proved easily by considering the set

    A={z:z is a positive real number and z^2=<p} Indeed, such a number will be x=supA, such that x^2=p.

    Replace, p=2, and you see, that there is actually some real number x, such that x*x=2.

    I don't see what is so irrational about this.

    EDIT: An even more generalized version:

    If p is any positive real number and n is a positive integer, there is a unique positive real number x such that x^n=p.
     
  20. Mar 24, 2010 #19

    Mark44

    Staff: Mentor

    OK, I'll bite. The square root of 2 doesn't actually exist in any physical or measurable sense. The example was given earlier of a square 1 unit on each side. It's not possible to make such a square, since we would not be able to measure exactly 1 unit, or even get the angles to be exactly 90 degrees.

    But so what? At its heart, mathematics is not about geometric figures that we physically create. It's about the perfect, idealized versions of these objects--points that have zero width and height, lines with zero width. If we could create a perfect square that was 1 foot by 1 foot, the length of each diagonal would be exactly sqrt(2) ft. If you want that as a decimal number, you'll have to settle for a finite length representation. One of the attributes of an irrational number is that its decimal representation is infinitely long, and doesn't have any finite-length repeating patterns.
    It doesn't make sense to you because your understanding of numbers is not very developed. That's something you can fix.
     
  21. Mar 24, 2010 #20
    Yes, but that just begs the question, how is 1 any different from [tex]\sqrt{2}[/tex] ? You can't even draw a line of length '1'.

    edit: how do I make my [tex]\sqrt{2}[/tex] show smaller?
     
    Last edited: Mar 24, 2010
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